Array String Hash Map Sliding Window
Intermediate Level

Implement the boundedMaximumSubarrayCount method that counts subarrays whose maximum value stays within a given range.

You are given an integer array and two boundary values left and right. Count how many continuous subarrays have a maximum element between left and right, inclusive.

A subarray is invalid if all its values are smaller than left, or if any value is greater than right.

Example 1
Input:
nums (int[]) = [2,1,4,3]
size (int) = 4
left (int) = 2
right (int) = 3
Return:
(int) 3
Example 2
Input:
nums (int[]) = [2,9,2,5,6]
size (int) = 5
left (int) = 2
right (int) = 8
Return:
(int) 7
Example 3
Input:
nums (int[]) = [1,1,1]
size (int) = 3
left (int) = 2
right (int) = 3
Return:
(int) 0

Scan the array and keep track of two positions: the last index where a value was greater than right, and the last index where a value was at least left.

For each index, any subarray ending there must start after the last too-large value. It is valid only if it includes at least one value inside the range. The number of valid subarrays ending at the current index is lastInRange - lastTooLarge.

Add this count for every index.

Pseudocode:

function boundedMaximumSubarrayCount(nums, size, left, right):
    lastTooLarge = -1
    lastInRange = -1
    answer = 0
    for i from 0 to size - 1:
        if nums[i] > right:
            lastTooLarge = i
        if nums[i] >= left and nums[i] <= right:
            lastInRange = i
        if lastInRange > lastTooLarge:
            answer += lastInRange - lastTooLarge
    return answer
Run your code to see the result.