# K-Concatenation Maximum Sum in C++

Suppose we have an integer array arr and one integer k, we have to change the array by repeating it k times. So if arr = [1, 2] and k = 3 then the modified array will be [1, 2, 1, 2, 1, 2].

Now we have to find the maximum sub-array sum in the modified array. Note that the length of the sub-array can be 0 and its sum in that case is 0. As the answer may be very large, find the answer modulo 10^9 + 7.

So if the input is like [1,-2,1] and k = 5, then the result will be 2.

To solve this, we will follow these steps −

• Define method called getKadane(), this will take array, this will work like −

• ret := -inf, sum := 0, all values of ret and sum will be of mod 10^9 + 7

• for i in range 0 to size of arr – 1

• sum := max of arr[i] and arr[i] + sum

• ret := max of ret, sum

• if ret is < 0, then return 0, otherwise ret

• Define method called getSum(), this will take array, this will work like −

• ret := 0, The ret value will be of mod 10^9 + 7

• for i in range 0 to size of arr – 1

• ret := ret + arr[i]

• return ret

• Define method called getPrefix(), this will take array, this will work like −

• ret := -inf, sum := 0, all values of ret and sum will be of mod 10^9 + 7

• for i in range 0 to size of arr – 1

• sum := sum + arr[i]

• ret := max of ret and sum

• if ret is < 0, then return 0, otherwise ret

• Define method called getSuffix(), this will take array, this will work like −

• ret := inf, sum := 0, all values of ret and sum will be of mod 10^9 + 7

• for i in range size of arr – 1 down to 0

• sum := sum + arr[i]

• ret := max of ret and sum

• if ret is < 0, then return 0, otherwise ret

• From the main method, do the following −

• kadane := getKadane(arr), sum := getSum(arr), prefix := getPrefix(arr), suffix := getSuffix(arr)

• if k is 1, then return kadane

• if sum > 1, then return max of (sum * (k - 2)) + prefix + suffix and kadane

• otherwise return max of (prefix + suffix) and kadane

## Example(C++)

Let us see the following implementation to get a better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
const int MOD = 1e9 + 7;
return ((a % MOD) + (b % MOD)) % MOD;
}
int mul(lli a, lli b){
return ((a % MOD) * (b % MOD)) % MOD;
}
class Solution {
public:
int ret = INT_MIN;
int sum = 0;
for(int i = 0; i < arr.size(); i++){
sum = max(arr[i], arr[i] + sum);
ret = max(ret, sum);
sum %= MOD;
ret %= MOD;
}
return ret < 0? 0 : ret;
}
int getSum(vector <int>& arr){
int ret = 0;
for(int i = 0; i < arr.size(); i++){
ret += arr[i];
ret %= MOD;
}
return ret;
}
int getPrefix(vector <int>& arr){
int ret = INT_MIN;
int sum = 0;
for(int i = 0; i <arr.size(); i++){
sum += arr[i];
sum %= MOD;
ret = max(ret, sum);
ret %= MOD;
}
return ret < 0 ? 0 : ret;
}
int getSuffix(vector <int>& arr){
int sum = 0;
int ret = INT_MIN;
for(int i = arr.size() - 1; i >= 0 ; i--){
sum += arr[i];
ret = max(ret, sum);
sum %= MOD;
ret %= MOD;
}
return ret < 0 ? 0 : ret;
}
int kConcatenationMaxSum(vector<int>& arr, int k) {
int sum = getSum(arr);
int prefix = getPrefix(arr);
int suffix = getSuffix(arr);
if(sum > 0){
return max((int)mul((k-2) , sum) + prefix % MOD + suffix % MOD, kadane);
} else {
}
}
};
main(){
vector<int> v1 = {1,-2,1};
Solution ob;
cout << (ob.kConcatenationMaxSum(v1, 5));
}

## Input

[1,-2,1]
5

## Output

2