<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 mod = 1e9+7;
    int kInversePairs(int n, int k) {
        if(k>(n*(n-1))/2) return 0;
        if(k==0 || k==(n*(n-1))/2) return 1;
        vector<vector<long long>> dp(n+1,vector<long long>(k+1,0));
        dp[1][0]=1;dp[2][0]=1;dp[2][1]=1;
        for(int row=3;row<=n;row++){
            for(int col=0;col<=k;col++){
                if(col>(row*(row-1))/2) break;
                if(col==0 || col==(row*(row-1))/2) {dp[row][col]=1;continue;}
                if(col<=row-1) dp[row][col] = (dp[row][col-1]+dp[row-1][col])%mod;
                else dp[row][col] = ((dp[row][col-1]+dp[row-1][col])%mod-dp[row-1][col-row]+mod)%mod;
            }
        }
        return dp[n][k];
    }
};
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