Rearrange Odd and Even values in Alternate Fashion in Ascending Order

Difficulty Level : Easy
Last Updated : 15 Sep, 2021

Given an array of integers (both odd and even), the task is to arrange them in such a way that odd and even values come in alternate fashion in non-decreasing order(ascending) respectively.

If the smallest value is Even then we have to print Even-Odd pattern.
If the smallest value is Odd then we have to print Odd-Even pattern.

Note: No. of odd elements must be equal to No. of even elements in the input array.
Examples:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: arr[] = {1, 3, 2, 5, 4, 7, 10}
Output: 1, 2, 3, 4, 5, 10, 7
Smallest value is 1(Odd) so output will be Odd-Even pattern.
Input: arr[] = {9, 8, 13, 2, 19, 14}
Output: 2, 9, 8, 13, 14, 19
Smallest value is 2(Even) so output will be Even-Odd pattern.

Asked In: Microsoft Tech-Set-Go-2018

Algorithm:

Sort the given array.
Insert Even values in List-1 and Odd values in List-2.
Now if the smallest value is even, then insert an even value from list 1 and odd value from list 2 to original array and so on.
But if the smallest value is odd, then insert an odd value from list 2 and even value from list 1 to original array and so on.

Below is the implementation of the above approach:

C++

// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
void AlternateRearrange(int arr[], int n)
{
    // Sort the array
    sort(arr, arr + n);
 
    vector<int> v1; // to insert even values
    vector<int> v2; // to insert odd values
 
    for (int i = 0; i < n; i++)
        if (arr[i] % 2 == 0)
            v1.push_back(arr[i]);
        else
            v2.push_back(arr[i]);
 
    int index = 0, i = 0, j = 0;
 
    bool flag = false;
 
    // Set flag to true if first element is even
    if (arr[0] % 2 == 0)
        flag = true;
 
    // Start rearranging array
    while (index < n) {
 
        // If first element is even
        if (flag == true) {
            arr[index++] = v1[i++];
            flag = !flag;
        }
 
        // Else, first element is Odd
        else {
            arr[index++] = v2[j++];
            flag = !flag;
        }
    }
 
    // Print the rearranged array
    for (i = 0; i < n; i++)
        cout << arr[i] << " ";
}
 
// Driver code
int main()
{
    int arr[] = { 9, 8, 13, 2, 19, 14 };
    int n = sizeof(arr) / sizeof(int);
    AlternateRearrange(arr, n);
    return 0;
}

Java

// Java implementation of the above approach
 
import java.util.* ;
 
class GFG
{
 
    static void AlternateRearrange(int arr[], int n)
    {
        // Sort the array
         
        // Collection.sort() sorts the
        // collection in ascending order
        Arrays.sort(arr) ;
         
        Vector v1 = new Vector(); // to insert even values
        Vector v2 = new Vector(); // to insert odd values
     
        for (int i = 0; i < n; i++)
            if (arr[i] % 2 == 0)
                v1.add(arr[i]);
            else
                v2.add(arr[i]);
     
        int index = 0, i = 0, j = 0;
     
        boolean flag = false;
     
        // Set flag to true if first element is even
        if (arr[0] % 2 == 0)
            flag = true;
     
        // Start rearranging array
        while (index < n)
        {
     
            // If first element is even
            if (flag == true)
            {
                arr[index] = (int)v1.get(i);
                i += 1 ;
                index += 1 ;
                flag = !flag;
            }
     
            // Else, first element is Odd
            else
            {
                arr[index] = (int)v2.get(j) ;
                j += 1 ;
                index += 1 ;
                flag = !flag;
            }
        }
     
        // Print the rearranged array
        for (i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
     
    // Driver code
    public static void main(String []args)
    {
        int arr[] = { 9, 8, 13, 2, 19, 14 };
        int n = arr.length ;
     
        AlternateRearrange(arr, n);
    }
}
 
// This code is contributed by aishwarya.27

Python3

# Python3 implementation of the above approach
def AlternateRearrange(arr, n):
 
    # Sort the array
    arr.sort()
 
    v1 = list()  # to insert even values
    v2 = list()  # to insert odd values
 
    for i in range(n):
        if (arr[i] % 2 == 0):
            v1.append(arr[i])
        else:
            v2.append(arr[i])
 
    index = 0
    i = 0
    j = 0
 
    flag = False
 
    # Set flag to true if first element is even
    if (arr[0] % 2 == 0):
        flag = True
 
    # Start rearranging array
    while (index < n):
 
        # If first element is even
        if (flag == True and i < len(v1)):
            arr[index] = v1[i]
            index += 1
            i += 1
            flag = ~flag
 
