Skip to content

Latest commit

 

History

History
327 lines (276 loc) · 9.39 KB

README_EN.md

File metadata and controls

327 lines (276 loc) · 9.39 KB
Raw
comments difficulty edit_url tags
true
Medium
Stack
Array
Dynamic Programming
Monotonic Stack

907. Sum of Subarray Minimums

中文文档

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

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
    }
}