<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) {
        // memset(dp, -1, sizeof(dp));
        vector<vector<int>> dp(n+1, vector<int>(k+1,0));
        dp[1][0] = 1;
        vector<int> pf(k+1,1);
        for(int i=2;i<=n;i++){
            
            for(int j=0;j<=k;j++){
                if(j==0){
                    dp[i][j] = 1;
                    continue;
                }

                int en=j,st=j+1-min(i,j+1);
                dp[i][j] = pf[en];
                if(st>0)dp[i][j] = (dp[i][j]-pf[st-1])%mod;

                // int val = 0;
                // for(int l=1;l<=(i,j+1);l++){
                //     val = (val + dp[i-1][j-l+1])%mod;
                // }
                // dp[i][j] = val;
            }

            pf[0] = dp[i][0];
            for(int c=1;c<=k;c++){
                pf[c] = (pf[c-1]+dp[i][c])%mod;
            }
        }

        // int fin = rec(n, k);
        if(dp[n][k] < 0)dp[n][k] += mod;
        return dp[n][k];
    }
};
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