We are given an array A of positive integers, and two positive integers L and R (L <= R).
Return the number of (contiguous, non-empty) subarrays such that the value of the maximum array element in that subarray is at least L and at most R.
Example :
Input:
A = [2, 1, 4, 3]
L = 2
R = 3
Output: 3
Explanation: There are three subarrays that meet the requirements: [2], [2, 1], [3].
Note:
L, R and A[i] will be an integer in the range [0, 10^9].
The length of A will be in the range of [1, 50000].
/*
The condition A[i]>=L && A[i]<=R,means that A[j:i] is a valid subarray and thus we can have (i-j+1) valid subarrays,
count is the valid subarrays between j to i at this point.
The condition A[i]<L means that A[j:i] is still a valid subarray but we need the last element (>=L and <=R) which is within A[j:i],
thus adding last valid number of subarrays which is count.
Else just move the back pointer forward
*/
class Solution {
public int numSubarrayBoundedMax(int[] A, int L, int R) {
int start = 0, count = 0, res = 0;
for (int end = 0; end < A.length; end++) {
if (A[end] >= L && A[end] <= R) {
res += end - start + 1;
count = end - start + 1;
} else if (A[end] < L)
res += count;
else {
start = end + 1;
count = 0;
}
}
return res;
}
}