<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:
    def kInversePairs(self, n: int, k: int) -> int:
        mod = 10**9 + 7
        dp = collections.Counter()

        # Thoughts
        # f(n, k) = f(n - 1, k)        -- n is at the last position 
        #           + f(n - 1, k - 1)  -- n is at the 2nd last position
        #           + f(n - 1, k - 2)  -- n is at the 3rd last position
        #           + ...
        #           + f(n - 1, k - n + 1)      -- n is at the 1st position
        # recall
        # f(n, k - 1) = f(n - 1, k - 1) + f(n - 1, k - 2) + ... + f(n - 1, k - n)
        # So f(n, k) = f(n - 1, k) + f(n, k - 1) if k - n < 0 because f(n - 1, k - n) would be 0
        # However, if n <= k, f(n - 1, k - n) is non zero, and we have to update the value
        for i in range(1, n + 1):
            dp[(i, 0)] = 1
            for j in range(1, min(k, i * (i - 1) // 2) + 1):
                dp[(i, j)] = (dp[(i, j)] + dp[(i - 1, j)] + dp[(i, j - 1)]) % mod
                if i <= j:
                    dp[(i, j)] = (dp[(i, j)] - dp[(i - 1, j - i)]) % mod

        return dp[(n, k)] % mod

212000