when we calculate dp[n][k] we need sum from last one, use sliding window to save time

```
int kInversePairs(int n, int k) {
int permutation = n*(n - 1) / 2;
if (k > permutation)return 0;
if (n == 1 || k == 0)return 1;
k = min(k, permutation - k);
vector<vector<int>> dp(n, vector<int>(k+1, 0));
dp[0][0] = 1;
for (int i = 1; i < n; i++)
{
int p = i*(i + 1) / 2+1;
long sum = 0;
for (int j = 0; j < min(p, k+1); j++)
{
sum += dp[i - 1][j];
if (j > i)sum -= dp[i - 1][j - i-1];
dp[i][j] = sum%1000000007;
}
}
return dp[n - 1][k];
}
```