K Inverse Pairs Array
<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>
Dynamic Programming
func kInversePairs(n int, k int) int {
	f := make([]int, k+1)
	s := make([]int, k+2)
	f[0] = 1
	for i, x := range f {
		s[i+1] = s[i] + x
	}
	const mod = 1e9 + 7
	for i := 1; i <= n; i++ {
		for j := 1; j <= k; j++ {
			f[j] = (s[j+1] - s[max(0, j-(i-1))] + mod) % mod
		}
		for j := 1; j <= k+1; j++ {
			s[j] = (s[j-1] + f[j-1]) % mod
		}
	}
	return f[k]
}