Given an array arr[] of integers, segregate even and odd numbers in the array such that all the even numbers should be present first, and then the odd numbers.
Examples:
Input: arr[] = {7, 2, 9, 4, 6, 1, 3, 8, 5}
Output: 2 4 6 8 7 9 1 3 5Input: arr[] = {1, 3, 2, 4, 7, 6, 9, 10}
Output: 2 4 6 10 7 1 9 3
We have discussed two different approaches in Segregate Even and Odd numbers
Here we are going to discuss Lomuto’s Partition Scheme to segregate even and odd.
- Start with an array of numbers.
For the sake of illustration, let's consider the following array: [7, 2, 9, 4, 6, 1, 3, 8, 5].
- Choose a pivot element from the array.
In this case, let's choose the last element, which is 5, as the pivot.
- Initialize two pointers, i and j.
- The pointer i will keep track of the boundary between the even and odd numbers, and
- the pointer j will iterate through the array.
- Iterate through the array using the pointer j.
- Compare each element with the pivot and perform the following steps:
- If the element is even, swap it with the element at index i and increment i.
- If the element is odd, do nothing and continue.
- Compare each element with the pivot and perform the following steps:
- Finally, swap the pivot element with the element at index I.
// CPP code to segregate even odd
// numbers in an array
#include <bits/stdc++.h>
using namespace std;
// Function to segregate even odd numbers
vector<int> arrayEvenAndOdd(vector<int>& arr)
{
int i = -1, j = 0;
int pivot = arr.back();
for (j = 0; j < arr.size() - 1; j++) {
if (arr[j] % 2 == 0) {
i++;
// Swapping even and odd numbers
swap(arr[i], arr[j]);
}
}
swap(arr[i + 1], arr.back());
return arr;
}
// Driver code
int main()
{
vector<int> arr = { 7, 2, 9, 4, 6, 1, 3, 8, 5 };
vector<int> updatedArr = arrayEvenAndOdd(arr);
// Function Call
cout << "Updated Array: ";
for (int num : updatedArr) {
cout << num << " ";
}
cout << endl;
return 0;
}
// java code to segregate even odd
// numbers in an array
public class GFG {
// Function to segregate even
// odd numbers
static void arrayEvenAndOdd(int arr[], int n)
{
int i = -1, j = 0;
while (j != n) {
if (arr[j] % 2 == 0) {
i++;
// Swapping even and
// odd numbers
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
j++;
}
// Printing segregated array
for (int k = 0; k < n; k++)
System.out.print(arr[k] + " ");
}
// Driver code
public static void main(String args[])
{
int arr[] = { 1, 3, 2, 4, 7, 6, 9, 10 };
int n = arr.length;
arrayEvenAndOdd(arr, n);
}
}
// This code is contributed by Sam007
# Python3 code to segregate even odd
# numbers in an array
# Function to segregate even odd numbers
def arrayEvenAndOdd(arr, n):
i = -1
j = 0
while (j != n):
if (arr[j] % 2 == 0):
i = i+1
# Swapping even and odd numbers
arr[i], arr[j] = arr[j], arr[i]
j = j+1
# Printing segregated array
for i in arr:
print(str(i) + " ", end='')
# Driver Code
if __name__ == '__main__':
arr = [1, 3, 2, 4, 7, 6, 9, 10]
n = len(arr)
arrayEvenAndOdd(arr, n)
# This code was contributed by
# Yatin Gupta
// C# code to segregate even odd
// numbers in an array
using System;
class GFG {
// Function to segregate even
// odd numbers
static void arrayEvenAndOdd(int[] arr, int n)
{
int i = -1, j = 0;
while (j != n) {
if (arr[j] % 2 == 0) {
i++;
// Swapping even and
// odd numbers
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
j++;
}
// Printing segregated array
for (int k = 0; k < n; k++)
Console.Write(arr[k] + " ");
}
// Driver code
static void Main()
{
int[] arr = { 1, 3, 2, 4, 7, 6, 9, 10 };
int n = arr.Length;
arrayEvenAndOdd(arr, n);
}
}
// This code is contributed by Sam007
<script>
// JavaScript code to segregate even odd
// numbers in an array
// Function to segregate even odd numbers
function arrayEvenAndOdd(arr, n)
{
let i = -1, j = 0;
let t;
while (j != n) {
if (arr[j] % 2 == 0) {
i++;
// Swapping even and odd numbers
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
j++;
}
// Printing segregated array
for (let i = 0; i < n; i++)
document.write(arr[i] + " ");
}
// Driver code
let arr = [ 1, 3, 2, 4, 7, 6, 9, 10 ];
let n = arr.length;
arrayEvenAndOdd(arr, n);
// This code is contributed by Surbhi Tyagi
</script>
<?php
// PHP code to segregate even odd
// numbers in an array
// Function to segregate
// even odd numbers
function arrayEvenAndOdd($arr, $n)
{
$i = -1;
$j = 0;
$t;
while ($j != $n)
{
if ($arr[$j] % 2 == 0)
{
$i++;
// Swapping even and
// odd numbers
$x = $arr[$i];
$arr[$i] = $arr[$j];
$arr[$j] = $x;
}
$j++;
}
// Printing segregated
// array
for ($i = 0; $i < $n; $i++)
echo $arr[$i] . " ";
}
// Driver code
$arr = array(1, 3, 2, 4, 7, 6, 9, 10);
$n = sizeof($arr);
arrayEvenAndOdd($arr, $n);
// This code is contributed by Anuj_67
?>
Output
Updated Array: 2 4 6 8 5 1 3 7 9
Illustration
Let's go through the steps:
- Initially, i = 0 and j = 0.
- The first element is 7 (odd),
- so we do nothing and move j to the next element.
- The second element is 2 (even),
- so we swap it with the element at index i (which is also 2) and increment i to 1.
- The third element is 9 (odd),
- so we do nothing and move j to the next element.
- The fourth element is 4 (even),
- so we swap it with the element at index i (which is 9) and increment i to 2.
- The fifth element is 6 (even),
- so we swap it with the element at index i (which is 9) and increment i to 3.
- The sixth element is 1 (odd),
- so we do nothing and move j to the next element.
- The seventh element is 3 (odd),
- so we do nothing and move j to the next element.
- The eighth element is 8 (even),
- so we swap it with the element at index i (which is 1) and increment i to 4.
- The ninth element is 5 (odd),
- so we do nothing and move j to the next element.
- After iterating through the array, we partitioned the even and odd numbers.
The array now looks like this: [2, 4, 6, 8, 5, 1, 3, 9, 7].
- Finally, swap the pivot element with the element at index I.
In this case, we swap 5 with 7. This step ensures that the pivot element is in its final sorted position.
The array after the final swap will be: [2, 4, 6, 8, 7, 1, 3, 9, 5].
Time Complexity: O(n)
Auxiliary Space: O(1), since no extra space has been taken.
