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Maximum subarray sum in array formed by repeating the given array k times in C++
In this problem, we are given an array and a number k. Our task is to create a program that will find the maximum subarray sum in an array formed by repeating the given array k time in c++.
Problem description − Here, we will find the maximum sum of the subarray formed from array formed by repeating the given array k times.
Let’s take an example to understand the problem,
Input − array = {3, 5, 1} k = 2
Output − 18
Explanation −
array formed by repeating k times,
array = {3, 5, 1, 3, 5, 1}
Maximum subarray sum = 3+5+1+3+5+1 = 18
To solve this problem, we will calculate the sum of all elements of the array.
Then based on the sum, we will calculate the subarray sum calculation.
If sum > 0, we will multiply the sum of array k times which given the actual sum.
If sum < 0, we will find the subarray with maxSum taking two array repetitions.
Example
Program to show the implementation of our solution,
#include<iostream>
using namespace std;
void repeatArray(int *arr, int *b, int k,int len) {
int j = 0;
while (k > 0){
for (int i = 0; i < len; i++)
b[j++] = arr[i];
k--;
}
}
long subArraySum(int *a,int len) {
int max = 0;
long newmax = 0;
for (int i = 0; i < len; i++) {
newmax = newmax + a[i];
if (max < newmax)
max = newmax;
if (newmax < 0)
newmax = 0;
}
return max;
}
long findMaxSubArraySum(int *arr, int k,int len) {
int arraySum = 0;
long maxSum = 0;
int b[(2 * len)]= {0};
repeatArray(arr, b, 2,len);
for (int i = 0; i < len; i++)
arraySum += arr[i];
maxSum = subArraySum(b,2*len);
if (arraySum > 0)
maxSum = subArraySum(b,2*len) + (k - 2) * arraySum;
return maxSum;
}
int main() {
int arr[] = { 3, 5, 1};
int length=sizeof(arr)/sizeof(arr[0]);
int k = 3;
cout<<"The maximum subarray sum in array formed by repeating the given array "<<k<<" times is "<<findMaxSubArraySum(arr, k,length);
return 0;
}
Output
The maximum subarray sum in array formed by repeating the given array 3 times is 27