<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>


/*Here are three solutions for the problem:


*Recursive Solution:*

```
class Solution {
public:
    int kInversePairs(int n, int k) {
        if (k == 0) return 1;
        if (n == 1) return 1;
        int mod = 1e9 + 7;
        return (n-1 * kInversePairs(n-1, k) + kInversePairs(n-2, k-1)) % mod;
    }
};
```

*Memoization Solution:*

```
class Solution {
public:
    int kInversePairs(int n, int k) {
        int mod = 1e9 + 7;
        memo = vector<vector<int>>(n+1, vector<int>(k+1, -1));
        return dp(n, k);
    }
    
    int dp(int n, int k) {
        if (k == 0) return 1;
        if (n == 1) return 1;
        if (memo[n][k] != -1) return memo[n][k];
        return memo[n][k] = (n-1 * dp(n-1, k) + dp(n-2, k-1)) % 1e9 + 7;
    }
    
    vector<vector<int>> memo;
};
```

*Tabular Solution (DP):*

```
class Solution {
public:
    int kInversePairs(int n, int k) {
        int mod = 1e9 + 7;
        vector<vector<int>> dp(n+1, vector<int>(k+1, 0));
        for (int i = 1; i <= n; i++) {
            dp[i][0] = 1;
            for (int j = 1; j <= k; j++) {
                dp[i][j] = (dp[i-1][j] * (i-1) + dp[i-2][j-1]) % mod;
            }
        }
        return dp[n][k];
    }
};
```

*Space Optimized Solution:*

```*/
class Solution {
public:
    int kInversePairs(int n, int k) {
        int mod = 1e9 + 7;
        vector<int> dp(k+1, 0);
        dp[0] = 1;
        for (int i = 1; i <= n; i++) {
            vector<int> temp(k+1, 0);
            for (int j = 0; j <= k; j++) {
                for (int l = 0; l <= min(i-1, j); l++) {
                    temp[j] = (temp[j] + dp[j-l]) % mod;
                }
            }
            dp = temp;
        }
        return dp[k];
    }
};
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