README_EN.md

2104. Sum of Subarray Ranges

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Description

You are given an integer array nums. The range of a subarray of nums is the difference between the largest and smallest element in the subarray.

Return the sum of all subarray ranges of nums.

A subarray is a contiguous non-empty sequence of elements within an array.

 

Example 1:

Input: nums = [1,2,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0 
[2], range = 2 - 2 = 0
[3], range = 3 - 3 = 0
[1,2], range = 2 - 1 = 1
[2,3], range = 3 - 2 = 1
[1,2,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 1 + 1 + 2 = 4.

Example 2:

Input: nums = [1,3,3]
Output: 4
Explanation: The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0
[3], range = 3 - 3 = 0
[3], range = 3 - 3 = 0
[1,3], range = 3 - 1 = 2
[3,3], range = 3 - 3 = 0
[1,3,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 2 + 0 + 2 = 4.

Example 3:

Input: nums = [4,-2,-3,4,1]
Output: 59
Explanation: The sum of all subarray ranges of nums is 59.

 

Constraints:

• 1 <= nums.length <= 1000
• -109 <= nums[i] <= 109

 

Follow-up: Could you find a solution with O(n) time complexity?

Solutions

Python3

class Solution:
    def subArrayRanges(self, nums: List[int]) -> int:
        ans, n = 0, len(nums)
        for i in range(n - 1):
            mi = mx = nums[i]
            for j in range(i + 1, n):
                mi = min(mi, nums[j])
                mx = max(mx, nums[j])
                ans += (mx - mi)
        return ans

Java

class Solution {
    public long subArrayRanges(int[] nums) {
        long ans = 0;
        int n = nums.length;
        for (int i = 0; i < n - 1; ++i) {
            int mi = nums[i], mx = nums[i];
            for (int j = i + 1; j < n; ++j) {
                mi = Math.min(mi, nums[j]);
                mx = Math.max(mx, nums[j]);
                ans += (mx - mi);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    long long subArrayRanges(vector<int>& nums) {
        long long ans = 0;
        int n = nums.size();
        for (int i = 0; i < n - 1; ++i)
        {
            int mi = nums[i], mx = nums[i];
            for (int j = i + 1; j < n; ++j)
            {
                mi = min(mi, nums[j]);
                mx = max(mx, nums[j]);
                ans += (mx - mi);
            }
        }
        return ans;
    }
};

Go

func subArrayRanges(nums []int) int64 {
	var ans int64
	n := len(nums)
	for i := 0; i < n-1; i++ {
		mi, mx := nums[i], nums[i]
		for j := i + 1; j < n; j++ {
			mi = min(mi, nums[j])
			mx = max(mx, nums[j])
			ans += (int64)(mx - mi)
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

TypeScript

function subArrayRanges(nums: number[]): number {
    const n = nums.length;
    let res = 0;
    for (let i = 0; i < n - 1; i++) {
        let min = nums[i];
        let max = nums[i];
        for (let j = i + 1; j < n; j++) {
            min = Math.min(min, nums[j]);
            max = Math.max(max, nums[j]);
            res += max - min;
        }
    }
    return res;
}

Rust

impl Solution {
    pub fn sub_array_ranges(nums: Vec<i32>) -> i64 {
        let n = nums.len();
        let mut res: i64 = 0;
        for i in 1..n {
            let mut min = nums[i - 1];
            let mut max = nums[i - 1];
            for j in i..n {
                min = min.min(nums[j]);
                max = max.max(nums[j]);
                res += (max - min) as i64;
            }
        }
        res
    }
}

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