<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 find(int i, int j, vector<vector<int>>& dp, int n, int k) {
        if (i <= 0 || j < 0) {
            // cout << i << " " << j << " 0\n";
            return 0;
        }
        if(dp[i][j] != -1) {
            // cout << i << " " << j << " " << dp[i][j] << "\n";
            return dp[i][j];
        }
        // cout << i << " " << j << "\n";
        if(j == 0) {
            dp[i][j] = 1;
            // cout << i << " " << j << " " << dp[i][j] << "\n";
            return dp[i][j];
        }
        if(i == 1) {
            dp[i][j] = 0;
            return dp[i][j];
        }
        int ans = 0;
        int mod = 1e9+7;
        // for(int k=0;k<min(i,j+1);k++) {
        //     ans = (ans + find(i-1,j-k,dp,n,k))%mod;
        // }
        ans = (ans + find(i,j-1,dp,n,k))%mod;
        ans = (ans + find(i-1,j,dp,n,k))%mod;
        if (i-1<j) {
            // cout << "Print: " << i << " " << j << " " << ans << "\n";
            // cout << "Tmp: " << find(i-1,j-i,dp,n,k) << "\n";
            ans = (ans - find(i-1,j-i,dp,n,k) + mod)%mod;
        }
        dp[i][j] = ans;
        // cout << i << " " << j << " " << dp[i][j] << "\n";
        return ans;
    }
    int kInversePairs(int n, int k) {
        vector<vector<int>> dp(n+1, vector<int>(k+1, -1));
        return find(n, k, dp, n, k);
    }
};
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