132 Pattern
10 Prerequisite LeetCode Problems
This involves pattern recognition and usage of data structures like stack to maintain the order of elements. Here are 10 problems to develop the necessary skills:
LeetCode 496: Next Greater Element I: This problem involves understanding the concept of the next greater element, which is similar to the 132 pattern.
LeetCode 739: Daily Temperatures: This problem also involves finding the next greater element and can reinforce your understanding of the concept.
LeetCode 84: Largest Rectangle in Histogram: This problem requires usage of a stack to keep track of the heights of the bars in the histogram.
LeetCode 42: Trapping Rain Water: This problem involves understanding the structure of an array to compute the trapped water. The concept of maintaining a stack of elements can be helpful.
LeetCode 121: Best Time to Buy and Sell Stock: This problem involves finding a pair of numbers (buying price and selling price) in an array to maximize the profit, which is a simpler version of the 132 pattern.
LeetCode 122: Best Time to Buy and Sell Stock II: This problem is a slight variation of the previous problem and can help reinforce your understanding of finding patterns in arrays.
LeetCode 167: Two Sum II  Input array is sorted: This problem involves finding two numbers in an array that add up to a target number. Understanding this problem can help you understand how to look for patterns involving two numbers.
LeetCode 15: 3Sum: This problem involves finding three numbers in an array that add up to zero. Understanding this problem can help you understand how to look for patterns involving three numbers.
LeetCode 16: 3Sum Closest: This problem is a variation of the previous problem and can help reinforce your understanding of finding patterns involving three numbers.
LeetCode 33: Search in Rotated Sorted Array: This problem involves searching for a target number in a sorted array that has been rotated. The concept of maintaining a certain order while dealing with rotations can be useful in understanding the 132 pattern.
These cover how to maintain the order of elements using a stack and how to look for patterns involving two or three numbers in an array, which are both crucial skills for solving the problem at hand.
Clarification Questions
Can the 132 pattern involve nonadjacent elements in the array? For example, in the array [1, 5, 2, 4, 3], can we consider the sequence [1, 4, 3] as a valid 132 pattern?
Can the array contain duplicate elements? If yes, how should they be handled?
Should we consider multiple occurrences of the 132 pattern in the array, or is finding a single occurrence sufficient to return true?
How should we handle an array with fewer than three elements? Can we assume that the input array always has at least three elements?
Can we assume that the input array will always be a list of integers? Will we need to handle other data types?
How large can the array be? Is there a limit on the size of the array?
Problem Analysis and Key Insights
The problem involves identifying a specific pattern (132 pattern) within an array. This means we need a way to compare three elements in the array to identify the pattern.
The pattern involves a specific order: the first number is smaller than the third, and the third is smaller than the second. This detail can help guide our approach for comparing elements.
The pattern doesn’t need to involve adjacent elements. This means we might need to compare elements that are not next to each other in the array.
We only need to find one occurrence of the pattern to return true. This means we can stop searching as soon as we find a pattern.
The constraints suggest that the array can be large (up to (2 \times 10^5) elements), so we should aim for an efficient solution.
The presence of negative numbers and the wide range of possible integer values (10^9 to 10^9) means we can’t make assumptions about the numbers being positive or within a certain range.
These insights can help guide the development of a solution strategy. For example, a brute force approach might involve three nested loops to check every possible triplet of numbers, but this would be too slow for large arrays. So, we need to find a more efficient way to identify the 132 pattern.
Problem Boundary
The scope of this problem includes:
Receiving an array of integers as input. The array can have a size ranging from 1 to (2 \times 10^5) and each integer can be anywhere between 10^9 to 10^9.
Implementing a function or method that checks for the existence of a 132 pattern in the given array. A 132 pattern is a subsequence of three integers such that the first integer is less than the third, which in turn is less than the second.
The function must return a boolean value 
True
if a 132 pattern exists in the array andFalse
if it does not.The solution must handle edge cases, such as arrays with fewer than three elements and arrays with duplicate elements.
The function needs to be efficient enough to handle the upper limits of the input size within a reasonable time frame.
Out of scope:
 Modifying the input array.
 Handling data types other than integers in the array.
 Handling arrays that exceed the given constraints.
The boundaries of this problem are defined by the constraints and requirements mentioned in the problem statement:
Input: The input is an array of integers. The size of the array can vary from 1 to (2 \times 10^5). Each integer in the array can range from 10^9 to 10^9. This restricts the input to a specific type (integer) and limits the size of the array.
Output: The output is a boolean value, either
True
orFalse
. This restricts the output to two possible values.Functionality: The function needs to check for a 132 pattern in the array, which is a sequence of three integers where the first is smaller than the third and the third is smaller than the second. The function doesn’t need to return the pattern itself or the indices of the pattern.
Performance: The function needs to perform this check within a reasonable time frame, even for arrays at the upper limit of the allowed size. This restricts the complexity of the solution and rules out certain approaches that would be too slow, such as a brute force solution with three nested loops.
These boundaries provide a clear understanding of what the function needs to do (identify a 132 pattern), what it needs to return (a boolean value), and what constraints it must operate under (specific input size and element range, performance requirements).
Problem Classification
Domainbased categorization: This problem falls under the domain of array manipulation and pattern detection. It involves scanning through a list of integers to identify a particular pattern.
‘What’ Components:
 We’re given an array of integers.
 We’re asked to find a specific pattern, known as the 132 pattern. This pattern involves three integers where the first one is smaller than the third one, and the third one is smaller than the second one.
 We need to return a Boolean value indicating whether such a pattern exists in the array.
Further Classification: This problem is a pattern detection problem. In essence, we are searching for a specific sequence of numbers (in this case, a 132 pattern) within a larger sequence (the given array). These types of problems involve iterating over the data set, comparing elements, and identifying patterns based on those comparisons.
