Sum of Variable Length Subarrays

Sum of Variable Length Subarrays - Problem

You are given an integer array nums of size n. For each index i where 0 <= i < n, define a subarray nums[start ... i] where start = max(0, i - nums[i]).

Return the total sum of all elements from the subarray defined for each index in the array.

Input & Output

Example 1 — Basic Case

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

› Output: 17

💡 Note: For i=0: start=max(0,0-1)=0, subarray=[1], sum=1. For i=1: start=max(0,1-2)=0, subarray=[1,2], sum=3. For i=2: start=max(0,2-3)=0, subarray=[1,2,3], sum=6. For i=3: start=max(0,3-2)=1, subarray=[2,3,2], sum=7. Total: 1+3+6+7=17.

Example 2 — Edge Case

$ Input: nums = [5,1,3,2]

› Output: 26

💡 Note: For i=0: start=max(0,0-5)=0, subarray=[5], sum=5. For i=1: start=max(0,1-1)=0, subarray=[5,1], sum=6. For i=2: start=max(0,2-3)=0, subarray=[5,1,3], sum=9. For i=3: start=max(0,3-2)=1, subarray=[1,3,2], sum=6. Total: 5+6+9+6=26.

Example 3 — Small Array

$ Input: nums = [2,1]

› Output: 5

💡 Note: For i=0: start=max(0,0-2)=0, subarray=[2], sum=2. For i=1: start=max(0,1-1)=0, subarray=[2,1], sum=3. Total: 2+3=5.

Constraints

1 ≤ nums.length ≤ 10³
0 ≤ nums[i] ≤ nums.length

Visualization

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Asked in

G Google 25 a Amazon 20

The key insight is that each element defines a variable-length subarray ending at its position. The prefix sum approach optimizes range sum queries from O(n) to O(1). Best approach uses prefix sums with Time: O(n), Space: O(n).

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(1)

For each index i, find the start position using max(0, i - nums[i]), then sum all elements from start to i. Repeat for all indices and add to total.

Prefix Sum Optimization

⏱️ Time: O(n) Space: O(n)

Build a prefix sum array first, then use it to calculate any subarray sum in constant time using the formula: sum[start...end] = prefix[end+1] - prefix[start].

Brute Force — Algorithm Steps

Step 1: For each index i, calculate start = max(0, i - nums[i])
Step 2: Sum elements from start to i
Step 3: Add this sum to total result

Visualization

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Step-by-Step Walkthrough

For each index

Calculate start position

Sum elements

Add all elements in range

Add to total

Accumulate result

Code -

solution.c — C

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

int solution(int* nums, int n) {
    int totalSum = 0;
    
    for (int i = 0; i < n; i++) {
        int start = (i - nums[i] > 0) ? i - nums[i] : 0;
        int subarraySum = 0;
        for (int j = start; j <= i; j++) {
            subarraySum += nums[j];
        }
        totalSum += subarraySum;
    }
    
    return totalSum;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int nums[1000];
    int n = 0;
    char* token = strtok(line + 1, ",]");
    while (token != NULL) {
        nums[n++] = atoi(token);
        token = strtok(NULL, ",]");
    }
    
    int result = solution(nums, n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops - outer loop n times, inner loop up to n times for each subarray sum

⚠ Quadratic Growth

Space Complexity

O(1)

Only using variables for calculations, no extra data structures

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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