        # Else, first element is Odd
        elif j < len(v2):
            arr[index] = v2[j]
            index += 1
            j += 1
 
            flag = ~flag
 
    # Print the rearranged array
    for i in range(n):
        print(arr[i], end=" ")
 
# Driver code
arr = [9, 8, 13, 2, 19, 14, 21, 23, 25]
n = len(arr)
 
AlternateRearrange(arr, n)
 
# This code is contributed
# by Mohit Kumar 29.

C#

// C# implementation of the above approach
 
using System;
using System.Collections;
class GFG
{
 
    static void AlternateRearrange(int []arr, int n)
    {
        // Sort the array
         
        // Collection.sort() sorts the
        // collection in ascending order
        Array.Sort(arr) ;
         
        ArrayList v1 = new ArrayList(); // to insert even values
        ArrayList v2 = new ArrayList(); // to insert odd values
     
        for (int j = 0; j < n; j++)
            if (arr[j] % 2 == 0)
                v1.Add(arr[j]);
            else
                v2.Add(arr[j]);
     
        int index = 0, i = 0, k = 0;
     
        bool flag = false;
     
        // Set flag to true if first element is even
        if (arr[0] % 2 == 0)
            flag = true;
     
        // Start rearranging array
        while (index < n)
        {
     
            // If first element is even
            if (flag == true)
            {
                arr[index] = (int)v1[i];
                i += 1 ;
                index += 1 ;
                flag = !flag;
            }
     
            // Else, first element is Odd
            else
            {
                arr[index] = (int)v2[k] ;
                k += 1 ;
                index += 1 ;
                flag = !flag;
            }
        }
     
        // Print the rearranged array
        for (i = 0; i < n; i++)
            Console.Write(arr[i] + " ");
    }
     
    // Driver code
    static void Main()
    {
        int []arr = { 9, 8, 13, 2, 19, 14 };
        int n = arr.Length ;
     
        AlternateRearrange(arr, n);
    }
}
 
// This code is contributed by mits

PHP

<?php
// PHP implementation of the above approach
function AlternateRearrange($arr, $n)
{
 
    // Sort the array
    sort($arr);
 
    $v1 = array(); // to insert even values
    $v2 = array(); // to insert odd values
 
    for ($i = 0; $i < $n; $i++)
        if ($arr[$i] % 2 == 0)
            array_push($v1, $arr[$i]);
        else
            array_push($v2, $arr[$i]);
 
    $index = 0;
    $i = 0;
    $j = 0;
 
    $flag = false;
 
    // Set flag to true if first element is even
    if ($arr[0] % 2 == 0)
        $flag = true;
 
    // Start rearranging array
    while ($index < $n)
    {
 
        // If first element is even
        if ($flag == true)
        {
            $arr[$index++] = $v1[$i++];
            $flag = !$flag;
        }
 
        // Else, first element is Odd
        else
        {
            $arr[$index++] = $v2[$j++];
            $flag = !$flag;
        }
    }
 
    // Print the rearranged array
    for ($i = 0; $i < $n; $i++)
        echo $arr[$i], " " ;
}
 
// Driver code
$arr = array( 9, 8, 13, 2, 19, 14 );
$n = sizeof($arr);
 
AlternateRearrange($arr, $n);
 
// This code is contributed by Ryuga
?>

Javascript

<script>
 
// Javascript implementation of the above approach
 
function AlternateRearrange(arr, n)
{
    // Sort the array
    arr.sort((a,b)=>a-b);
 
    var v1 = []; // to insert even values
    var v2 = []; // to insert odd values
 
    for (var i = 0; i < n; i++)
        if (arr[i] % 2 == 0)
            v1.push(arr[i]);
        else
            v2.push(arr[i]);
 
    var index = 0, i = 0, j = 0;
 
    var flag = false;
 
    // Set flag to true if first element is even
    if (arr[0] % 2 == 0)
        flag = true;
 
    // Start rearranging array
    while (index < n) {
 
        // If first element is even
        if (flag == true) {
            arr[index++] = v1[i++];
            flag = !flag;
        }
 
        // Else, first element is Odd
        else {
            arr[index++] = v2[j++];
            flag = !flag;
        }
    }
 
    // Print the rearranged array
    for (i = 0; i < n; i++)
        document.write( arr[i] + " ");
}
 
// Driver code
var arr = [9, 8, 13, 2, 19, 14];
var n = arr.length;
AlternateRearrange(arr, n);
 
</script>

Output:

2 9 8 13 14 19

Time Complexity: O(N * logN)
Auxiliary Space: O(N)

My Personal Notes

Related Articles

Table of Contents

Rearrange Odd and Even values in Alternate Fashion in Ascending Order

C++

Java

Python3

C#

PHP

Javascript

Start Your Coding Journey Now!