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 * 1041 <= arr[i] <= 3 * 104
Solutions
Solution 1: Monotonic Stack
The problem asks for the sum of the minimum values of each subarray, which is equivalent to finding the number of subarrays for which each element $arr[i]$ is the minimum, then multiplying by $arr[i]$, and finally summing these up.
Therefore, the focus of the problem is to find the number of subarrays for which $arr[i]$ is the minimum. For $arr[i]$, we find the first position $left[i]$ to its left that is less 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 for which $arr[i]$ is the minimum is $(i - left[i]) \times (right[i] - i)$.
Note, why do we find the first position $right[i]$ to the right that is less than or equal to $arr[i]$, rather than less than $arr[i]$? This is because if we find the first position $right[i]$ to the right that is less than $arr[i]$, it will lead to duplicate calculations.
Let's take an example to illustrate. For the following array:
The element at index $3$ is $2$, the first element to its left that is less than $2$ is at index $0$. If we find the first element to its right that is less than $2$, we get index $7$. That is, the subarray interval is $(0, 7)$. Note that this is an open interval.
0 4 3 2 5 3 2 1
* ^ *
In the same way, we can find the subarray interval for the element at index $6$, and find that its subarray interval is also $(0, 7)$. That is, the subarray intervals for the elements at index $3$ and index $6$ are the same. This leads to duplicate calculations.
0 4 3 2 5 3 2 1
* ^ *
If we find the first element to its right that is less than or equal to its value, there will be no duplication, because the subarray interval for the element at index $3$ becomes $(0, 6)$, and the subarray interval for the element at index $6$ is $(0, 7)$, which are not the same.
Back to this problem, we just need to traverse the array, for each element $arr[i]$, use a monotonic stack to find the first position $left[i]$ to its left that is less 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 for which $arr[i]$ is the minimum is $(i - left[i]) \times (right[i] - i)$, then multiply by $arr[i]$, and finally sum these up.
Be aware of data overflow and modulo operations.
The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $arr$.
Python3
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
Java
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;
}
}
C++
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;
}
};
Go
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
}
TypeScript
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;
}
Rust
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
}
}