K-Concatenation Maximum Sum

/*
Logic:
case 1: if k==1,then maximum sum subarray will be answer.
case 2: if sum<0 then answer will be extracted by two array only because if sum<0 for whole then why will
        you take whole array in sum.so answer will be maximum sum subarray of two array  which will be
        partial sum of two array definitely.
case 3: if sum>0 then find maxium sum of two array and sum of rest array*k-2 because 2 array you alread used
        so only k-2 array left.
for eg. [-1,6,-4], k=3  then 6 is the maximum sum subarray.
-1,6,-4,-1,6,-4,-1,6,-4 maxium sum subarray of two combined array is [6,-4,-1,6] ie.7
so we are including elements of second array after the 6 if we have to combine with the third array but in answer , [6,-4,-1,6,-4,-1,6] we compensate it by leaving the same elements of last array.
*/

int m=pow(10,9)+7;
int kadanes(vector <int> arr){
    // for(auto itr:arr) cout<<itr<<" ";
    int cur_max=0,_max=INT_MIN,sum=0;
    for(int i=0;i<arr.size();i++){
            cur_max+=arr[i];
            _max=max(cur_max,_max);
            if(cur_max<0) cur_max=0;
        }
    return max(0,_max);
}
int mul(int n1,int n2){
    int ans=0;
    for(int i=0;i<n2;i++){
        ans+=n1;
        ans%=m;
    }
    return ans;
}
class Solution {
public:
    int kConcatenationMaxSum(vector<int>& a, int k) {
        vector<int> arr=a;
        int sum=0;
        for(auto itr:a){ arr.push_back(itr); sum+=itr;}
        if(k==1) return kadanes(a);
        else if(sum<0) return kadanes(arr);
        else return (kadanes(arr)+mul(sum,k-2))%m;
    }
};