LeetCode 2552: Count Increasing Quadruplets

Explanation

To solve this problem, we can iterate through the given array nums and maintain two arrays left and right. The left array stores the count of increasing subarrays ending at index i such that nums[i] is the smallest element, and the right array stores the count of increasing subarrays starting at index i such that nums[i] is the largest element.

We then iterate through the array to calculate the total count of increasing quadruplets. For each index i, we calculate the count of increasing quadruplets where nums[i] is the second smallest element. This count is the product of left[i] * right[i].

Finally, we sum up the counts for all indices i to get the total count of increasing quadruplets.

• Time complexity: O(n^2) where n is the length of the input array nums.
• Space complexity: O(n) to store the left and right arrays.

class Solution {
    public int countQuadruplets(int[] nums) {
        int n = nums.length;
        int[] left = new int[n];
        int[] right = new int[n];
        
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[j] < nums[i]) {
                    left[i]++;
                }
            }
        }
        
        for (int i = n - 1; i >= 0; i--) {
            for (int j = n - 1; j > i; j--) {
                if (nums[i] < nums[j]) {
                    right[i]++;
                }
            }
        }
        
        int quadruplets = 0;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                quadruplets += left[i] * right[j];
            }
        }
        
        return quadruplets;
    }
}