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
class Solution {
public:
    int kInversePairs(int n, int k) {
        int f[k + 1];
        int s[k + 2];
        memset(f, 0, sizeof(f));
        f[0] = 1;
        fill(s, s + k + 2, 1);
        s[0] = 0;
        const int mod = 1e9 + 7;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                f[j] = (s[j + 1] - s[max(0, j - (i - 1))] + mod) % mod;
            }
            for (int j = 1; j <= k + 1; ++j) {
                s[j] = (s[j - 1] + f[j - 1]) % mod;
            }
        }
        return f[k];
    }
};