LeetCode 2444: Count Subarrays With Fixed Bounds Solution

Master LeetCode problem 2444 (Count Subarrays With Fixed Bounds), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2444. Count Subarrays With Fixed Bounds

Hard

Array Queue Sliding Window Monotonic Queue

Problem Explanation

Explanation:

To solve this problem, we can use a two-pointer sliding window approach. We will iterate through the array and keep track of the indices of the minimum and maximum values within the current subarray. Whenever we encounter a subarray where the minimum value is equal to minK and the maximum value is equal to maxK, we can count all the valid subarrays that end at the right pointer.

Here are the detailed steps:

Initialize two pointers left and right to track the current subarray.
Iterate through the array using the right pointer.
Update the minimum and maximum values in the current subarray.
If the current subarray satisfies the conditions, we can calculate the number of valid subarrays ending at the right pointer.
Increment the right pointer and update the count of valid subarrays.
Continue the process until we reach the end of the array.

Time Complexity: O(N) where N is the number of elements in the given array. Space Complexity: O(1) since we are using a constant amount of extra space.

Solution Code

class Solution {
    public int numSubarrayBoundedMax(int[] nums, int minK, int maxK) {
        int count = 0;
        int left = 0, right = 0;
        int validSubarrays = 0;
        int prevCount = 0;
        
        while (right < nums.length) {
            if (nums[right] >= minK && nums[right] <= maxK) {
                prevCount = right - left + 1;
                validSubarrays += prevCount;
                count += prevCount;
            } else if (nums[right] < minK) {
                validSubarrays += prevCount;
            } else {
                prevCount = 0;
                left = right + 1;
            }
            right++;
        }
        
        return count;
    }
}

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