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