<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>

MOD = 10**9 + 7
class Solution:
    def countOfPairs(self, nums: List[int]) -> int:
        d = {} # (lb, ub, index)
        # print("FINALE", self.recur(nums, d, 0, nums[0], 0))
        # for k, v in d.items():
        #     print(k, v)
        return self.recur(nums, d, 0, nums[0], 0)
        
    def recur(self, nums, d, lb, ub, index):
        if (lb, ub, index) in d:
            return d[(lb, ub, index)]
        lb_copy = lb
        ub_copy = ub
        if index == len(nums) - 1:
            # print("HELLO")
#             can optimize
            # if nums[-1] > lb:
            #     lb = nums[-1]
            # elif ub < nums[-1]:
            #     ub = nums[-1]
            # print(lb, ub)
            count = 0
            while ub >= 0 and lb <= nums[-1]:
                if lb + ub == nums[-1]:
                    count+=1
                    lb+=1
                    ub-=1
                elif lb + ub < nums[-1]:
                    lb+=1
                else:
                    ub-=1
            count = count % MOD
            d[(lb_copy, ub_copy, index)] = count
            return count
        count = 0
        while ub >= 0 and lb <= nums[index]:
            if lb + ub == nums[index]:
                count+=self.recur(nums, d, lb, ub, index + 1)
                lb+=1
                ub-=1
            elif lb + ub < nums[index]:
                lb+=1
            else:
                ub-=1
        count = count % MOD
        d[(lb_copy, ub_copy, index)] = count
        return count




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