<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 kInversePairs(int n, int k) {
        // dp[i][j]: number of inverse pairs for numbers 1,...,i and j pairs
        // dp[i][j] = sum_{j-(n-1)<=k<=j} dp[i-1][k]
        // let dp[i][:] be a row, just need to keep track of the previous row
        int ub = 1e9+7;

        vector<int> prev_row(k+1, 0);
        prev_row[0] = 1;
        vector<int> row(prev_row);
        for (int i=2; i<=n; ++i) {
            prev_row = row;
            long long cur_sum = 1;
            for (int j=1; j<=k; ++j) {
                cur_sum += prev_row[j] % ub;
                cur_sum = cur_sum % ub;
                if (j>(i-1)) cur_sum -= (prev_row[j-(i-1)-1] % ub);
                cur_sum = cur_sum % ub;
                row[j] = cur_sum;
            }
        }
        if (row[k]>=0) return row[k];
        else return (row[k]+ub);
    }
};
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