Introduction

In this blog post, we will explore how to solve the LeetCode problem "1425" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given an integer array nums and an integer k, return the maximum sum of a non-empty subsequence of that array such that for every two consecutive integers in the subsequence, nums[i] and nums[j], where i < j, the condition j - i <= k is satisfied. A subsequence of an array is obtained by deleting some number of elements (can be zero) from the array, leaving the remaining elements in their original order. Example 1: Input: nums = [10,2,-10,5,20], k = 2 Output: 37 Explanation: The subsequence is [10, 2, 5, 20]. Example 2: Input: nums = [-1,-2,-3], k = 1 Output: -1 Explanation: The subsequence must be non-empty, so we choose the largest number. Example 3: Input: nums = [10,-2,-10,-5,20], k = 2 Output: 23 Explanation: The subsequence is [10, -2, -5, 20]. Constraints: 1 <= k <= nums.length <= 105 -104 <= nums[i] <= 104

Explanation

Here's the breakdown of the solution:

• Dynamic Programming with a Deque: The problem can be solved using dynamic programming. dp[i] stores the maximum sum of a subsequence ending at index i. We use a deque to efficiently maintain a sliding window of the best dp values within the allowed range k.
• Deque for Optimization: The deque stores indices of potentially optimal dp values. It ensures that the deque only contains indices j such that dp[j] is greater than or equal to dp[x] for all x in the deque where j < x. This allows us to quickly find the best preceding element to extend the subsequence.

• Handling All Negative Numbers: If all numbers are negative, the subsequence must consist of only the largest negative number.

• Runtime Complexity: O(N)

• Storage Complexity: O(N)

Code

    from collections import deque

def constrainedSubsetSum(nums, k):
    n = len(nums)
    dp = [0] * n
    dp[0] = nums[0]
    deque_obj = deque([0])
    max_sum = nums[0]

    for i in range(1, n):
        while deque_obj and deque_obj[0] < i - k:
            deque_obj.popleft()

        if deque_obj:
            dp[i] = max(nums[i], nums[i] + dp[deque_obj[0]])
        else:
            dp[i] = nums[i]

        max_sum = max(max_sum, dp[i])

        while deque_obj and dp[deque_obj[-1]] <= dp[i]:
            deque_obj.pop()

        deque_obj.append(i)

    return max_sum