3250. Find the Count of Monotonic Pairs I

Problem Statement

You are given an array of positive integers nums of length n.

We call a pair of non-negative integer arrays (arr1, arr2) monotonic if:

The lengths of both arrays are n.
arr1 is monotonically non-decreasing, in other words, arr1[0] <= arr1[1] <= ... <= arr1[n - 1].
arr2 is monotonically non-increasing, in other words, arr2[0] >= arr2[1] >= ... >= arr2[n - 1].
arr1[i] + arr2[i] == nums[i] for all 0 <= i <= n - 1.

Return the count of monotonic pairs.

Since the answer may be very large, return it modulo 109 + 7.

Example 1:

Input: nums = [2,3,2]

Output: 4

Explanation:

The good pairs are:

([0, 1, 1], [2, 2, 1])
([0, 1, 2], [2, 2, 0])
([0, 2, 2], [2, 1, 0])
([1, 2, 2], [1, 1, 0])

Example 2:

Input: nums = [5,5,5,5]

Output: 126

Constraints:

1 <= n == nums.length <= 2000
1 <= nums[i] <= 50

Intuition

At each index
Keep track of Prev_of array A and B

And select the numbers accordingly


Since, Prev_a <= NExt num

For loop should start from NExt_num to nums[index]

To take all positbilites

Now, 

At initial stages,

Prev A = 0, Prev B = 50
Take max and min, 

So initial will not affect

Links

https://leetcode.com/problems/find-the-count-of-monotonic-pairs-i/

Video Links

https://www.youtube.com/watch?v=W9Job3hNmvA

Approach 1:

C++

#define MOD 1000000007

class Solution {
public:
    int dp[2002][52][52];
    int find_ans(int index, vector<int>& nums, int a_prev, int b_prev){
        if(index == nums.size()){
            return 1;
        }

        if(dp[index][a_prev][b_prev] != -1)
            return dp[index][a_prev][b_prev];

        int ans = 0;
        for(int i=a_prev; i<=nums[index]; i++){
            int x = i, y = nums[index] - i;
            if(a_prev <= x and b_prev>=y)
                ans = ((ans%MOD) + (find_ans(index+1, nums, x, y)%MOD)%MOD);
        }

        return dp[index][a_prev][b_prev] = ans;
    }

    int countOfPairs(vector<int>& nums) {
        memset(dp, -1, sizeof(dp));

        return find_ans(0, nums, 0, 50);
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++