Understanding these aspects of the problem helps us recognize that we need to traverse the array and compare elements in a specific way to identify the 132 pattern.
Identifying Problem Isomorphism
“132 Pattern” asks you to identify if there exists a sequence i < j < k with nums[i] < nums[k] < nums[j] in a given array nums.
An approximate isomorphic problem to this could be “Increasing Triplet Subsequence”. The problem statement asks if there exists i < j < k such that nums[i] < nums[j] < nums[k].
The core structure of both problems involves finding three indices in an array such that they form an increasing or nonincreasing sequence. However, the conditions differ: in the “132 Pattern” problem, you’re looking for a pattern where the third number is less than the second but greater than the first, whereas in the “Increasing Triplet Subsequence” problem, you are looking for a strictly increasing sequence.
Both are similar in complexity, as they both require a strategy to traverse the array while maintaining some kind of state. “132 Pattern” is more complex due to the less straightforward pattern that needs to be identified.
“Valid Mountain Array” (#941 on LeetCode) could be considered another related problem, as it also involves identifying a specific pattern in a sequence. This problem asks whether there exists i < j < k with nums[i] < nums[j] > nums[k]  a pattern of strictly increasing and then strictly decreasing. The problem adds an extra layer of complexity as you’re required to find this pattern across the entire array.
Distilling the Problem to Its Core Elements
Fundamental Concept: This problem is based on the concept of pattern recognition in a sequence or array. It involves understanding the specific pattern to look for (a 132 pattern) and designing an efficient algorithm to find it. The problem also leverages the principle of comparison and the concept of order in a sequence.
Simplest Description: Imagine you’re given a long list of numbers. Your task is to find out if there are any three numbers in the list such that the first number is smaller than the third number and the third number is smaller than the second number. And importantly, these three numbers must appear in the list in the same order. If you can find such a trio of numbers anywhere in the list, you say “yes”, otherwise “no”.
Core Problem: The core problem is to efficiently find a 132 pattern in a given list of numbers.
Key Components: The key components of the problem are:
 The input array of integers
 The definition of the 132 pattern
 The requirement to return a boolean value
 The performance requirement due to the potential size of the input array
Minimal Set of Operations: At a high level, the minimal set of operations needed to solve this problem would be:
 Traverse the array
 For each element, compare it with other elements in the array to check for the 132 pattern
 If the pattern is found, return
True
 If the end of the array is reached without finding the pattern, return
False
However, due to the size of the input array, we would need to find a way to perform these operations more efficiently than a simple brute force approach.
Visual Model of the Problem
To visualize the problem, let’s consider an example array: [3, 1, 4, 2].
We can depict this as a sequence of numbers on a number line:
3142
We’re looking for a pattern where one number (1) is less than another number (4), and this second number is less than a third number (3). Importantly, these numbers must appear in the same order in the sequence: the first number before the second and the second before the third.
Visualizing this, we could underline the sequence that forms the 132 pattern:
3142
 (1, 4, 3 forms a 132 pattern)
This visual representation helps clarify what the 132 pattern looks like within a sequence of numbers. We can see that the numbers involved in the pattern don’t have to be adjacent in the sequence, and they don’t have to be the only numbers in the sequence.
Problem Restatement
You’re given a list of numbers. Your task is to figure out if there’s a specific pattern within this list. The pattern is called a “132” pattern, which means you’re looking for three numbers where the first number is smaller than the third number, and the third number is smaller than the second number. Also, these three numbers should maintain their order in the list.
For instance, if the list of numbers is [3, 1, 4, 2], the numbers 1, 4, and 3 form a 132 pattern because 1 is less than 4, and 4 is less than 3. Also, 1 appears before 4, and 4 appears before 3 in the list.
If you find this pattern in the list, you should return true. If not, return false.
The size of the list can be as small as 1 or as large as 200,000. The numbers in the list can range from 1 billion to 1 billion. Your solution should be efficient enough to handle a list at the maximum size within a reasonable time frame.
Abstract Representation of the Problem
Consider a sequence S
of n
elements, where each element e
is an integer within a large range. We are interested in identifying a specific pattern P
within this sequence.
Pattern P
is defined as a subsequence of three elements [a, b, c]
such that a < c < b
and a
, c
, and b
maintain their relative order in the sequence. In other words, a
appears before c
, and c
appears before b
in the sequence.
The problem is to devise a function F
that takes the sequence S
as input and returns a boolean value. This function F
should return true
if pattern P
exists in S
, and false
otherwise.
The function F
should be efficient enough to handle a sequence S
with a size at the upper limit of n
within a reasonable time frame.
The abstract representation helps us focus on the structure of the problem and the relationships between the key elements, rather than getting bogged down in specific realworld details or examples.
Terminology
Here are a few technical concepts and terms that are crucial to understanding this problem:
Array/List: An array (or a list, as it’s called in Python) is a fundamental data structure that holds a collection of elements. In this problem, we’re given an array of integers.
Subsequence: A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. In this problem, the 132 pattern is a subsequence of the given array.
Pattern Recognition: Pattern recognition refers to the identification of repeated or recurring entities, themes, or trends within a dataset. In this case, we’re searching for a specific pattern (132 pattern) within the array.
Boolean: A boolean is a data type that has two possible values,
true
orfalse
. In this problem, we’re asked to return a boolean value indicating whether the 132 pattern exists in the given array.Brute Force Approach: A brute force approach to a problem involves checking all possible solutions to find the correct one. This is typically not efficient for large input sizes, but it can sometimes be a starting point for understanding the problem and working towards a more efficient solution. In this problem, a brute force approach might involve checking all possible subsequences of three elements to find a 132 pattern, but we would need to find a more efficient solution to meet the performance requirements.
