<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:
        memo = {}
        def dp(i, arr1_constraint, arr2_constraint):
            if i == len(nums):
                return 1

            curr_key = "{}-{}-{}".format(i, arr1_constraint, arr2_constraint)
            if curr_key in memo:
                return memo[curr_key]

            arr1_val = arr1_constraint # always going up
            
            arr2_val = min(arr2_constraint, nums[i]) # always going down

            curr_total_pairs = 0
            while arr1_val <= nums[i] and arr2_val >= 0:
                curr_sum = arr1_val + arr2_val
                
                if curr_sum > nums[i]:
                    arr2_val -= 1
                elif curr_sum < nums[i]:
                    arr1_val += 1
                else:
                    curr_total_pairs = (curr_total_pairs + dp(i + 1, arr1_val, arr2_val)) % 1000000007
                    arr1_val += 1
                    arr2_val -= 1
            memo[curr_key] = curr_total_pairs
            return memo[curr_key]
    
        return dp(0, 0, 50)
84053