Leetcode 629: K Inverse Pairs Array

https://leetcode.com/problems/k-inverse-pairs-array/description/

Given two integers n and k, find how many different arrays consist of numbers from 1 to n such that there are exactly k inverse pairs.

We define an inverse pair as following: For ith and jth element in the array, if i < j and a[i] > a[j] then it's an inverse pair; Otherwise, it's not.

Since the answer may be very large, the answer should be modulo 109 + 7.

Example 1:

Input: n = 3, k = 0
Output: 1
Explanation: 
Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pair.

Example 2:

Input: n = 3, k = 1
Output: 2
Explanation: 
The array [1,3,2] and [2,1,3] have exactly 1 inverse pair.

Note:

1. The integer n is in the range [1, 1000] and k is in the range [0, 1000].

Notes:

The question can be solved by dp.

Let dp[i][j] represents the number of arrays with the the first i numbers with j inverse pairs.

Then,

dp[i][j] = dp[i-1][j] + dp[i-1][j-1] + ... + dp[i-1][j - (i-1)]

The first term on the left means we place the ith element in the last position;

The second term means that we place the ith element in the second last position;

...;

The last term means that we place the ith element in the first position.

One improvement is that we can use a sum to memorize the sum of the previous dp[i-1][j], ... dp[i-1][j-(i-1)], which essentially is a "sliding window" sum. This improvement can reduce the time complexity from O(NK^2) to O(NK).

See the code below:

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