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 * 10^4
- 1 <= arr[i] <= 3 * 10^4
Solution
class Solution {
public int sumSubarrayMins(int[] arr) {
// increasing stack
Stack<Integer> prevLess = new Stack<>();
Stack<Integer> nextLess = new Stack<>();
int[] left = new int[arr.length];
int[] right = new int[arr.length];
for (int i = 0; i < arr.length; i++) {
while (prevLess.size() > 0 && arr[i] <= arr[prevLess.peek()]) {
prevLess.pop();
}
left[i] = prevLess.size() == 0 ? i + 1: i - prevLess.peek();
prevLess.push(i);
}
for (int i = arr.length - 1; i >=0; i--) {
while (nextLess.size() > 0 && arr[i] < arr[nextLess.peek()]) {
nextLess.pop();
}
right[i] = nextLess.size() == 0 ? arr.length - i: nextLess.peek() - i;
nextLess.push(i);
}
long MOD = (long) 1e9 + 7;
long res = 0;
for (int i = 0; i < arr.length; i++) {
res = (res + (long)arr[i] * left[i] * right[i]) % MOD;
}
return (int) res;
}
}