Understanding these terms and concepts can help you better understand the problem and how to approach solving it.
Problem Simplification and Explanation
You’re given a list of numbers. Think of this list as a group of students in a line for a school activity. Each student has a numbered tag.
Now, you’re looking for a specific pattern among these students  you want to find three students such that the one with the lowest number (let’s call him Adam) is standing somewhere before a student with a higher number (let’s call him Charlie), and Charlie is standing somewhere before a student whose number is higher than Adam’s but lower than Charlie’s (let’s call her Beth).
So, the pattern you’re looking for is Adam  Charlie  Beth. It doesn’t matter if there are other students standing between Adam, Beth, and Charlie; they just need to be standing in the line in this order.
If you can find such a group of three students in the line, you say “Yes”, otherwise “No”.
The key concept in this problem is the idea of order  Adam needs to be somewhere before Charlie, and Charlie needs to be somewhere before Beth in the line. You’re looking through the line of students (the list of numbers) to find a group of three students who meet this criterion.
This metaphor helps illustrate the idea of the 132 pattern and the concept of order within a sequence. It’s not about the absolute values of the numbers, but rather about their relative values and positions within the sequence.
Constraints
Here are some specific characteristics or conditions from the problem statement and constraints that can be leveraged to find an efficient solution:
Order Matters: The problem is looking for a specific pattern of three numbers, where the first number is less than the third number, and the third number is less than the second number. This pattern must maintain its order in the sequence. This means we don’t need to consider every combination of three numbers, but only the ones that maintain their relative order in the sequence, which can significantly reduce the number of possibilities we need to check.
Element Range: Each number in the array can be as low as 10^9 or as high as 10^9. While this wide range doesn’t provide a specific advantage in terms of reducing possibilities, it does mean we need to consider negative numbers and very large numbers when thinking about the solution.
Sequence Length: The size of the array can be as large as 200,000. This large upper limit means a brute force approach that checks every possible subsequence of three numbers will likely be too slow. This constraint hints that we need to find a more efficient algorithm, perhaps one that scans the sequence only once or a limited number of times.
At least one solution: The problem doesn’t specify this, but if we know that there’s at least one 132 pattern in the sequence, we could potentially stop the search as soon as we find the first pattern. However, without this guarantee, we need to check the entire sequence.
Identifying these characteristics helps us recognize what parts of the problem can be exploited to find an efficient solution, and what aspects of the problem we need to keep in mind as we design the solution.
Analyzing the constraints of this problem provides several key insights:
Scale of Input: The size of the array can go up to 200,000 elements, which means that our solution needs to be highly efficient. A simple brute force solution that checks every possible subsequence of three elements for the 132 pattern would likely be too slow. This suggests that we should be looking for a solution with a time complexity better than O(n^3), where n is the size of the input array.
Range of Elements: The elements in the array can range from 1 billion to 1 billion. This means that we need to consider both negative and positive numbers, and very large numbers. However, the wide range of possible element values does not give us an immediate advantage in terms of simplifying the problem or reducing the search space.
Existence of Pattern: The problem does not guarantee that a 132 pattern exists in the array. Therefore, our solution must be able to handle cases where no such pattern exists, and should return
false
in such cases.Order of Pattern: The 132 pattern must maintain its relative order in the sequence. This is a crucial aspect of the problem, and it means that we don’t need to consider all possible combinations of three elements, but only those that maintain their relative order in the sequence. This could potentially help us in reducing the search space and improving the efficiency of our solution.
These insights help us understand the nature of the problem better and guide us in formulating an efficient solution strategy.
Case Analysis
Here are some additional examples and test cases:
1. Case of Increasing Numbers  No 132 Pattern:
Input: nums = [1, 2, 3, 4, 5]
Output: false
Explanation: In this case, all numbers are in increasing order. So, there is no way to find three numbers where the first is less than the third and the third is less than the second. Hence, no 132 pattern exists.
2. Case of Decreasing Numbers  No 132 Pattern:
Input: nums = [5, 4, 3, 2, 1]
Output: false
Explanation: Here, all numbers are in decreasing order. Similar to the previous case, it’s impossible to find a 132 pattern when all numbers are in decreasing order.
3. Case of Same Numbers  No 132 Pattern:
Input: nums = [2, 2, 2, 2, 2]
Output: false
Explanation: When all numbers are the same, there are no three numbers where the first is less than the third and the third is less than the second. Hence, no 132 pattern exists.
4. Minimal Input Case  No 132 Pattern:
Input: nums = [1]
Output: false
Explanation: With only one element, it’s impossible to find a 132 pattern which requires at least three elements.
5. Case with Negative Numbers  132 Pattern Exists:
Input: nums = [1, 2, 3, 1]
Output: true
Explanation: Here, 3 < 1 < 2 forms a 132 pattern.
6. Case with Large Numbers  132 Pattern Exists:
Input: nums = [1000000000, 1, 1000000000]
Output: true
Explanation: In this case, 1 < 1000000000 < 1000000000 forms a 132 pattern.
These test cases cover a wide range of input scenarios, including edge and boundary conditions, and help ensure that the solution is robust and handles all possible cases. Analyzing these cases provides insights into the problem and helps identify key aspects to consider when devising a solution.
Visualizing these cases can help to better understand the pattern we’re looking for. Here’s how we can visualize these cases:
Case of Increasing Numbers  No 132 Pattern:
We can visualize this case as a straight line on a graph that continuously increases.
1  2  3  4  5
There’s no 132 pattern here because there’s no place where a number dips below the previous one and then rises again.
Case of Decreasing Numbers  No 132 Pattern:
This is a straight line on a graph that continuously decreases.
5  4  3  2  1
Like the increasing case, there’s no 132 pattern because there’s no place where a number rises above the previous one and then dips again.
