<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:
        '''
        2, 3, 2
        Generate pairs of numbers  that can add to nums[i]
        [0, 1, 2] [2, 2, 0]
        [0, 1, 1] [2, 2, 1]
        [0, 1, 0] --> NO CAN DO
        [0, 2, 2] [2, 1, 0]
        [0, 2, 1] --> NO CAN DO
        [0, 3] [2, 0] --> NO CAN DO
        [1, 1] [1, 2] --> NO CAN DO
        [1, 2, 2] [1, 1, 0]
        '''
        @cache
        def calculate(idx, last_1, last_2):
            if idx == len(nums):
                return 1
            # Generate valid pairs
            pairs = [(i, nums[idx]-i) for i in range(0, nums[idx]+1)]
            res = 0
            for a, b in pairs:
                if last_1 <= a and last_2 >= b:
                    res+=calculate(idx+1, a, b)
            return res

        result = calculate(0, -math.inf, math.inf)
        return result%(10**9 + 7)
        
47096