<p>For an integer array <code>nums</code>, an <strong>inverse pair</strong> is a pair of integers <code>[i, j]</code> where <code>0 &lt;= i &lt; j &lt; nums.length</code> and <code>nums[i] &gt; nums[j]</code>.</p>

<p>Given two integers n and k, return the number of different arrays consisting of numbers from <code>1</code> to <code>n</code> such that there are exactly <code>k</code> <strong>inverse pairs</strong>. Since the answer can be huge, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 3, k = 0
<strong>Output:</strong> 1
<strong>Explanation:</strong> Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pairs.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 3, k = 1
<strong>Output:</strong> 2
<strong>Explanation:</strong> The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
	<li><code>0 &lt;= k &lt;= 1000</code></li>
</ul>

class Solution:
    def kInversePairs(self, n: int, K: int) -> int:
        MOD = 10 ** 9 + 7
        #dfs(i,j) = sum() j-i+k+1 >= 
        # j-i+k+1 >= 0
        # k >= -j+i-1
        # j-(i-k-1) = j-i+i-1 +1

        dp = [[0] * (K + 1) for _ in range(n + 1)]
        pre = [[0] * (K + 2) for _ in range(n + 1)]
        dp[1][0] = 1
        pre[1] = [1] * (K + 2)
        pre[1][0] = 0
        for i in range(1, n):
            for j in range(K+1):
                dp[i+1][j] = pre[i][j+1] if j < i else (pre[i][j+1] - pre[i][j - i]) % MOD
                pre[i+1][j+1] = dp[i+1][j] if j == 0 else (pre[i+1][j] + dp[i+1][j]) % MOD

        return dp[n][K]


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