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Java-based LeetCode algorithm problem solutions, regularly updated

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package g2501_2600.s2552_count_increasing_quadruplets;

// #Hard #Array #Dynamic_Programming #Prefix_Sum #Enumeration #Binary_Indexed_Tree
// #2023_08_18_Time_48_ms_(97.29%)_Space_43_MB_(93.41%)

import java.util.Arrays;

/**
 * 2552 - Count Increasing Quadruplets\.
 *
 * Hard
 *
 * Given a **0-indexed** integer array `nums` of size `n` containing all numbers from `1` to `n`, return _the number of increasing quadruplets_.
 *
 * A quadruplet `(i, j, k, l)` is increasing if:
 *
 * *   `0 <= i < j < k < l < n`, and
 * *   `nums[i] < nums[k] < nums[j] < nums[l]`.
 *
 * **Example 1:**
 *
 * **Input:** nums = [1,3,2,4,5]
 *
 * **Output:** 2
 *
 * **Explanation:**
 *
 * - When i = 0, j = 1, k = 2, and l = 3, nums[i] < nums[k] < nums[j] < nums[l].
 *
 * - When i = 0, j = 1, k = 2, and l = 4, nums[i] < nums[k] < nums[j] < nums[l]. There are no other quadruplets, so we return 2. 
 *
 * **Example 2:**
 *
 * **Input:** nums = [1,2,3,4]
 *
 * **Output:** 0
 *
 * **Explanation:**
 *
 * There exists only one quadruplet with i = 0, j = 1, k = 2, l = 3, but since nums[j] < nums[k], we return 0. 
 *
 * **Constraints:**
 *
 * *   `4 <= nums.length <= 4000`
 * *   `1 <= nums[i] <= nums.length`
 * *   All the integers of `nums` are **unique**. `nums` is a permutation.
**/
public class Solution {
    public long countQuadruplets(int[] nums) {
        int n = nums.length;
        long[] dp = new long[n];
        Arrays.fill(dp, 0);
        long ret = 0;
        for (int i = 1; i < n; i++) {
            int choice = 0;
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    choice++;
                    ret += dp[j];
                } else if (nums[i] < nums[j]) {
                    dp[j] += choice;
                }
            }
        }
        return ret;
    }
}