# 3250. Find the Count of Monotonic Pairs I

## Problem Statement

<br>

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`.

&#x20;

**Example 1:**

**Input:** nums = \[2,3,2]

**Output:** 4

**Explanation:**

The good pairs are:

1. `([0, 1, 1], [2, 2, 1])`
2. `([0, 1, 2], [2, 2, 0])`
3. `([0, 2, 2], [2, 1, 0])`
4. `([1, 2, 2], [1, 1, 0])`

**Example 2:**

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

**Output:** 126

&#x20;

**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:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
#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);
    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###


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