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LeetCode 795: Number of Subarrays with Bounded Maximum Solution

Master LeetCode problem 795 (Number of Subarrays with Bounded Maximum), a medium 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.

795. Number of Subarrays with Bounded Maximum

Medium
ArrayTwo Pointers

Problem Explanation

Explanation:

To solve this problem, we can use a sliding window approach. We iterate through the array and maintain two pointers, left and right, representing the left and right boundaries of the current subarray. We also keep track of the count of valid subarrays where the maximum element falls within the given range [left, right].

Here are the steps:

  1. Initialize variables count and result to track the total count and the count of valid subarrays, respectively.
  2. Iterate through the array:
    • If the current element is within the range [left, right], update the count of valid subarrays by the length of the current subarray.
    • If the current element is less than left, reset the count of valid subarrays to 0 as the current subarray cannot contribute to the count.
    • If the current element is greater than right, update the last invalid index to the current index.
    • Update the result by adding the count of valid subarrays.
  3. Return the final result.

The time complexity of this approach is O(N), where N is the number of elements in the input array. The space complexity is O(1) as we are using a constant amount of space.

:

Solution Code

class Solution {
    public int numSubarrayBoundedMax(int[] nums, int left, int right) {
        int count = 0;
        int result = 0;
        int lastInvalid = -1;
        
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] >= left && nums[i] <= right) {
                count = i - lastInvalid;
                result += count;
            } else if (nums[i] < left) {
                count = 0;
            } else {
                lastInvalid = i;
            }
            result += count;
        }
        
        return result;
    }
}

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