<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:
    // for a no n only n-1 inversions are possible, and if I place nth element at any previous 
    //index then (n-i) inversion pairs will be there
    // EX= 1,2,3,4,5- 1,2,5,3,4--(2 inversion pairs)
    //tabulation is required
    int M= 1e9+7;
  /*  long long int t[1001][1001];
   long long int f(int n, int k){
       
        if(n==0){
            return 0;
        }
         if(k==0){
            return 1;
        }
        if(t[n][k]!=-1){
            return t[n][k];
        }
        long long int ans=0;
        for(int i=0; i<=min(k,n-1); i++){
          ans= (ans+ f(n-1,k-i))%M;
        }
        return t[n][k]= (ans)%M;
    } */
    int kInversePairs(int n, int k) {
        //tabulation -optimised tablulation
        vector<vector<int>>t(n+1,vector<int>(k+1,0));
        for(int i=0; i<=n; i++){
            t[i][0]=1;
        }
        for(int i=1; i<=n; i++){
            long long int cumsum=1;
            for(int j=1; j<=k; j++){
                 cumsum+= t[i-1][j];
                 if(j>=i){
                    cumsum-= t[i-1][j-i];
                 }
                 t[i][j]= (cumsum%M);
            }
        }
        

        return t[n][k];
    }
};
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