Arithmetic Slices II - Subsequence

The problem for the number of all arithmetic subsequences in the given integer array nums. A sequence is called arithmetic if it consists of at least three elements and the difference between any two consecutive elements is the same.

We can solve this problem by using a dynamic programming approach.

  1. Initialize a DP Table: We will use a list of dictionaries, where dp[i][diff] represents the number of arithmetic subsequences that end at index i with a common difference of diff.
  2. Iterate Through the Array: We will iterate through the array, and for each pair of elements, calculate the difference and update the DP table.
  3. Count the Arithmetic Sequences: While updating the DP table, we also keep track of the count of valid arithmetic sequences.

Here’s the code:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15

class Solution:
    def numberOfArithmeticSlices(self, nums: List[int]) -> int:
        n = len(nums)
        dp = [collections.defaultdict(int) for _ in range(n)]
        count = 0

        for i in range(n):
            for j in range(i):
                diff = nums[i] - nums[j]
                # Increment the count of valid arithmetic sequences ending at index i with difference diff
                dp[i][diff] += dp[j][diff] + 1
                # If there was a valid sequence ending at index j with difference diff, it contributes to the count
                count += dp[j][diff]

        return count

The time complexity of this code is (O(n^2)), and the space complexity is also (O(n^2)), where (n) is the length of the nums array.

Example Explanation:

  • For nums = [2,4,6,8,10], the output will be 7, as there are seven arithmetic subsequences: [2,4,6], [4,6,8], [6,8,10], [2,4,6,8], [4,6,8,10], [2,4,6,8,10], [2,6,10].
  • For nums = [7,7,7,7,7], the output will be 16, as any subsequence of this array is arithmetic.
 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27

from collections import defaultdict
from typing import List

class Solution:
    def numberOfArithmeticSlices(self, nums: List[int]) -> int:
        # Using dictionary to keep track of the number of subsequences ending at each element
        # with a given difference
        dp = [defaultdict(int) for _ in nums]
        count = 0

        for i in range(len(nums)):
            for j in range(i):
                diff = nums[i] - nums[j]  # Finding the common difference
                
                # Increment the count for the current element with this difference
                dp[i][diff] += 1

                # If the previous element had any arithmetic subsequences ending with this difference
                # we can extend them by the current element
                if diff in dp[j]:
                    # Increment the count for the current element with this difference
                    dp[i][diff] += dp[j][diff]
                    
                    # Add the count of these extended subsequences to the result
                    count += dp[j][diff]

        return count

Identifying Problem Isomorphism

“Arithmetic Slices II - Subsequence” asks you to count the number of arithmetic subsequences in an array, a bit more complicated than simply identifying them.

A simpler problem is “Arithmetic Slices”, which asks you to count the number of arithmetic subarrays. This problem doesn’t require handling the additional complexity of subsequences, which do not have to contain adjacent elements.

An approximately isomorphic problem is “Number of Longest Increasing Subsequence”. It’s not exactly the same, but similar in the sense that you’re looking for certain types of subsequences.

A more complex problem is “Distinct Subsequences II”. It requires you to count subsequences under different conditions, and the sequences don’t necessarily need to be arithmetic.

Here is the ordered list of problems from simplest to most complex:

  1. “Arithmetic Slices” - Count the number of arithmetic subarrays.
  2. “Arithmetic Slices II - Subsequence” - Count the number of arithmetic subsequences.
  3. “Number of Longest Increasing Subsequence” - Count subsequences that have a different condition.
  4. “Distinct Subsequences II” - Count subsequences under more complex conditions.

“446. Arithmetic Slices II - Subsequence” is about finding arithmetic subsequences in an array. It involves understanding of dynamic programming (DP) and hashing. Here are 10 problems to prepare for it:

  1. 413. Arithmetic Slices: This is a simpler version of the problem where you need to find arithmetic subarrays instead of subsequences.

  2. 300. Longest Increasing Subsequence: This problem is a classic dynamic programming problem about subsequences.

  3. 673. Number of Longest Increasing Subsequence: This is a more complex version of problem 300 where you need to count the number of longest increasing subsequences.

  4. 646. Maximum Length of Pair Chain: This problem is also about finding the longest chain (subsequence) with certain properties.

  5. 152. Maximum Product Subarray: This problem asks for the contiguous subarray which has the largest product, which will give you good practice working with subarrays.

  6. 198. House Robber: A classic dynamic programming problem that helps build a foundation for understanding how to approach dynamic programming problems.

  7. 560. Subarray Sum Equals K: This problem will help you understand how to handle the cumulative sum concept which is useful for this problem.

  8. 718. Maximum Length of Repeated Subarray: This problem can help you get familiar with working with arrays and finding subsequences.

  9. 494. Target Sum: This problem is about finding all the subsets of an array that sum to a particular target, which can help you understand how to traverse all subsequences of an array.

