<p>You are given an array of <strong>positive</strong> integers <code>nums</code> of length <code>n</code>.</p>

<p>We call a pair of <strong>non-negative</strong> integer arrays <code>(arr1, arr2)</code> <strong>monotonic</strong> if:</p>

<ul>
	<li>The lengths of both arrays are <code>n</code>.</li>
	<li><code>arr1</code> is monotonically <strong>non-decreasing</strong>, in other words, <code>arr1[0] &lt;= arr1[1] &lt;= ... &lt;= arr1[n - 1]</code>.</li>
	<li><code>arr2</code> is monotonically <strong>non-increasing</strong>, in other words, <code>arr2[0] &gt;= arr2[1] &gt;= ... &gt;= arr2[n - 1]</code>.</li>
	<li><code>arr1[i] + arr2[i] == nums[i]</code> for all <code>0 &lt;= i &lt;= n - 1</code>.</li>
</ul>

<p>Return the count of <strong>monotonic</strong> pairs.</p>

<p>Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [2,3,2]</span></p>

<p><strong>Output:</strong> <span class="example-io">4</span></p>

<p><strong>Explanation:</strong></p>

<p>The good pairs are:</p>

<ol>
	<li><code>([0, 1, 1], [2, 2, 1])</code></li>
	<li><code>([0, 1, 2], [2, 2, 0])</code></li>
	<li><code>([0, 2, 2], [2, 1, 0])</code></li>
	<li><code>([1, 2, 2], [1, 1, 0])</code></li>
</ol>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [5,5,5,5]</span></p>

<p><strong>Output:</strong> <span class="example-io">126</span></p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n == nums.length &lt;= 2000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 50</code></li>
</ul>

class Solution:
    def countOfPairs(self, nums: List[int]) -> int:
        MOD = 10**9 + 7
        n = len(nums)
        
        # dp[i][j] - number of ways to form arr1 and arr2 until index i with arr1[i] = j
        dp = [[0] * 51 for _ in range(n)]
        
        # Initialize dp for index 0
        for arr1_0 in range(nums[0] + 1):
            arr2_0 = nums[0] - arr1_0
            dp[0][arr1_0] = 1  # Only one way to split nums[0] into (arr1[0], arr2[0])
        
        # Fill dp for the rest of the array
        for i in range(1, n):
            # For every possible arr1[i-1]
            for arr1_prev in range(nums[i-1] + 1):
                # For every possible arr1[i] >= arr1_prev
                for arr1_curr in range(arr1_prev, nums[i] + 1):
                    arr2_curr = nums[i] - arr1_curr
                    # Check if arr2_curr is valid (it must be <= arr2_prev which is nums[i-1] - arr1_prev)
                    if arr2_curr <= nums[i-1] - arr1_prev:
                        dp[i][arr1_curr] = (dp[i][arr1_curr] + dp[i-1][arr1_prev]) % MOD
        
        # The result is the sum of all valid dp[n-1][j] values for the last index
        result = 0
        for j in range(51):
            result = (result + dp[n-1][j]) % MOD
        
        return result
21090