LeetCode 3113: Find the Number of Subarrays Where Boundary Elements Are Maximum Solution

Master LeetCode problem 3113 (Find the Number of Subarrays Where Boundary Elements Are Maximum), 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.

3113. Find the Number of Subarrays Where Boundary Elements Are Maximum

Hard

Array Binary Search Stack Monotonic Stack

Problem Explanation

Explanation

To solve this problem, we can iterate through the given array nums and keep track of the largest element seen so far. For each element in nums, we find the maximum element on its left side and the maximum element on its right side. If the current element is equal to the largest element, we count the number of subarrays that can be formed based on the positions of the maximum elements on the left and right sides. We accumulate this count for each element in the array.

Algorithm:

Initialize variables result to 0, n to the length of nums, and maxVal to the minimum integer value.
Initialize an array left of size n to store the maximum element on the left side of each element.
Initialize an array right of size n to store the maximum element on the right side of each element.
Iterate from left to right in nums and update left and maxVal accordingly.
Iterate from right to left in nums and update right.
Iterate over nums and for each element, if it equals maxVal, calculate the number of subarrays possible based on the positions of maximum elements on the left and right.
Add this count to result.
Return result.

The time complexity of this approach is O(n) where n is the length of the input array. The space complexity is O(n) for the left and right arrays.

Solution Code

class Solution {
    public int countSubarrays(int[] nums) {
        int result = 0;
        int n = nums.length;
        int maxVal = Integer.MIN_VALUE;
        int[] left = new int[n];
        int[] right = new int[n];

        for (int i = 0; i < n; i++) {
            maxVal = Math.max(maxVal, nums[i]);
            left[i] = maxVal;
        }

        maxVal = Integer.MIN_VALUE;

        for (int i = n - 1; i >= 0; i--) {
            maxVal = Math.max(maxVal, nums[i]);
            right[i] = maxVal;
        }

        for (int i = 0; i < n; i++) {
            if (nums[i] == left[i] && nums[i] == right[i]) {
                result += (i + 1) * (n - i);
            }
        }

        return result;
    }
}

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