  10. 53. Maximum Subarray: This problem is a basic one in dealing with subarrays, and will help you get comfortable with the concept.

Clarification Questions

What are the clarification questions we can ask about this problem?

 1
 2
 3
 4
 5
 6
 7
 8
 9
10

class Solution:
    def numberOfArithmeticSlices(self, nums: List[int]) -> int:
        total, n = 0, len(nums)
        dp = [defaultdict(int) for _ in nums]
        for i in range(1, n):
            for j in range(i):
                diff = nums[j] - nums[i]
                dp[i][diff] += dp[j][diff] + 1
                total += dp[j][diff]
        return total

The task is to count the number of arithmetic subsequences in an array. This is a Arithmetic Subsequence Counting Problem.

Language Agnostic Coding Drills

This problem is about counting all the arithmetic sequences in a list of numbers (also known as arithmetic slices). An arithmetic sequence is a sequence of numbers in which the difference between consecutive numbers is constant.

Drills:

  1. Understanding Lists, Arrays or Vectors: This is the fundamental data structure used in this problem. The input is a list of integers, and the goal is to find patterns within this list.

  2. Understanding Loops and Nested Loops: The main logic of the solution involves two nested loops that iterate through the list. Understanding how to use loops to iterate through a list, and particularly how to use one loop inside another, is crucial.

  3. Understanding Arithmetic Sequences: An arithmetic sequence is a sequence of numbers such that the difference of any two successive members is a constant. In this problem, you’re asked to find all such sequences within the given list.

  4. Understanding Dictionaries or Hash Maps: Dictionaries are used to keep track of the arithmetic slices found so far. This is an important part of the solution, as it allows us to effectively count the number of slices.

  5. Understanding Dynamic Programming: The solution to this problem involves a dynamic programming approach, where you store solutions to sub-problems (in this case, the number of arithmetic slices ending at each position in the list) and use them to solve the overall problem.

  6. Mathematics: This problem requires a clear understanding of arithmetic progression and summation of series. You need to calculate the difference between elements and check if it’s consistent.

  7. Problem Solving: This problem requires building up a solution incrementally, understanding the relationship between a smaller problem (the number of arithmetic slices ending at a particular position) and the overall problem.

Approach to the solution:

  • Create a list of dictionaries to track the number of arithmetic slices ending at each position in the list.
  • Iterate through the list twice using nested loops. In the inner loop, calculate the difference between the current element of the outer loop and the current element of the inner loop.
  • Update the dictionary at the outer loop’s current position by incrementing the count for this difference by one, and also add the count for this difference at the inner loop’s current position.
  • Increase the total count of arithmetic slices by the count for this difference at the inner loop’s current position.
  • Return the total count of arithmetic slices.

Targeted Drills in Python

Coding Drill 1: Understanding Lists, Arrays or Vectors

Create a list and perform various operations like appending elements, accessing elements, and iterating through the list.

1
2
3
4
5
6
7
8

numbers = [1, 2, 3, 4, 5]
print(numbers)  # Print the list

numbers.append(6)  # Append an element
print(numbers)

for number in numbers:  # Iterate through the list
    print(number)

Coding Drill 2: Understanding Loops and Nested Loops

Use nested loops to iterate through a list of lists.

1
2
3
4
5

matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

for row in matrix:
    for element in row:
        print(element)

Coding Drill 3: Understanding Dictionaries or Hash Maps

Create a dictionary, add key-value pairs to it, and iterate through it.

1
2
3
4
5
6
7

dictionary = {}  # Create a dictionary

dictionary['key1'] = 'value1'  # Add key-value pairs
dictionary['key2'] = 'value2'

for key, value in dictionary.items():  # Iterate through the dictionary
    print(key, value)

Coding Drill 4: Understanding Dynamic Programming

Implement a dynamic programming solution to the Fibonacci sequence.

1
2
3
4
5
6
7

def fibonacci(n):
    dp = [0, 1] + [0]*(n-1)
    for i in range(2, n+1):
        dp[i] = dp[i-1] + dp[i-2]
    return dp[n]

print(fibonacci(10))

Coding Drill 5: Mathematics - Arithmetic Progression

Write a function to generate an arithmetic progression.

1
2
3
4

def arithmetic_progression(start, diff, n):
    return [start + diff*i for i in range(n)]

print(arithmetic_progression(1, 2, 10))

Problem Specific Drill:

This problem involves identifying and counting arithmetic slices within a list. The following drill builds on the skills practiced in the previous drills and applies them to a simpler version of the problem: identifying and counting arithmetic slices of exactly length 3 within a list.

1
2
3
4
5
6
7
8

def count_arithmetic_slices(nums):
    count = 0
    for i in range(len(nums) - 2):
        if nums[i+1] - nums[i] == nums[i+2] - nums[i+1]:
            count += 1
    return count

print(count_arithmetic_slices([1, 2, 3, 4, 6, 8, 10, 12, 15]))