# 907. Sum of Subarray Minimums

## Description

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

## Solutions

The problem asks for the sum of the minimum values of each subarray, which is actually equivalent to finding the number of subarrays for each element $arr[i]$ where $arr[i]$ is the minimum, multiplying each by $arr[i]$, and then summing these products.

Thus, the focus of the problem is translated to finding the number of subarrays for which $arr[i]$ is the minimum.

For each $arr[i]$, we identify the first position $left[i]$ to its left that is smaller than $arr[i]$ and the first position $right[i]$ to its right that is less than or equal to $arr[i]$.

The number of subarrays where $arr[i]$ is the minimum can then be given by $(i - left[i]) \times (right[i] - i)$.

It’s important to note why we are looking for the first position $right[i]$ that is less than or equal to $arr[i]$ and not less than $arr[i]$.

If we were to look for the first position less than $arr[i]$, we would end up double-counting.

For instance, consider the following array:

The element at index $3$ is $2$, and the first element less than $2$ to its left is at index $0$. If we find the first element less than $2$ to its right, we would end up at index $7$. That means the subarray interval is $(0, 7)$. Note that this is an open interval.

0 4 3 2 5 3 2 1
*     ^       *


If we calculate the subarray interval for the element at index $6$ using the same method, we would find that its interval is also $(0, 7)$.

0 4 3 2 5 3 2 1
*           ^ *


Therefore, the subarray intervals of the elements at index $3$ and $6$ are overlapping, leading to double-counting.

If we were to find the first element less than or equal to $arr[i]$ to its right, we wouldn’t have this problem.

The subarray interval for the element at index $3$ would become $(0, 6)$ and for the element at index $6$ it would be $(0, 7)$, and these two are not overlapping.

To solve this problem, we just need to traverse the array.

For each element $arr[i]$, we use a monotonic stack to find its $left[i]$ and $right[i]$.

Then the number of subarrays where $arr[i]$ is the minimum can be calculated by $(i - left[i]) \times (right[i] - i)$. Multiply this by $arr[i]$ and sum these values for all $i$ to get the final answer.

Remember to take care of data overflow and modulus operation.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $arr$.

• 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;
}
}

• class Solution {
public:
int sumSubarrayMins(vector<int>& arr) {
int n = arr.size();
vector<int> left(n, -1);
vector<int> right(n, n);
stack<int> stk;
for (int i = 0; i < n; ++i) {
while (!stk.empty() && arr[stk.top()] >= arr[i]) {
stk.pop();
}
if (!stk.empty()) {
left[i] = stk.top();
}
stk.push(i);
}
stk = stack<int>();
for (int i = n - 1; i >= 0; --i) {
while (!stk.empty() && arr[stk.top()] > arr[i]) {
stk.pop();
}
if (!stk.empty()) {
right[i] = stk.top();
}
stk.push(i);
}
long long ans = 0;
const int mod = 1e9 + 7;
for (int i = 0; i < n; ++i) {
ans += 1LL * (i - left[i]) * (right[i] - i) * arr[i] % mod;
ans %= mod;
}
return ans;
}
};

• 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


• 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
}

• 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;
}


• 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
}
}