<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:
#         n = len(nums)
#         mod = 10**9 + 7
#         mx = max(nums)
        
#         dp = [[0]*(mx+1) for _ in range(n)]
#         for num1 in range(mx+1):
#             num2 = nums[0] - num1
#             if num2 >= 0:
#                 dp[0][num1] = 1  
#         print(dp[0])
#         for i in range(1, n):

#             for num1 in range(nums[i]+1):
#                 num2 = nums[i]-num1
#                 if num2 < 0: break

#                 for prev_num1 in range(min(nums[i-1]-num2, num1)+1):
#                     dp[i][num1] += dp[i-1][prev_num1]
#                     dp[i][num1] %= mod
#             print(dp[i])
#         return sum(dp[n-1]) % mod

class Solution:
    def countOfPairs(self, nums: List[int]) -> int:
        n = len(nums)
        mod = 10**9 + 7
        mx = max(nums)
        
        dp = [[0]*(mx+1) for _ in range(n)]
        for num1 in range(mx+1):
            num2 = nums[0] - num1
            if num2 >= 0:
                dp[0][num1] = 1
            else:
                break
        presum_dp = [[0]*(mx+1) for _ in range(n)]
        presum_dp[0][0] = dp[0][0]
        for num in range(1, mx+1):
            presum_dp[0][num] = presum_dp[0][num-1] + dp[0][num]

        for i in range(1, n):

            for num1 in range(nums[i]+1):
                num2 = nums[i]-num1
                if num2 < 0: break

                upperbound = min(nums[i-1]-num2, num1)
                # for prev_num1 in range(upperbound+1):
                #     dp[i][num1] += dp[i-1][prev_num1]
                #     dp[i][num1] %= mod

                if upperbound >= 0:
                    dp[i][num1] += presum_dp[i-1][upperbound]
                    dp[i][num1] %= mod
            
            presum_dp[i][0] = dp[i][0]
            for num in range(1, mx+1):
                presum_dp[i][num] = presum_dp[i][num-1] + dp[i][num]

        return sum(dp[n-1]) % mod

# [1, 1, 1, 0]
# [0, 1, 2, 3]
# [0, 1, 3, 0]
182727