629. K Inverse Pairs Array

class Solution {
public:
int kInversePairs(int n, int k) {
//if(n<=0) return;
int dp[n+1][k+1];
int mod = 1000000007;

memset(dp,0,sizeof(dp));
for(int i = 1 ; i <= n ;i++){
dp[i][0] = 1 ;
}
for(int i = 1 ; i <= n ; i++){

//j是不能超过k，和最大可能的逆序数二者中最小的的
for(int j = 1 ; j <=min(k, ((1+i-1)*(i-1)/2)) ;j++){
if(j>=i){

//减去的是dp[i-1][j-i]
dp[i][j] = (dp[i][j-1] + dp[i-1][j] - dp[i-1][j-i])%mod;
dp[i][j]=(dp[i][j-1]-dp[i-1][j-i])<0?(dp[i][j]+mod)%mod:dp[i][j];
}
else{
dp[i][j] =(dp[i][j-1] + dp[i-1][j])%mod ;
}

}
}
return dp
[k];
}
};