Case of Same Numbers  No 132 Pattern:
Here, the line on a graph is flat.
2  2  2  2  2
There’s no 132 pattern because all the numbers are the same.
Minimal Input Case  No 132 Pattern:
With just a single point, there’s no possibility of a pattern.
1
Case with Negative Numbers  132 Pattern Exists:
This case can be visualized as a line that decreases, then increases.
1  2  3  1
The sequence 3 < 1 < 2 forms a 132 pattern.
Case with Large Numbers  132 Pattern Exists:
This can be visualized as a line that starts high, dips low, and then rises high again.
1000000000  1  1000000000
The sequence 1 < 1000000000 < 1000000000 forms a 132 pattern.
Each visualization gives a clear idea of the structure of the array and how it relates to the possibility of a 132 pattern existing.
Analyzing different cases provides several key insights:
Increasing or Decreasing Order: If the input numbers are in strictly increasing or decreasing order, a 132 pattern cannot exist. In an increasing sequence, there’s no number that’s larger than a later number. In a decreasing sequence, there’s no number that’s smaller than a later number.
Identical Numbers: If all the numbers are identical, a 132 pattern cannot exist, as there are no two distinct numbers that can be used for the 1 and 3 of the pattern.
Negative Numbers: The existence of a 132 pattern doesn’t depend on whether the numbers are positive or negative. A sequence of negative numbers can have a 132 pattern, as long as the relative order of the numbers meets the 132 condition.
Large Numbers: Large numbers do not inherently prevent or guarantee a 132 pattern. The existence of a 132 pattern depends on the relative order of the numbers, not their absolute values.
Single Element: If the input array has fewer than three elements, a 132 pattern cannot exist, as the pattern requires at least three elements.
Pattern Identification: The 132 pattern is characterized by a sequence of numbers that first increases, then decreases, then increases again. It’s important to note that the 1, 3, and 2 in the pattern do not refer to the actual numbers, but to their relative magnitudes.
These insights guide us in understanding the conditions under which a 132 pattern can exist and in formulating a strategy to efficiently identify such a pattern.
Identification of Applicable Theoretical Concepts
This problem requires identifying a specific pattern within an array of integers. Here are a few mathematical and algorithmic concepts that can be applied to make the problem more manageable:
Monotonicity: Monotonicity refers to a property of a sequence to be entirely nonincreasing or nondecreasing. We can use this property to eliminate certain sequences where a 132 pattern cannot exist, like when the sequence is strictly increasing or decreasing.
Stack Data Structure: This problem can be solved efficiently using a stack data structure. A stack allows us to keep track of elements in a way that we can easily access the last element we added (the top of the stack), and remove it if necessary. This is particularly useful for this problem where we want to keep track of potential 2’s and 3’s for our 132 pattern.
Running Minimum or Maximum: Keeping track of the running minimum or maximum can be used to identify potential candidates for the 1 in our 132 pattern.
Element Pairing: The task of identifying a 132 pattern can be seen as a process of forming pairs of potential 2’s and 3’s and checking if there exists a 1 for each pair.
Iterative Scanning: Given the nature of this problem, an iterative approach to scan the array from left to right or right to left can be used.
These can help simplify the problem and make it more manageable. They provide a way to handle the problem in a structured and efficient manner, reducing the problem from a search problem in a 3dimensional space to one in a 2dimensional space.
Simple Explanation
Let’s use an analogy with a hill to explain this problem.
Imagine you’re a bird flying over a range of hills. Each hill has a certain height, and these heights are represented by the numbers in our array.
The 132 pattern is like looking for a specific kind of shape on the ground while you’re flying. You want to find a small hill (number 1), followed by a taller hill (number 3), and then another hill that’s taller than the first one but shorter than the second one (number 2). They need to appear in this exact order.
So, you’re basically looking for a hill that’s taller than the first one you saw, but not as tall as the highest one you’ve seen so far. If you can find such a sequence of hills while flying from one direction to the other, then we say that there’s a 132 pattern. If not, then there isn’t one.
In more general terms, we’re looking for a specific pattern in a sequence of numbers. This pattern involves three numbers where the first is smaller than the third, and the third is smaller than the second. We need to find these three numbers in the correct order in our sequence to say that a 132 pattern exists.
Problem Breakdown and Solution Methodology
Let’s break it down:
Understand the Problem: We are given an array of numbers and we are asked to find if there is any subsequence of three numbers in the array that matches the 132 pattern. This means we need to find three numbers, (a < c < b), where ‘a’ comes before ‘b’ and ‘b’ comes before ‘c’ in the array.
Initial Approach: A naive approach would be to try all possible triplets of numbers in the array. For each triplet, check if it follows the 132 pattern. However, this approach would take cubic time complexity, which is not efficient for large arrays.
Optimized Approach: To optimize the solution, we can use a stack data structure. We start by initializing a variable, call it ’third’, to the smallest possible integer. This ’third’ variable is used to keep track of the potential ‘2’ in our 132 pattern. The stack is used to maintain potential pairs of ‘2’ and ‘3’.
Now, we iterate through the array from right to left. For each number:
If it’s smaller than ’third’, then we’ve found a 132 pattern because ’third’ is ‘2’ and there’s a ‘3’ in the stack that’s larger than ‘2’, and the current number is ‘1’.
If it’s not, then while there are numbers in the stack that are smaller or equal to the current number, we pop them out and update ’third’ to the popped number.
Finally, we push the current number into the stack.
This way, for each number, we maintain a ’third’ that’s the largest number that’s smaller than it.
Impact of Changing Parameters: If the input array is sorted in increasing order, then no 132 pattern exists, so the function will return false. If the array is sorted in decreasing order, there’s still no 132 pattern. But if the array has a mix of increasing and decreasing subsequences, a 132 pattern may exist.
