<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 {
 public:
  int kInversePairs(int n, int k) {
    constexpr int kMod = 1'000'000'007;
    // dp[i][j] := the number of permutations of numbers 1..i with j inverse
    // pairs
    vector<vector<int>> dp(n + 1, vector<int>(k + 1));

    // If there's no inverse pair, the permutation is unique "123..i".
    for (int i = 0; i <= n; ++i)
      dp[i][0] = 1;

    for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= k; ++j) {
        dp[i][j] = (dp[i][j - 1] + dp[i - 1][j]) % kMod;
        if (j - i >= 0)
          dp[i][j] = (dp[i][j] - dp[i - 1][j - i] + kMod) % kMod;
      }

    return dp[n][k];
  }
};
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