<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) {
        int dp[n+1][k+1];
        int mod = 1e9+7;
        if(n == 1 && k >= 1)
        {
            return 0;
        }
        for(int i= 0;i<=n;i++)
        {
            dp[i][0] = 1;
        }
        for(int i = 0;i<=k;i++)
        {
            dp[1][i] = 1;
        }
        dp[1][0] = 1;
        for(int i = 2;i<n;i++)
        {
            for(int j = 1;j<=k;j++)
            {
                dp[i][j] = 0;
                // for(int t = 0;t<=min(i-1,j);t++)
                // {
                //     dp[i][j] += (dp[i-1][j-t])%mod;
                //     dp[i][j]= (dp[i][j])%mod;  
                // }
                // cout << dp[i][j] << " ";
                int x =0, y = dp[i-1][j];
                if(j-min(i-1,j) > 0)
                {
                    x = dp[i-1][j-min(i-1,j)-1]; 
                }
                dp[i][j] = (y-x+mod)%mod;
                dp[i][j] %= mod;
            }
            for(int j=1;j<=k;j++)
            {
                dp[i][j] += dp[i][j-1]%mod;
                dp[i][j] %= mod;
            }
        }
        for(int j = 1;j<=k;j++)
        {
            dp[n][j] =0;
            int x =0,y = dp[n-1][j];
                if(j-min(n-1,j) > 0)
                {
                    x = dp[n-1][j-min(n-1,j)-1]; 
                }
                dp[n][j] = (y-x+mod)%mod;
                dp[n][j] %= mod;
            
        }

        return dp[n][k];

    }
};
12000