LeetCode 3251: Find the Count of Monotonic Pairs II Solution

Master LeetCode problem 3251 (Find the Count of Monotonic Pairs II), 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.

3251. Find the Count of Monotonic Pairs II

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], and then recursively calculate the counts for the rest of the elements in arr1 and arr2 while making sure that the conditions for monotonic pairs hold. We can use dynamic programming to store and reuse the counts for subproblems.

Initialize a 2D array dp of size n x 1000 x 1000 to store the counts.
Define a recursive function countPairs which takes the current index idx, the current values of arr1 and arr2, and returns the count of monotonic pairs that can be formed.
Inside the function, if the current index is the last index, check if the values satisfy the conditions and return 1 if true, else return 0.
Otherwise, iterate through all possible values for arr1[idx] and arr2[idx], and recursively call the function for the next index.
Store the count in the dp array to avoid recomputation.
Finally, return the total count of monotonic pairs.

Time complexity: O(n * 1000^2) where n is the length of the input array. Space complexity: O(n * 1000^2) for the dp array.

Solution Code

class Solution {
    int MOD = 1000000007;
    
    public int countPairs(int[] nums) {
        int n = nums.length;
        long[][][] dp = new long[n][1001][1001];
        return (int) countPairs(0, nums, 0, 0, dp);
    }
    
    private long countPairs(int idx, int[] nums, int val1, int val2, long[][][] dp) {
        if (idx == nums.length) {
            return 1;
        }
        
        if (dp[idx][val1][val2] != 0) {
            return dp[idx][val1][val2];
        }
        
        long count = 0;
        
        for (int i = val1; i <= nums[idx]; i++) {
            for (int j = val2; j >= i; j--) {
                count = (count + countPairs(idx + 1, nums, i, j, dp)) % MOD;
            }
        }
        
        dp[idx][val1][val2] = count;
        return count;
    }
}

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