<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:
        # f(0, k) = 0, f(n, 0) = 1
        # suppose we had n numbers, and now we add a number which is greater than all existing number.
        # if we add it to end, no additional arr meets condition
        # if we add it to second last, one additional arr meets condition (i.e. the pair of this max and last elem)
        #     which is equivalent to f(n, k-1), that is, when we had n numbers and k-1 arr meets condition
        # now if we add it to third last, two additional arr, i.e. equivalent to f(n, k-2)
        # f(n+1, k) = f(n, k) + f(n, k-1) + f(n, k-2) + ... + f(n, k-n), by replacing n with n-1, we have:
        # f(n, k) = f(n-1, k) + f(n-1, k-1) + f(n-1, k-2) + ... + f(n-1, k-n+1)
        dp = [[0 for _ in range(k+1)] for _ in range(n+1)]
        for i in range(1, n+1):
            dp[i][0] = 1
        for i in range(1, n+1):
            curr = dp[i-1][0]
            for j in range(1, k+1):
                curr += dp[i-1][j]
                if j > i-1:
                    curr -= dp[i-1][j-i]
                dp[i][j] = curr
        return dp[n][k] % (1_000_000_007)

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