C++ program to separate even and odd numbers from an array
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C++ program to separate even and odd numbers from an array of integers. Put all even numbers first, and then odd numbers.
Given: An array having even and odd integers. We have to segregate even and odd integers and put even numbers first and then odd numbers.
Example:
Input: {2, 5, 1, 10, 3, 6, 7, 8}
Output: 2 8 6 10 3 1 7 5
# Algorithm
- Initialize two index variables
left
andright
at the beginning and end of the array, respectively. - Increment the left index until an odd number is encountered.
- Decrement the right index until an even number is encountered.
- If
left
is less thanright
, swap the elements at the indicesleft
andright
. - Repeat the above steps until
left
is greater than or equal toright
. - After the loop finishes, the array will be divided into two parts: even numbers at the beginning and odd numbers at the end.
Code:
#include<iostream>
using namespace std;
void swap(int *x, int *y)
{
int t = *x;
*x = *y;
*y = t;
}
void segregate(int array[], int n)
{
int left = 0, right = n - 1;
while (left < right)
{
while (array[left] % 2 == 0 && left < right)
left++;
while (array[right] % 2 == 1 && left < right)
right--;
if (left < right)
{
swap(&array[left], &array[right]);
left++;
right--;
}
}
}
int main()
{
int array[100], n, i;
cout << "Enter number of elements: ";
cin >> n;
cout << "\nEnter elements: ";
for (i = 0; i < n; i++)
cin >> array[i];
cout << "Original array: ";
for (int i = 0; i < n; i++)
cout << array[i] << " ";
segregate(array, n);
cout << "\nArray after divided: ";
for (int i = 0; i < n; i++)
cout << array[i] << " ";
return 0;
}
Output
Enter number of elements: 5
Enter elements: 1 2 3 4 5
Original array: 1 2 3 4 5
Array after divided: 2 4 1 3 5