Rearrange array such that all even-indexed elements in the Array is even

Given an array arr[], the task is to check if it is possible to rearrange the array in such a way that every even index(1-based indexing) contains an even number. If such a rearrangement is not possible, print “No”. Otherwise, print “Yes” and print a possible arrangement

Examples:

Input: arr[] = {2, 4, 8, 3, 1}
Output:
Yes
3 4 8 2 1

Input: arr[] = {3, 3, 11, 8}
Output: No
Explanation: Since, the array contains only one even element, all even indices cannot be filled with a even elements.

Approach:
Follow the steps below to solve the problem:

Count the total number of even elements present in the given array. If the count exceeds the total number of even indices in the given array, print “No”.
Otherwise, print “Yes”. Now, traverse the array using two pointers, i and j, pointing to even and odd indices respectively.
For any i^th index containing an odd element, iterate odd indices using j until an even element is encountered.
Swap a[i] and a[j]
Repeat the above steps until all even indices are filled with an even element.

Below is the implementation of the above approach:

C++

// C++ Program to implement 
// the above approach 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to check if it the 
// array can be rearranged such 
// such that every even indices 
// contains an even element 
void checkPossible(int a[], int n) 
{ 
    // Stores the count of even elements 
    int even_no_count = 0; 
  
    // Traverse array to count even numbers 
    for (int i = 0; i < n; i++)  
    { 
        if (a[i] % 2 == 0) 
            even_no_count++; 
    } 
  
    // If even_no_count exceeds 
    // count of even indices 
    if (n / 2 > even_no_count)  
    { 
        cout << "No" << endl; 
        return; 
    } 
  
    cout << "Yes" << endl; 
    int j = 0; 
    for (int i = 1; i < n; i += 2)  
    { 
        if (a[i] % 2 == 0) 
            continue; 
        else 
        { 
            while (j < n && a[j] % 2 != 0) 
                j += 2; 
  
            a[i] += a[j]; 
            a[j] = a[i] - a[j]; 
            a[i] -= a[j]; 
        } 
    } 
  
    for (int i = 0; i < n; i++)  
    { 
        cout << a[i] << " "; 
    } 
} 
  
// Driver Code 
int main() 
{ 
  
    int arr[] = { 2, 3, 4, 5, 6, 7 }; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    checkPossible(arr, n); 
    return 0; 
} 
  
// This code is contributed by gauravrajput1

Java

// Java Program to implement 
// the above approach 
import java.io.*; 
import java.util.*; 
  
class GFG { 
  
    // Function to check if it the 
    // array can be rearranged such 
    // such that every even indices 
    // contains an even element 
    static void checkPossible(int a[]) 
    { 
        // Stores the count of even elements 
        int even_no_count = 0; 
  
        // Traverse array to count even numbers 
        for (int i = 0; i < a.length; i++) { 
  
            if (a[i] % 2 == 0) 
                even_no_count++; 
        } 
  
        // If even_no_count exceeds 
        // count of even indices 
        if (a.length / 2 > even_no_count) { 
            System.out.println("No"); 
            return; 
        } 
  
        System.out.println("Yes"); 
        int j = 0; 
        for (int i = 1; i < a.length; i += 2) { 
            if (a[i] % 2 == 0) 
                continue; 
            else { 
                while (j < a.length && a[j] % 2 != 0) 
                    j += 2; 
  
                a[i] += a[j]; 
                a[j] = a[i] - a[j]; 
                a[i] -= a[j]; 
            } 
        } 
  
        for (int i = 0; i < a.length; i++) { 
            System.out.print(a[i] + " "); 
        } 
    } 
  
    // Driver Code 
    public static void main(String args[]) 
    { 
  
        int arr[] = { 2, 3, 4, 5, 6, 7 }; 
  
        checkPossible(arr); 
    } 
} 

Python3

# Python3 program to implement 
# the above approach 
  
# Function to check if it the 
# array can be rearranged such 
# such that every even indices 
# contains an even element 
def checkPossible(a, n): 
      
    # Stores the count of even elements 
    even_no_count = 0
  
    # Traverse array to count even numbers 
    for i in range(n): 
        if (a[i] % 2 == 0): 
            even_no_count += 1
      
    # If even_no_count exceeds 
    # count of even indices 
    if (n // 2 > even_no_count): 
        print("No") 
        return
      
    print("Yes")  
    j = 0
      
    for i in range(1, n, 2): 
        if (a[i] % 2 == 0): 
            continue
              
        else: 
            while (j < n and a[j] % 2 != 0): 
                j += 2
  
            a[i] += a[j] 
            a[j] = a[i] - a[j] 
            a[i] -= a[j] 
          
    for i in range(n):  
        print(a[i], end = " ") 
      
# Driver Code 
arr = [ 2, 3, 4, 5, 6, 7 ] 
n = len(arr) 
  
checkPossible(arr, n) 
  
# This code is contributed by code_hunt 

C#

// C# Program to implement 
// the above approach 
using System; 
class GFG{ 
  
  // Function to check if it the 
  // array can be rearranged such 
  // such that every even indices 
  // contains an even element 
  static void checkPossible(int []a) 
  { 
    // Stores the count of even elements 
    int even_no_count = 0; 
  
    // Traverse array to count even numbers 
    for (int i = 0; i < a.Length; i++)  
    { 
      if (a[i] % 2 == 0) 
        even_no_count++; 
    } 
  
    // If even_no_count exceeds 
    // count of even indices 
    if (a.Length / 2 > even_no_count)  
    { 
      Console.WriteLine("No"); 
      return; 
    } 
  
    Console.WriteLine("Yes"); 
    int j = 0; 
    for (int i = 1; i < a.Length; i += 2) 
    { 
      if (a[i] % 2 == 0) 
        continue; 
      else 
      { 
        while (j < a.Length && a[j] % 2 != 0) 
          j += 2; 
  
        a[i] += a[j]; 
        a[j] = a[i] - a[j]; 
        a[i] -= a[j]; 
      } 
    } 
  
    for (int i = 0; i < a.Length; i++) 
    { 
      Console.Write(a[i] + " "); 
    } 
  } 
  
  // Driver Code 
  public static void Main(String []args) 
  { 
    int []arr = { 2, 3, 4, 5, 6, 7 }; 
  
    checkPossible(arr); 
  } 
} 
  
// This code is contributed by Rajput-Ji

Output:

Yes
3 2 5 4 7 6

Time Complexity: O(N)
Auxiliary Space: O(1)

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My Personal Notes

C++

Java

Python3

C#

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