Constrained Subsequence Sum - Practice Coding Problems

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: 6

💡 Note: The subsequence is [4, 2, 3]. We skip -1. Distance: 2-0=2≤k, 3-2=1≤k. Sum = 4+2+3 = 9. Wait, let's recalculate: actually [4,3] gives 7, [2,3] gives 5, single element max is 4. Best is actually [4,2,3] = 9. But we need to check constraint carefully.

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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Understanding the Visualization

Input

Array [10,2,-10,5,20] with constraint k=2

Process

Find subsequence where consecutive elements are at most k positions apart

Output

Maximum sum 37 from subsequence [10,2,5,20]

Key Takeaway

🎯 Key Insight: Use dynamic programming with sliding window maximum to efficiently find the best previous position within distance k

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

Approach	Time	Space	Notes
✓ Monotonic Deque Optimization	O(n)	O(k)	Use monotonic deque to efficiently find maximum in sliding window
Dynamic Programming	O(n × k)	O(n)	Use DP to track maximum sum ending at each position
Brute Force - Try All Subsequences	O(2^n)	O(n)	Generate all valid subsequences and find the maximum sum

Monotonic Deque Optimization — Algorithm Steps

Use DP array to track maximum sum ending at each position
Maintain a monotonic deque of indices with decreasing DP values
For each position, remove outdated indices and get maximum from deque front

Visualization

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Step-by-Step Walkthrough

Initialize

Start with first element in deque

Slide Window

Remove outdated indices, add current index

Get Maximum

Front of deque gives maximum in current window

Code -

solution.c — C

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

int max(int a, int b) {
    return a > b ? a : b;
}

int solution(int* nums, int n, int k) {
    int* dp = (int*)malloc(n * sizeof(int));
    dp[0] = nums[0];
    
    // Simple array as deque (for this problem size)
    int* dq = (int*)malloc(n * sizeof(int));
    int front = 0, back = 0;
    dq[back++] = 0;
    
    for (int i = 1; i < n; i++) {
        // Remove indices that are out of range
        while (front < back && dq[front] < i - k) {
            front++;
        }
        
        // Current dp[i] is at least nums[i]
        dp[i] = nums[i];
        if (front < back) {
            dp[i] = max(dp[i], dp[dq[front]] + nums[i]);
        }
        
        // Maintain monotonic property
        while (front < back && dp[dq[back-1]] <= dp[i]) {
            back--;
        }
        
        dq[back++] = i;
    }
    
    int result = dp[0];
    for (int i = 1; i < n; i++) {
        result = max(result, dp[i]);
    }
    
    free(dp);
    free(dq);
    return result;
}

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(n)

Each element is added and removed from deque at most once

✓ Linear Growth

Space Complexity

O(k)

Deque stores at most k indices, plus O(n) for DP array

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Monotonic Deque Optimization — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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