2104. Sum of Subarray Ranges

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

• 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;
}
}

• 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;
}
};

• 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


• 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
}

• 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;
}


• 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
}
}