Let’s consider an example:
Suppose we have the array [3, 5, 1, 4].
 Start with an empty stack and ’third’ as ∞.
 The first number from the right is 4. Push it into the stack.
 The second number is 1, which is smaller than ’third’, so continue to the next number.
 The third number is 5. It’s not smaller than ’third’, and it’s larger than the top of the stack, so pop 4 from the stack and set ’third’ to 4. Then push 5 into the stack.
 The fourth number is 3, which is smaller than ’third’ (which is 4), so we’ve found a 132 pattern.
The pattern is 3 < 4 < 5, so we return true.
This approach efficiently checks for a 132 pattern in linear time complexity and space complexity proportional to the size of the input array.
Inference of ProblemSolving Approach from the Problem Statement
There are several key terms and concepts in this problem that guide the approach to solving it:
Subsequence: A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements. In this problem, we are looking for a subsequence of three elements that follow a specific pattern. This suggests that our solution should involve some form of sequence traversal.
132 Pattern: The problem specifically asks for a 132 pattern in the sequence. This pattern involves three numbers where the first is smaller than the third, and the third is smaller than the second. Identifying this pattern guides us to look for a solution that can find such a triplet in the sequence.
i, j, k: These are indices in the sequence with the condition that (i < j < k). The conditions on these indices suggest a possible order of traversal or a way to structure our search for the 132 pattern.
Array or List: The problem input is an array (or list in Python). This informs us that we could use arrayspecific operations or characteristics in our solution. For example, we can easily access elements at any position, which may not be the case with other data structures like a linked list.
These key concepts guide the choice of data structure (an array and a stack in this case) and the choice of algorithmic technique (an iteration from right to left in the array while maintaining a stack of potential 2’s and 3’s for our 132 pattern).
The problem deals with identifying patterns in an array of numbers, which is often best visualized through a line graph. Each number in the array can be represented as a point on the graph, with its position in the array on the xaxis and its value on the yaxis.
Let’s consider the example array [3, 5, 1, 4].
Start by drawing a point for each number in the array:
 Draw a point at (1,3) for the first number.
 Draw a point at (2,5) for the second number.
 Draw a point at (3,1) for the third number.
 Draw a point at (4,4) for the fourth number.
Connect the points in the order they appear in the array to form a line graph. This line graph visually represents the sequence of numbers in the array.
To find a 132 pattern, look for a path in the graph where it goes up, then down, then up again.
In our example, you can see such a pattern: start at 3 (up to 5), down to 1, and up again to 4. This visually represents the 132 pattern in the sequence.
This visual representation provides a way to recognize the properties related to the problem. It helps us understand the pattern we are looking for and allows us to visualize how the numbers in the array relate to each other in terms of this pattern.
This diagram might not help directly in coding the solution but it is a good way to understand the problem and the patterns we are looking for.
How did you infer from the problem statement that this problem can be solved using ?
Simple Explanation of the Proof
Let’s break down how the algorithm works and why it’s correct.
The algorithm uses a stack and a variable called ’third’ to keep track of the ‘2’ and ‘3’ in our 132 pattern. We start by setting ’third’ to negative infinity and the stack to empty.
We then iterate over the array from right to left. For each number, we check:
If it’s less than ’third’, we’ve found a 132 pattern. This is because ’third’ holds the value of ‘2’ in our pattern, and there’s a ‘3’ in the stack that’s larger than ‘2’. Since the current number is less than ‘2’ and it comes before ‘2’ and ‘3’ in the array, it’s our ‘1’ in the pattern.
If it’s not less than ’third’, we pop elements from the stack that are less than or equal to the current number and set ’third’ to the popped value. This is done because we are looking for a ‘3’ that’s larger than ‘2’, and the current number could potentially be our new ‘3’. So we pop all numbers from the stack that are smaller or equal to the current number as they can’t be our ‘3’. At the same time, we update ’third’ to the popped number, which is the largest number that’s smaller than ‘3’, making it our potential ‘2’.
Finally, we push the current number into the stack. This is done because the current number could potentially be our ‘3’ in the future.
By the end of the iteration, if we’ve found a 132 pattern, we return true. If not, we return false.
The correctness of this algorithm is guaranteed by its careful management of the ’third’ variable and the stack. ’third’ is always the largest number we’ve seen so far that’s smaller than all numbers in the stack. If we find a number smaller than ’third’, it must come before both ‘2’ and ‘3’ in the array, making it the ‘1’ in the pattern.
Stepwise Refinement
Could you please provide a stepwise refinement of our approach to solving this problem?
How can we take the highlevel solution approach and distill it into more granular, actionable steps?
Could you identify any parts of the problem that can be solved independently?
Are there any repeatable patterns within our solution?
Solution Approach and Analysis


Here’s a stepbystep breakdown of how the code works using the second example [3,1,4,2]:
Initialize ’third’ as negative infinity and the ‘stack’ as an empty list.
Start iterating from the last element of the list.
For the first iteration, ‘2’ is not less than ’third’ (negative infinity). So, we skip the ‘if’ condition. In the ‘while’ loop, since the stack is empty, we skip the loop. We then push ‘2’ to the stack.
For the second iteration, ‘4’ is not less than ’third’. In the ‘while’ loop, ‘2’ (from the stack) is less than ‘4’, so we pop ‘2’ from the stack and update ’third’ to ‘2’. We then push ‘4’ to the stack.
For the third iteration, ‘1’ is less than ’third’ (‘2’), so we return True. We’ve found a 132 pattern: ‘1’ is our ‘1’, ‘2’ (stored in ’third’) is our ‘2’, and ‘4’ (in the stack) is our ‘3’.
