Max Chunks To Make Sorted II

Identifying Problem Isomorphism

“Max Chunks To Make Sorted II” is about splitting an array into the largest possible number of chunks, each of which can be individually sorted to result in a completely sorted array.

An isomorphic problem to this is “Partition Labels”. This problem requires partitioning a string into as many parts as possible, so that each letter appears in at most one part. Both problems involve finding a way to divide the input (array or string) into the largest number of parts according to a specific rule.

A simpler problem is “Best Time to Buy and Sell Stock”. It’s a simpler problem because it only involves finding the maximum difference between two elements in an array, considering their order.

A more complex problem is “Best Time to Buy and Sell Stock III”, which extends the above problem by allowing up to two non-overlapping transactions, adding a layer of complexity. In this context, a “transaction” can be seen as analogous to a “chunk” in the original problem.

These problems are connected by the common theme of partitioning an input based on certain conditions, but they differ in their specific rules and constraints, and in the types of input they deal with (arrays of numbers vs. strings).

10 Prerequisite LeetCode Problems

Here are 10 problems as a preparation for tackling the “Max Chunks To Make Sorted II” problem:

    1. Move Zeroes: It’s about manipulating arrays in place.
    1. Merge Sorted Array: This problem is about merging two sorted arrays.
    1. Kth Largest Element in an Array: This problem is about sorting and finding elements in a sorted array.
    1. Find All Duplicates in an Array: This problem also requires manipulation of arrays.
    1. Shortest Unsorted Continuous Subarray: It’s about finding subarrays in an array.
    1. Sort Colors: This is a variation of sorting which might give you insight on the “chunks”.
    1. Rotate Array: This is about performing operations on arrays.
    1. Missing Number: This is a bit more complex and works on finding elements in a modified sorted array.
    1. Find the Duplicate Number: This problem helps you understand better about finding specific patterns in arrays.
    1. Maximum Product of Three Numbers: This is a problem about finding specific patterns in a sorted array.
class Solution:
    def maxChunksToSorted(self, arr: List[int]) -> int:
        n = len(arr)
        maxOfLeft = [0]*n
        minOfRight = [0]*n

        maxOfLeft[0] = arr[0]
        for i in range(1, n):
            maxOfLeft[i] = max(maxOfLeft[i-1], arr[i])

        minOfRight[n - 1] = arr[n - 1]
        for i in range(n - 2, -1, -1):
            minOfRight[i] = min(minOfRight[i + 1], arr[i])

        res = 0
        for i in range(0, n - 1):
            if maxOfLeft[i] <= minOfRight[i + 1]:
                res += 1

        return res + 1

Problem Classification

This problem belongs to the domain of ‘Array Manipulation’ and ‘Sorting’.

‘What’ Components:

  1. An integer array arr.
  2. A task to split arr into some number of chunks and individually sort each chunk.
  3. The requirement that after concatenating sorted chunks, the result should equal the sorted array.
  4. The requirement to return the largest number of chunks we can make to sort the array.

This is an ‘Optimization Problem’ as the goal is to maximize the number of chunks. It involves the ‘Greedy Algorithm’ strategy since the best choice at each step (i.e., choosing the maximum number of chunks) should lead to the optimal global solution.

It is also a ‘Combinatorial Problem’ as it explores different ways to partition an array, and ‘Sorting’ as it involves sorting individual partitions. The problem is a ‘Single-objective Optimization’ as we have only one goal to achieve - maximizing the number of chunks.

Thought Process

Language Agnostic Coding Drills

  1. Dissecting the Code

The code provided can be deconstructed into the following distinct concepts or coding drills:

a. Variables and Lists: The code defines several variables and lists to keep track of relevant information as it processes the input.

b. Loops: The code uses for loops to traverse through the array.

c. Conditionals: The if statement is used to compare values and decide whether to increase the count of chunks.

d. List indexing: The code uses Python’s list indexing feature to access and update elements in the list.

e. Function definitions: The code is encapsulated in a function definition.

f. Mathematical operations: Basic mathematical operations such as max and min are used.

  1. Coding Concepts/Drills Classification

In increasing order of difficulty:

a. Variables and Lists: The simplest concept, this forms the base for data storage and manipulation in any coding task.

b. Mathematical operations: Basic use of max and min functions in Python.

c. List indexing: Understanding how to properly index and manipulate elements in a list.

d. Loops: More complex than the previous concepts, requires understanding of iteration over lists.

e. Conditionals: Used to perform action based on a condition, needs understanding of logical statements.

f. Function definitions: Requires understanding of how to encapsulate code in functions for modularity and reusability.

  1. Problem-Solving Approach

To solve the problem statement using the identified drills, the following step-by-step process can be used:

a. First, understand the problem statement and identify what needs to be done - in this case, finding maximum chunks that can be sorted individually to result in a sorted array.

b. Define variables and lists - Initialize a list for maxOfLeft and minOfRight to keep track of maximum and minimum values.

c. Loop over the input list from the start to the end, fill up maxOfLeft list such that every index i in maxOfLeft represents the maximum value from the start of the input list to index i.

d. Loop over the input list from the end to the start, fill up minOfRight list such that every index i in minOfRight represents the minimum value from the end of the input list to index i.

e. Finally, loop over the input list and increase the count of chunks when the maximum value from the start of the list to the index i is less than or equal to the minimum value from the index i+1 to the end of the list.

This process combines all the coding drills to form a complete solution for the problem statement.

Targeted Drills in Python

a. Variables and Lists
variable = 10
my_list = [1, 2, 3, 4, 5]

b. Mathematical operations
max_value = max(2, 3, 4)
min_value = min(2, 3, 4)

c. List Indexing
my_list = [1, 2, 3, 4, 5]
first_element = my_list[0]
last_element = my_list[-1]

d. Loops
for i in range(5):

e. Conditionals
if 5 > 3:
    print("5 is greater than 3")
    print("5 is not greater than 3")

f. Function definitions
def my_function():
    print("Hello, world!")

g. Creating auxiliary arrays python original_array = [5,4,3,2,1] aux_array = [0]*len(original_array) # initializing auxiliary array with zeroes

h. Filling auxiliary arrays using loops and conditionals ```python original_array = [5,4,3,2,1] max_left_array = [0]*len(original_array) max_left_array[0] = original_array[0]

for i in range(1, len(original_array)): 
    max_left_array[i] = max(max_left_array[i-1], original_array[i]) # filling max_of_left array

i. Comparing elements from two arrays ```python array1 = [5,4,3,2,1] array2 = [1,2,3,4,5]

for i in range(len(array1)):
    if array1[i] > array2[i]: 
        print(f"Element {array1[i]} in array1 is larger than element {array2[i]} in array2.")
        print(f"Element {array1[i]} in array1 is smaller or equal to element {array2[i]} in array2.")

j. Initializing auxiliary arrays python arr = [5,4,3,2,1] n = len(arr) max_of_left = [0]*n min_of_right = [0]*n

k. Filling auxiliary arrays ```python max_of_left[0] = arr[0] for i in range(1, n): max_of_left[i] = max(max_of_left[i-1], arr[i])

min_of_right[n-1] = arr[n-1]
for i in range(n-2, -1, -1):
    min_of_right[i] = min(min_of_right[i+1], arr[i])

l. Counting chunks python chunks = 0 for i in range(n-1): if max_of_left[i] <= min_of_right[i+1]: chunks += 1 chunks += 1 # adding the last chunk print(chunks)