<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 dp[1001][1001];
    int mod=1e9+7;
    int helper(int i,int inv,int k,int n){
        if(i==n&&inv==k){
            return 1;
        }if(inv>k){
            return 0;
        }
        if(i==n){
            return 0;
        }
        if(dp[i][inv]!=-1){
            return dp[i][inv];
        }
        int sum=0;
        for(int j=inv;j<=min(inv+i,k);j++){
           sum=(sum+helper(i+1,j,k,n))%mod;
        }
        return dp[i][inv]=sum;
    }
    int kInversePairs(int n, int k) {
        memset(dp,0,sizeof(dp));
         int ans=helper(0,0,k,n);
        
        dp[0][0]=1;
        for(int i=1;i<=n;i++){
            for(int j=0;j<=k;j++){
                for(int l=0;l<=min(j,i-1);l++){
                    dp[i][j]=(dp[i][j]+dp[i-1][j-l])%mod;
                }
            }
        }
        return dp[n][k];
    }
};
11600