As for how changes in the problem’s parameters would affect the solution, the time complexity of this solution is O(n), where n is the length of the input list. So if the list becomes longer, the time it takes to solve the problem will increase linearly. The space complexity is also O(n) in the worst case as we might end up pushing all elements to the stack. Thus, a longer list would also require more memory to solve the problem.
Identify Invariant
In the context of this problem, the invariant is the property that for any number we identify as the “2” in our 132 pattern (stored in the ’third’ variable), all numbers in the stack are greater than this “2”. This invariant is maintained throughout the entire execution of the algorithm.
This is crucial because it ensures that for any potential “1” we encounter in the array (which must be less than “2”), there exists a “3” (in the stack) that is greater than “2” and appears after “2” in the original array. This upholds the condition we need for the 132 pattern: “1” < “2” < “3”.
During the execution of the algorithm, we continuously update “2” and make sure that the stack holds numbers that are greater than the updated “2”, thereby maintaining the invariant.
Identify Loop Invariant
In the context of this problem, the loop invariant is the following property: At the start of each iteration of the for loop, all numbers in the stack are greater than the current “third” number. This is the same as the overall invariant in this problem.
This loop invariant is crucial for the following reasons:
It maintains the order of elements as per the original sequence. This is important because the 132 pattern is ordersensitive. By iterating from the end of the array, we ensure that all elements in the stack always appear after the current “third” number in the original sequence.
It ensures that if we find a number smaller than “third” (which becomes our “1”), we know there is a “3” that is larger than “2” and appears after “2” in the sequence, as all numbers in the stack are larger than “third” (our “2”).
This loop invariant is maintained by popping elements from the stack that are less than or equal to the current number and updating “third” to the popped value, and then pushing the current number to the stack. This ensures that all numbers in the stack remain greater than the “third” number at the start of each loop iteration.
In the context of this specific problem, the overall invariant and the loop invariant happen to be the same. The property that all numbers in the stack are greater than the current “third” number is maintained throughout the entire execution of the algorithm (making it an overall invariant) and is also maintained at the start of each iteration of the loop (making it a loop invariant).
However, it’s important to note that this isn’t always the case. Loop invariants and overall invariants are two distinct concepts:
A loop invariant is a condition that is initially true and remains true after each iteration of a loop. It’s used to prove the correctness of an algorithm.
An overall invariant is a condition that remains true throughout the entire execution of a program or a specific algorithm, not just a loop.
In some problems, the loop invariant and the overall invariant may differ. For instance, a loop invariant might pertain to a property that holds true during a specific loop in the algorithm, while the overall invariant might refer to a broader property that the algorithm maintains from start to end.
Thought Process
The problem is asking us to find a specific pattern in the input list: a “132” pattern, which is a subsequence of three integers such that the first is less than the third, which is less than the second. The challenge here is that these three numbers don’t have to be contiguous, but they do need to maintain their order in the original sequence.
Here’s a stepbystep thought process for solving this problem:
Understanding the ‘132’ pattern: The first step is understanding what a ‘132’ pattern is. It’s a sequence where the first number is less than the third, which is less than the second. The numbers don’t have to be next to each other, but they must maintain their order.
How to find the pattern: One way to find the pattern is to keep track of potential “2"s and “3"s. If we find a number that can be a potential “1” (meaning it’s less than our current “2”), we return True.
Iterating through the array: We iterate through the array from right to left. We do this because we want to keep track of potential “2"s and “3"s that come after the “1” in the pattern.
Updating ’third’ and the stack: For each number, if it’s less than ’third’, we return True. Otherwise, we update ’third’ and the stack. We keep popping from the stack while the current number is greater than the top of the stack, and we update ’third’ to be the popped number. Then we add the current number to the stack. This ensures that all numbers in the stack are greater than ’third’.
Returning the result: If we finish iterating through the array without finding a “1”, we return False.
Here’s the Python code implementing this thought process:


This solution uses a stack to keep track of potential “2"s and “3"s. The ’third’ variable keeps track of the current “2”. If we ever find a number less than ’third’, we know we have found a ‘132’ pattern and return True. If we go through the entire list without finding such a number, we return False. The time complexity is O(n), and the space complexity is O(n) in the worst case.
Establishing Preconditions and Postconditions
Parameters:
 The method takes one input:
nums
which is a List of integers.  This list represents the sequence of numbers in which we need to find the 132 pattern.
 The method takes one input:
Preconditions:
 Before this method is called, it must be true that the list
nums
is initialized and contains at least one element.  The list can contain a maximum of 2 * 10^5 elements.
 Each element of
nums
is an integer that can range from 10^9 to 10^9.  There’s no precondition about the state of the program.
 Before this method is called, it must be true that the list
Method Functionality:
 The method is expected to check whether there is a 132 pattern in
nums
.  The 132 pattern is a subsequence of three integers nums[i], nums[j] and nums[k] such that i < j < k and nums[i] < nums[k] < nums[j].
 The method interacts with the inputs by iterating over the elements in
nums
from right to left and maintaining a stack of elements that could potentially form a 132 pattern with the current element.
 The method is expected to check whether there is a 132 pattern in
Postconditions:
 After the method has been called, the state of the program doesn’t change as the method doesn’t have any side effects. The list
nums
remains the same.  The return value is a boolean that indicates whether a 132 pattern is found in the list
nums
.  There are no side effects as the method does not modify the input list or any global variables.
 After the method has been called, the state of the program doesn’t change as the method doesn’t have any side effects. The list
Error Handling:
 If the preconditions are not met (i.e.,
nums
is not initialized or does not contain at least one element), the method does not handle these cases explicitly. In Python, this would typically lead to a runtime error.  It does not throw any exceptions or return special values. The return type is always a boolean.
