LeetCode #907 — MEDIUM

Sum of Subarray Minimums

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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 10⁹ + 7.

Example 1:

Input: arr = [3,1,2,4]
Output: 17
Explanation: 
Subarrays are [3], [1], [2], [4], [3,1], [1,2], [2,4], [3,1,2], [1,2,4], [3,1,2,4]. 
Minimums are 3, 1, 2, 4, 1, 1, 2, 1, 1, 1.
Sum is 17.

Example 2:

Input: arr = [11,81,94,43,3]
Output: 444

Constraints:

1 <= arr.length <= 3 * 10⁴
1 <= arr[i] <= 3 * 10⁴

Step 01

Brute Force Baseline

Problem summary: 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 109 + 7.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming · Stack

Example 1

[3,1,2,4]

Example 2

[11,81,94,43,3]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #907: Sum of Subarray Minimums
class Solution {
    public int sumSubarrayMins(int[] arr) {
        int n = arr.length;
        int[] left = new int[n];
        int[] right = new int[n];
        Arrays.fill(left, -1);
        Arrays.fill(right, n);
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = 0; i < n; ++i) {
            while (!stk.isEmpty() && arr[stk.peek()] >= arr[i]) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                left[i] = stk.peek();
            }
            stk.push(i);
        }
        stk.clear();
        for (int i = n - 1; i >= 0; --i) {
            while (!stk.isEmpty() && arr[stk.peek()] > arr[i]) {
                stk.pop();
            }
            if (!stk.isEmpty()) {
                right[i] = stk.peek();
            }
            stk.push(i);
        }
        final int mod = (int) 1e9 + 7;
        long ans = 0;
        for (int i = 0; i < n; ++i) {
            ans += (long) (i - left[i]) * (right[i] - i) % mod * arr[i] % mod;
            ans %= mod;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #907: Sum of Subarray Minimums
func sumSubarrayMins(arr []int) (ans int) {
	n := len(arr)
	left := make([]int, n)
	right := make([]int, n)
	for i := range left {
		left[i] = -1
		right[i] = n
	}
	stk := []int{}
	for i, v := range arr {
		for len(stk) > 0 && arr[stk[len(stk)-1]] >= v {
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			left[i] = stk[len(stk)-1]
		}
		stk = append(stk, i)
	}
	stk = []int{}
	for i := n - 1; i >= 0; i-- {
		for len(stk) > 0 && arr[stk[len(stk)-1]] > arr[i] {
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			right[i] = stk[len(stk)-1]
		}
		stk = append(stk, i)
	}
	const mod int = 1e9 + 7
	for i, v := range arr {
		ans += (i - left[i]) * (right[i] - i) * v % mod
		ans %= mod
	}
	return
}

# Accepted solution for LeetCode #907: Sum of Subarray Minimums
class Solution:
    def sumSubarrayMins(self, arr: List[int]) -> int:
        n = len(arr)
        left = [-1] * n
        right = [n] * n
        stk = []
        for i, v in enumerate(arr):
            while stk and arr[stk[-1]] >= v:
                stk.pop()
            if stk:
                left[i] = stk[-1]
            stk.append(i)

        stk = []
        for i in range(n - 1, -1, -1):
            while stk and arr[stk[-1]] > arr[i]:
                stk.pop()
            if stk:
                right[i] = stk[-1]
            stk.append(i)
        mod = 10**9 + 7
        return sum((i - left[i]) * (right[i] - i) * v for i, v in enumerate(arr)) % mod

// Accepted solution for LeetCode #907: Sum of Subarray Minimums
use std::collections::VecDeque;

impl Solution {
    pub fn sum_subarray_mins(arr: Vec<i32>) -> i32 {
        let n = arr.len();
        let mut left = vec![-1; n];
        let mut right = vec![n as i32; n];
        let mut stk: VecDeque<usize> = VecDeque::new();

        for i in 0..n {
            while !stk.is_empty() && arr[*stk.back().unwrap()] >= arr[i] {
                stk.pop_back();
            }
            if let Some(&top) = stk.back() {
                left[i] = top as i32;
            }
            stk.push_back(i);
        }

        stk.clear();
        for i in (0..n).rev() {
            while !stk.is_empty() && arr[*stk.back().unwrap()] > arr[i] {
                stk.pop_back();
            }
            if let Some(&top) = stk.back() {
                right[i] = top as i32;
            }
            stk.push_back(i);
        }

        let MOD = 1_000_000_007;
        let mut ans: i64 = 0;
        for i in 0..n {
            ans += ((((right[i] - (i as i32)) * ((i as i32) - left[i])) as i64) * (arr[i] as i64))
                % MOD;
            ans %= MOD;
        }
        ans as i32
    }
}

// Accepted solution for LeetCode #907: Sum of Subarray Minimums
function sumSubarrayMins(arr: number[]): number {
    const n: number = arr.length;
    const left: number[] = Array(n).fill(-1);
    const right: number[] = Array(n).fill(n);
    const stk: number[] = [];
    for (let i = 0; i < n; ++i) {
        while (stk.length > 0 && arr[stk.at(-1)] >= arr[i]) {
            stk.pop();
        }
        if (stk.length > 0) {
            left[i] = stk.at(-1);
        }
        stk.push(i);
    }

    stk.length = 0;
    for (let i = n - 1; ~i; --i) {
        while (stk.length > 0 && arr[stk.at(-1)] > arr[i]) {
            stk.pop();
        }
        if (stk.length > 0) {
            right[i] = stk.at(-1);
        }
        stk.push(i);
    }

    const mod: number = 1e9 + 7;
    let ans: number = 0;
    for (let i = 0; i < n; ++i) {
        ans += ((((i - left[i]) * (right[i] - i)) % mod) * arr[i]) % mod;
        ans %= mod;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.