<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;= 1000</code></li>
</ul>

class Solution:
    def countOfPairs(self, nums: List[int]) -> int:
        MOD = 10**9 + 7
        # print(len(nums))
        # @cache
        # def dp(i, s):
        #     # print(f'i = {i} mn = {mn} mx = {mx}')
        #     if i == len(nums):
        #         return 1
        #     if i == 0:
        #         mn, mx = 0, nums[i]
        #     else:
        #         mn, mx = s, nums[i-1]-s
        #     ans = 0
        #     for j in range(mn, nums[i]+1):
        #         arr1 = j
        #         arr2 = nums[i] - arr1
        #         if 0 <= arr2 <= mx:
        #             ans += dp(i+1, arr1)
        #     return ans%MOD
        # return dp(0, 0)

        n = len(nums)
        dp = [[0 for _ in range(1001)] for _ in range(n+1)]
        pre = [[0 for _ in range(1001)] for _ in range(n+1)]
        for s in range(nums[-1], -1, -1):
            dp[n-1][s] += 1
        pre[n-1] = list(accumulate(dp[n-1]))
        
        for i in range(n-2, -1, -1):
            for s in range(nums[i], -1, -1):
                dp[i][s] += pre[i+1][nums[i+1]]
                b = max(0, nums[i+1]-nums[i]) + s
                if b:
                    dp[i][s] -= pre[i+1][b-1]
                dp[i][s]%=MOD
                pre[i][s] = dp[i][s]
            
            pre[i] = list(accumulate(dp[i]))
    
        return sum(dp[0])%MOD
202587