 If the preconditions are not met (i.e.,
Problem Decomposition
Problem Understanding:
 The problem is about finding a specific pattern, called the 132 pattern, in a given list of numbers. The 132 pattern means that there are three numbers such that the first number is smaller than the third number which in turn is smaller than the second number. The first number must appear before the second and third in the list, and the third number must appear before the second.
Initial Breakdown:
 We can break the problem into three major parts: a. Iterating through the list from right to left. b. Maintaining a stack of potential second numbers. c. Checking if the current number is smaller than the top of the stack.
Subproblem Refinement:
 Each subproblem is atomic and cannot be further broken down into smaller tasks.
Task Identification:
 The tasks that are repeated are iterating through the list, comparing the current number with the top of the stack, and updating the stack.
Task Abstraction:
 The tasks identified are abstracted enough to be clear and reusable.
Method Naming:
 The tasks can be named as follows:
a. Iterating through the list:
iterate_list
b. Maintaining a stack of potential second numbers:update_stack
c. Checking if the current number is smaller than the top of the stack:check_pattern
 The tasks can be named as follows:
a. Iterating through the list:
Subproblem Interactions:
 These subproblems interact in the main loop of the solution. We start by iterating through the list. For each number, we check if it is smaller than the top of the stack. If it is, we have found a 132 pattern and return true. Otherwise, we update the stack with the current number and continue to the next number. These subproblems need to be performed in this order and the updating of the stack is dependent on the check for the pattern.
From Brute Force to Optimal Solution
A simple brute force solution to this problem would be to generate all possible subsequences of three numbers in the list and check each one to see if it forms a 132 pattern. We can do this by using three nested loops, where the outer loop represents the first number, the middle loop represents the third number, and the innermost loop represents the second number. For each combination, we check if the numbers form a 132 pattern and return true if we find any such pattern.
This brute force solution would be very inefficient because it tries out all possible combinations of three numbers, leading to a time complexity of (O(n^3)), where (n) is the size of the list. This is not practical for large lists, as the problem statement allows for a list of up to 2 * 10^5 elements.
Optimization Steps:
Use a Stack for Potential Second Numbers: Instead of trying out all possible second numbers for each pair of first and third numbers, we can maintain a stack of potential second numbers. When iterating through the list from right to left, any number that is larger than the current number could potentially be the second number in a 132 pattern with the current number as the first number. Therefore, we push such numbers onto the stack.
Iterate from Right to Left: We iterate from right to left so that when we are at a certain number, we have already seen all the numbers to its right. This ensures that the potential second numbers in the stack are all numbers that are to the right of the current number.
Early Return: If we find a 132 pattern at any point, we can return true immediately without needing to check the rest of the list.
The optimized solution has a time complexity of (O(n)), where (n) is the size of the list, because we only do a constant amount of work for each element in the list. This is a significant improvement over the brute force solution. The space complexity is also (O(n)) in the worst case, because in the worst case scenario, we might end up pushing all elements onto the stack. However, this is usually acceptable as we often prioritize reducing time complexity over space complexity.
Code Explanation and Design Decisions
Initial Parameters: The initial parameter is an integer array
nums
. The array represents a sequence of numbers in which we are looking for a 132 pattern.Primary Loop: The primary loop in this problem is iterating over the
nums
array in reverse order (from right to left). Each iteration represents the potential first number of a 132 pattern. By iterating in reverse, we can maintain a stack of potential second numbers that are all to the right of the current number.Conditions or Branches: Inside the loop, there is a condition to check whether the stack is not empty and the top element of the stack is less than the current number. This condition is checking whether we have found a potential second number for the current first number. There’s another condition to check whether the third number (
third
) is less than the current number. This condition checks whether we have found a 132 pattern.Updates or Modifications: Inside the loop, we update the third number if the stack is not empty and the top element of the stack is less than the current number. The third number represents the maximum second number we have found so far that can form a 132 pattern with a smaller first number. We also push the current number onto the stack if it is larger than the third number, because it could potentially be the second number in a 132 pattern with a smaller first number.
Invariant: An invariant in this problem is that the elements in the stack are always in decreasing order from bottom to top. This is maintained by always pushing a number onto the stack only if it is larger than the current third number, and always popping from the stack when the top element is less than the current number.
Final Output: The final output is a boolean value indicating whether a 132 pattern exists in the
nums
array. It represents the answer to the problem and satisfies the requirement by correctly indicating the presence or absence of a 132 pattern.
Coding Constructs
HighLevel Strategies: The code uses a combination of a stack and a greedy approach to find a 132 pattern in the input array. The stack is used to maintain potential second numbers of the pattern, and the greedy approach is used to find the maximum second number that can form a 132 pattern with a smaller first number.
NonProgrammer Explanation: This code is like trying to find a specific pattern of three numbers in a list. The pattern is like a valley where the first number is smaller than the second number, and the third number is in between the first and second. The code looks at the list from the end to the start, keeping track of potential valleys it can form with the numbers it has already seen.
Logical Elements: The logical elements used in this code include iteration (a reversed loop), conditional branching (
if
statements), and a data structure (a stack) to keep track of the potential second numbers of the 132 pattern.Algorithmic Approach: Starting from the end of the list, the code checks each number to see if it can be the first number of a 132 pattern with the potential second numbers it has seen so far. If it can, the code returns
True
. If the code finishes checking all the numbers and hasn’t found a 132 pattern, it returnsFalse
.Key Steps: The key steps include iterating through the input array in reverse, using a stack to keep track of potential second numbers, maintaining the maximum second number (
third
) that can form a 132 pattern with a smaller first number, and checking for the existence of a 132 pattern.Algorithmic Patterns: The algorithmic pattern used by this code is a combination of a stack for maintaining potential second numbers in descending order and a greedy approach for maintaining the maximum second number. This pattern of using a stack with a greedy approach is common in problems where we need to find a specific pattern or sequence in an array.
