Sum of Subarray Minimums

Sum of Subarray Minimums - Problem

Given an array of integers arr, find the sum of min(b), where b ranges over every (contiguous) subarray of arr.

Since the answer may be large, return the answer modulo 10^9 + 7.

A subarray is a contiguous part of an array.

Input & Output

Example 1 — Basic Case

$ Input: arr = [3,1,2,4]

› Output: 17

💡 Note: Subarrays: [3]→3, [1]→1, [2]→2, [4]→4, [3,1]→1, [1,2]→1, [2,4]→2, [3,1,2]→1, [1,2,4]→1, [3,1,2,4]→1. Sum = 3+1+2+4+1+1+2+1+1+1 = 17

Example 2 — Small Array

$ Input: arr = [11,81,94,43,3]

› Output: 444

💡 Note: All 15 possible subarrays are considered, with their minimum values summed up to get 444

Example 3 — Single Element

$ Input: arr = [5]

› Output: 5

💡 Note: Only one subarray [5] with minimum 5

Constraints

1 ≤ arr.length ≤ 3 × 10⁴
1 ≤ arr[i] ≤ 3 × 10⁴

Visualization

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Asked in

G Google 45 a Amazon 38 M Microsoft 32

The key insight is to count how many subarrays each element is the minimum for, rather than generating all subarrays. Use a monotonic stack to efficiently find the range where each element is minimum. Best approach: Monotonic Stack. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force - Generate All Subarrays

⏱️ Time: O(n³) Space: O(1)

For each starting position, generate all subarrays beginning at that position. Find the minimum element in each subarray and add to total sum.

Monotonic Stack - Count Contributions

⏱️ Time: O(n) Space: O(n)

Instead of generating all subarrays, count how many subarrays each element will be the minimum for. Use a monotonic stack to efficiently find the range where each element is the minimum.

Brute Force - Generate All Subarrays — Algorithm Steps

Iterate through each starting position i
For each i, iterate through ending positions j
Find minimum in subarray arr[i:j+1]
Add minimum to running sum

Visualization

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Step-by-Step Walkthrough

Generate Subarrays

Create all possible contiguous subarrays

Find Minimums

Find minimum element in each subarray

Sum Results

Add all minimums together

Code -

solution.c — C

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

int solution(int* arr, int n) {
    const int MOD = 1000000007;
    long long total = 0;
    
    for (int i = 0; i < n; i++) {
        for (int j = i; j < n; j++) {
            int minVal = arr[i];
            for (int k = i; k <= j; k++) {
                if (arr[k] < minVal) {
                    minVal = arr[k];
                }
            }
            total = (total + minVal) % MOD;
        }
    }
    
    return total;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    int arr[1000];
    int n = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        if (token[0] >= '0' && token[0] <= '9' || token[0] == '-') {
            arr[n++] = atoi(token);
        }
        token = strtok(NULL, "[,]");
    }
    
    printf("%d\n", solution(arr, n));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

Three nested loops: n starting positions × n ending positions × n elements to find minimum

✓ Linear Growth

Space Complexity

O(1)

Only using constant extra variables

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Generate All Subarrays — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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