LeetCode 3250: Find the Count of Monotonic Pairs I Solution

Master LeetCode problem 3250 (Find the Count of Monotonic Pairs I), 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.

3250. Find the Count of Monotonic Pairs I

Hard

Array Math Dynamic Programming Combinatorics Prefix Sum

Problem Explanation

Explanation:

To solve this problem, we can iterate through all possible values for arr1[0] and arr2[0], then recursively calculate the count of monotonic pairs with remaining elements. We can use dynamic programming to avoid recalculating the same subproblems. The total count of monotonic pairs will be the sum of counts for each possible arr1[0] and arr2[0].

Algorithm:

Initialize a 2D array dp to store the count of monotonic pairs for each position in arr1 and arr2.
Iterate through all possible values for arr1[0] and arr2[0].
Recursively calculate the count of monotonic pairs for remaining elements based on the constraints.
Update dp with the calculated count.
Return the total count modulo 10^9 + 7.

Time Complexity: O(n^2) where n is the size of the input array. Space Complexity: O(n^2) for the dynamic programming array.

Solution Code

class Solution {
    public int countMonotonicPairs(int[] nums) {
        int n = nums.length;
        int MOD = 1000000007;
        int[][] dp = new int[n + 1][n + 1];
        
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= n; j++) {
                if (i == 0 && j == 0) {
                    dp[i][j] = 1;
                } else {
                    for (int k = 1; k <= n; k++) {
                        if (i >= k && nums[i - k] >= k) {
                            dp[i][j] = (dp[i][j] + dp[i - k][j]) % MOD;
                        }
                        if (j >= k && nums[j - k] >= k) {
                            dp[i][j] = (dp[i][j] + dp[i][j - k]) % MOD;
                        }
                    }
                }
            }
        }
        
        return dp[n][n];
    }
}

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