<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;
    long long dp[1001][1001];
    long long  res[1001][1001];

    int helper(int n,int k){

        if(n==1){
        return (k==0);
        }
        if(dp[n][k]!=-1)return dp[n][k];
        long long ans =0;

        for(int i=0;i<=min(n-1,k);i++){
            ans += helper(n-1,k-i);
            ans %= mod;
        }

    return dp[n][k]=(ans+mod)%mod;
    }

    int kInversePairs(int n, int k) {

        memset(dp,0,sizeof dp);
        memset(res,0,sizeof res);
        // return helper(n,k);
        // dp[1][0]=1;
        // res[1][0]=1;
        dp[0][0]=1;

        for(int i=0;i<=n;i++){
            for(int j=0;j<=k;j++){
               if(i>0) dp[i][j]=(res[i-1][j]-((j<i)?0:res[i-1][j-i]));
                res[i][j]=(((j<1)?0:res[i][j-1])+dp[i][j])%mod;
                dp[i][j]%=mod;
            }
        }

        return (dp[n][k]+mod)%mod;
      

     
    }
};
25100