<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:
    const int MOD = 1e9+7;
    int kInversePairs(int n, int k) {
        int dp[n+1][k+1];
        memset(dp, 0, sizeof(dp));
        vector<int> presum(k+1, 1);
        dp[1][0] = 1;
        for (int i=2; i<=n; i++) {
            vector<int> new_presum(k+1);
            for (int j=0; j<=k; j++) {
                dp[i][j] = presum[j] - (j-i>=0 ? presum[j-i] : 0);
                dp[i][j] += MOD;
                dp[i][j] %= MOD;
                new_presum[j] = ((j-1>=0?new_presum[j-1]:0) + dp[i][j]) % MOD;
                // for (int t=0; t<=i-1; t++) {
                //     if (j-t >= 0) {
                //         // dp[i-1][j-i+1] +...+ dp[i-1][j]
                //         dp[i][j] += dp[i-1][j-t];
                //         dp[i][j] %= MOD;
                //     }
                // }
            }
            swap(presum, new_presum);
        }
        return dp[n][k];
    }
};
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