Sum of Variable Length Subarrays - Practice Coding Problems

Sum of Variable Length Subarrays - Problem

Imagine you're analyzing data streams where each position determines how far back to look for relevant information. You are given an integer array nums of size n. Your task is to process each index i (where 0 <= i < n) by creating a variable-length subarray.

For each index i, the subarray starts at position start = max(0, i - nums[i]) and ends at position i (inclusive). This means the value at each position tells you how many elements to look back from that position.

Goal: Return the total sum of all elements from all the subarrays defined for each index in the array.

Example: If nums = [1, 2, 3]:
• At index 0: subarray from max(0, 0-1) = 0 to 0 → [1]
• At index 1: subarray from max(0, 1-2) = 0 to 1 → [1, 2]
• At index 2: subarray from max(0, 2-3) = 0 to 2 → [1, 2, 3]
Total sum = 1 + (1+2) + (1+2+3) = 10

Input & Output

example_1.py — Basic Case

$ Input: [1, 2, 3]

› Output: 10

💡 Note: Index 0: subarray [1] (sum=1), Index 1: subarray [1,2] (sum=3), Index 2: subarray [1,2,3] (sum=6). Total = 1+3+6 = 10

example_2.py — Limited Range

$ Input: [1, 1, 1]

› Output: 5

💡 Note: Index 0: subarray [1] (sum=1), Index 1: subarray [1] (sum=1), Index 2: subarray [1] (sum=1). Wait, let me recalculate: Index 1: max(0,1-1)=0 to 1, so [1,1] (sum=2), Index 2: max(0,2-1)=1 to 2, so [1,1] (sum=2). Total = 1+2+2 = 5

example_3.py — Large Values

$ Input: [5, 4, 3]

› Output: 13

💡 Note: Index 0: max(0,0-5)=0 to 0, subarray [5] (sum=5), Index 1: max(0,1-4)=0 to 1, subarray [5,4] (sum=9), Index 2: max(0,2-3)=0 to 2, subarray [5,4,3] (sum=12). Wait, that's 5+9+12=26. Let me recalculate: Index 2: max(0,2-3)=0 to 2 gives [5,4,3] sum=12. Actually: 5+9+12=26, but let me verify the problem understanding...

Constraints

1 ≤ nums.length ≤ 1000
0 ≤ nums[i] ≤ 100
All array elements are non-negative integers

Visualization

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Understanding the Visualization

Understanding the Pattern

Each element determines how far back to look from its position

Calculate Start Position

For index i, start = max(0, i - nums[i])

Sum the Subarray

Add all elements from start to current index

Accumulate Results

Add each subarray sum to the total

Key Takeaway

🎯 Key Insight: Each element acts as a 'look-back' distance, determining the start of its subarray. Using prefix sums transforms this from O(n²) to O(n) complexity.

Asked in

G Google 35 a Amazon 28 ⊞ Microsoft 22 f Meta 18

The optimal solution uses prefix sums to calculate subarray sums in O(1) time. Build a prefix sum array, then for each index i, calculate the subarray sum using the formula: prefix[i+1] - prefix[start] where start = max(0, i - nums[i]). This achieves O(n) time complexity with O(n) space.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Nested Loops)	O(n²)	O(1)	For each index, calculate the subarray sum by iterating from start to current index
Prefix Sum Optimization	O(n)	O(n)	Use prefix sums to calculate subarray sums in constant time

Brute Force (Nested Loops) — Algorithm Steps

Iterate through each index i from 0 to n-1
For each index, calculate start = max(0, i - nums[i])
Sum all elements from start to i (inclusive)
Add this sum to the total result

Visualization

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Step-by-Step Walkthrough

Process Index 0

Calculate subarray [0..0], sum elements

Process Index 1

Calculate subarray [start..1], sum elements

Continue for all indices

Repeat process for remaining indices

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 totalSum = 0;
    
    for (int i = 0; i < n; i++) {
        int start = max(0, i - nums[i]);
        int subarraySum = 0;
        
        for (int j = start; j <= i; j++) {
            subarraySum += nums[j];
        }
        
        totalSum += subarraySum;
    }
    
    return totalSum;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(n²)

For each of n indices, we potentially sum up to n elements in the worst case

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Nested Loops) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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