Input:The first line of the input consists of an integer num,representing the size of the list (N) the second line of the input consists of N space seperated integers representing the values of the list
Answers
Explanation:
before the changes the elements of the list are
are[0]
Answer:
#include<iostream>
using namespace std;
void swap(int *i, int *j)
{
int temp = *i;
*i = *j;
*j = temp;
}
int seperateEvenAndOdd(int arr[], int size)
{
int left = 0;
int right = size - 1;
while(left < right)
{
while(arr[left]%2 == 0 && left < right)
{
left++;
}
while(arr[right]%2 == 1 && left < right)
{
right--;
}
if(left < right)
{
swap(&arr[left], &arr[right]);
left++;
right--;
}
}
}
int main()
{
int arr_size;
cin >> arr_size;
int arr[arr_size];
for(int i=0;i<arr_size;i++)
{
cin >> arr[i];
}
int i=0;
seperateEvenAndOdd(arr,arr_size);
for(i=0;i<arr_size;i++)
cout << arr[i] << " ";
return 0;
}
Explanation:
- It is given that the size of the list is taken as the input.
- Then the integers separated by space are taken as input.
- The output prints the space-separated integers such that all the odd numbers of the list come after the even numbers.
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