Language Agnostic Coding Drills
Concept Identification:
 Basic Syntax and Operations (Easy): Understanding of basic programming syntax and operations like variable assignments, comparison operators, loops, and conditional statements.
 Arrays (Easy): Understanding of how to work with arrays, including accessing elements and traversing them in a particular order (in this case, reverse order).
 Stacks (Intermediate): Knowledge of stack data structure and its operations like push and pop. In this problem, a stack is used to keep track of potential second numbers in the 132 pattern.
 Greedy Algorithms (Intermediate): Understanding of greedy algorithms, where we make the locally optimal choice at each stage with the hope of finding a global optimum. In this case, we’re maintaining the maximum second number (
third
) that can form a 132 pattern with a smaller first number.
Difficulty Order and Descriptions:
 Basic Syntax and Operations: These are fundamental to any programming language and serve as the building blocks for solving problems. They are the easiest as they form the basic requirement for understanding any code.
 Arrays: Slightly more complex, they require understanding of data structures and how to manipulate them, including traversing in reverse order, which is a bit more complex than regular forward traversal.
 Stacks: This is an intermediate concept that requires understanding of the stack data structure and its LIFO (Last In, First Out) property. Stacks are not as commonly used as arrays, making them a bit more challenging.
 Greedy Algorithms: This is also an intermediate concept that involves decisionmaking for optimal results. It requires a deeper understanding of the problem and ability to identify that a greedy approach can be used.
ProblemSolving Approach:
 Start by understanding the problem and identifying that we need to find a 132 pattern in the array.
 Recognize that we can iterate through the array in reverse order and use a stack to keep track of potential second numbers.
 Identify that we can use a greedy approach to maintain the maximum second number that can form a 132 pattern with a smaller first number.
 Assemble these concepts to form the overall solution. Start iterating from the end of the array, and for each number, check if it can be the first number of a 132 pattern with the maximum second number seen so far. If it can, return
True
. If no 132 pattern is found after checking all numbers, returnFalse
.
Targeted Drills in Python
Python Coding Drills:
a. Basic Syntax and Operations:
Here’s a simple exercise to understand variable assignments, comparison operators, loops, and conditional statements.
1 2 3 4 5 6
numbers = [1, 2, 3, 4, 5] for number in numbers: if number < 3: print(f"{number} is less than 3") else: print(f"{number} is greater than or equal to 3")
b. Arrays:
This drill helps in understanding how to traverse arrays in reverse order.
1 2 3
numbers = [1, 2, 3, 4, 5] for number in reversed(numbers): print(number)
c. Stacks:
This drill illustrates how to use a stack to store and retrieve elements.
1 2 3 4 5
stack = [] stack.append(1) # Push 1 stack.append(2) # Push 2 top = stack.pop() # Pop from stack print(top) # Outputs: 2
d. Greedy Algorithms:
This drill demonstrates a simple case of a greedy algorithm where we’re trying to find the largest number.
1 2 3 4 5 6
numbers = [1, 2, 3, 4, 5] max_number = numbers[0] for number in numbers: if number > max_number: max_number = number print(max_number) # Outputs: 5
ProblemSpecific Concepts:
No problemspecific concepts are needed for this problem beyond the general concepts identified above. The solution uses array traversal, stack, and a greedy approach, which are all covered in the general concepts.
Integration of Drills:
We start by initializing an empty stack and a variable
third
with negative infinity. We then iterate through the array in reverse order. For each number, we first check if it’s less thanthird
. If it is, we returnTrue
because we’ve found a 132 pattern. If it’s not, we pop elements from the stack while they’re less than the current number, and then push the current number onto the stack. We also updatethird
to be the last popped number. After checking all numbers, if no 132 pattern is found, we returnFalse
.The key is understanding how the stack and the
third
variable are used to find a potential 132 pattern. The stack stores potential second numbers, andthird
stores the maximum second number that can form a 132 pattern with a smaller first number. This greedy approach of maintaining the maximum second number is the key to solving this problem.
Q&A
Similar Problems
Here are 10 problems that use similar underlying concepts:
Daily Temperatures (LeetCode 739): This problem requires maintaining a stack of indices and using a similar logic of popping elements from the stack when a larger element is encountered.
Trapping Rain Water (LeetCode 42): Here, we use a similar strategy to maintain a stack of heights and calculate the trapped water based on the heights in the stack.
Next Greater Element II (LeetCode 503): This problem is similar because it also involves iterating over an array and using a stack to find the next greater element, akin to finding the second number in our 132 pattern.
Largest Rectangle in Histogram (LeetCode 84): This problem uses a stack to track the heights of the bars in the histogram and calculates the maximum area in a way that’s similar to how we track the second number in the 132 pattern.
Maximum Binary Tree (LeetCode 654): The construction of the maximum binary tree uses a stack and a similar logic of maintaining the largest element, akin to the 132 pattern problem.
Online Stock Span (LeetCode 901): This problem involves a stack and the idea of maintaining elements in a decreasing order, which is similar to the concept used in the 132 pattern problem.
Remove Duplicate Letters (LeetCode 316): This problem uses a stack to maintain the lexicographical order of characters in a similar way to how we maintain the 132 pattern.
Next Greater Element I (LeetCode 496): This problem involves finding the next greater element in an array using a stack, similar to finding the second number in the 132 pattern.
Asteroid Collision (LeetCode 735): This problem uses a stack to handle collisions between asteroids, similar to handling the comparisons between numbers to find the 132 pattern.
Valid Parentheses (LeetCode 20): Although this problem is about validating parentheses in a string, it uses a stack to keep track of the opening parentheses and matches them with the closing ones. This mechanism of using a stack to maintain some order or matching is similar to the way we used a stack to find the 132 pattern.