Problem
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 <= 1000-10^9 <= nums[i] <= 10^9
Follow-up: Could you find a solution with O(n) time complexity?
Solution (Java)
class Solution {
public long subArrayRanges(int[] nums) {
int n = nums.length;
long sum = 0;
Deque<Integer> q = new ArrayDeque<>();
q.add(-1);
for (int i = 0; i <= n; i++) {
while (q.peekLast() != -1 && (i == n || nums[q.peekLast()] <= nums[i])) {
int cur = q.removeLast();
int left = q.peekLast();
int right = i;
sum += 1L * (cur - left) * (right - cur) * nums[cur];
}
q.add(i);
}
q.clear();
q.add(-1);
for (int i = 0; i <= n; i++) {
while (q.peekLast() != -1 && (i == n || nums[q.peekLast()] >= nums[i])) {
int cur = q.removeLast();
int left = q.peekLast();
int right = i;
sum -= 1L * (cur - left) * (right - cur) * nums[cur];
}
q.add(i);
}
return sum;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).