Constrained Subsequence Sum

Constrained Subsequence Sum - Problem

Array Dynamic Programming Queue Sliding Window Heap (Priority Queue) Monotonic Queue Hard

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [10,2,-10,5,20], k = 2

› Output: 37

💡 Note: The subsequence is [10, 2, 5, 20]. Distance between consecutive elements: 2-0=2≤k, 3-1=2≤k, 4-3=1≤k. Sum = 10+2+5+20 = 37

Example 2 — Constraint Limitation

$ Input: nums = [4,-1,2,3], k = 2

› Output: 9

💡 Note: The subsequence is [4, 2, 3] at indices [0, 2, 3]. Distance between consecutive elements: 2-0=2≤k, 3-2=1≤k. Sum = 4+2+3 = 9

Example 3 — Single Element

$ Input: nums = [-1,-2,-3], k = 1

› Output: -1

💡 Note: All elements are negative. The best subsequence is just [-1] with sum -1

Constraints

1 ≤ nums.length ≤ 10⁵
-10⁴ ≤ nums[i] ≤ 10⁴
1 ≤ k ≤ nums.length

Visualization

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Asked in

G Google 35 a Amazon 28 f Facebook 22 M Microsoft 18

The key insight is to use dynamic programming where dp[i] represents the maximum sum of a valid subsequence ending at position i. The optimal approach uses a monotonic deque to efficiently find the maximum DP value in the sliding window of size k. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force - Try All Subsequences

⏱️ Time: O(2^n) Space: O(n)

For each element, try including it or skipping it. When including an element, ensure the distance constraint is satisfied. This explores all possible valid subsequences.

Dynamic Programming

⏱️ Time: O(n × k) Space: O(n)

For each position i, calculate the maximum sum of a valid subsequence ending at position i. The answer is the maximum value across all positions.

Monotonic Deque Optimization

⏱️ Time: O(n) Space: O(k)

Optimize the DP approach by using a monotonic deque to maintain the maximum DP value in the current window of size k. This reduces the time complexity from O(nk) to O(n).

Brute Force - Try All Subsequences — Algorithm Steps

Generate all possible subsequences
Check if each subsequence satisfies the distance constraint
Calculate sum for valid subsequences and track maximum

Visualization

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Step-by-Step Walkthrough

Start

Begin with empty subsequence

Branch

For each element, try including or skipping it

Validate

Check distance constraint j - i <= k

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>

int maxSum = INT_MIN;

void backtrack(int* nums, int n, int k, int index, int currentSum, int lastIndex) {
    if (currentSum > maxSum) {
        maxSum = currentSum;
    }
    
    for (int i = index; i < n; i++) {
        if (lastIndex == -1 || i - lastIndex <= k) {
            backtrack(nums, n, k, i + 1, currentSum + nums[i], i);
        }
    }
}

int solution(int* nums, int n, int k) {
    maxSum = INT_MIN;
    backtrack(nums, n, k, 0, 0, -1);
    return maxSum;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int nums[100];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int k;
    scanf("%d", &k);
    
    int result = solution(nums, n, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

We generate 2^n possible subsequences and check each one

✓ Linear Growth

Space Complexity

O(n)

Recursion stack depth can go up to n levels

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Try All Subsequences — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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