# 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 ith 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++ Program to implement ` `// the above approach ` `#include ` `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 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 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# 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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