Even numbers at even index and odd numbers at odd index
Given an array of size n containing equal number of odd and even numbers. The problem is to arrange the numbers in such a way that all the even numbers get the even index and odd numbers get the odd index. Required auxiliary space is O(1).
Examples :
Input : arr[] = {3, 6, 12, 1, 5, 8} Output : 6 3 12 1 8 5 Input : arr[] = {10, 9, 7, 18, 13, 19, 4, 20, 21, 14} Output : 10 9 18 7 20 19 4 13 14 21
Source: Amazon Interview Experience | Set 410.
Approach :
- Start from the left and keep two index one for even position and other for odd positions.
- Traverse these index from left.
- At even position there should be even number and at odd positions, there should be odd number.
- Whenever there is mismatch , we swap the values at odd and even index.
Below is the implementation of the above approach :
CPP
// C++ implementation to arrange // odd and even numbers #include <bits/stdc++.h> using namespace std; // function to arrange odd and even numbers void arrangeOddAndEven( int arr[], int n) { int oddInd = 1; int evenInd = 0; while ( true ) { while (evenInd < n && arr[evenInd] % 2 == 0) evenInd += 2; while (oddInd < n && arr[oddInd] % 2 == 1) oddInd += 2; if (evenInd < n && oddInd < n) swap (arr[evenInd], arr[oddInd]); else break ; } } // function to print the array void printArray( int arr[], int n) { for ( int i = 0; i < n; i++) cout << arr[i] << " " ; } // Driver program to test above int main() { int arr[] = { 3, 6, 12, 1, 5, 8 }; int n = sizeof (arr) / sizeof (arr[0]); cout << "Original Array: " ; printArray(arr, n); arrangeOddAndEven(arr, n); cout << "\nModified Array: " ; printArray(arr, n); return 0; } |
Java
// Java implementation to // arrange odd and even numbers import java.util.*; import java.lang.*; class GfG { // function to arrange // odd and even numbers public static void arrangeOddAndEven( int arr[], int n) { int oddInd = 1 ; int evenInd = 0 ; while ( true ) { while (evenInd < n && arr[evenInd] % 2 == 0 ) evenInd += 2 ; while (oddInd < n && arr[oddInd] % 2 == 1 ) oddInd += 2 ; if (evenInd < n && oddInd < n) { int temp = arr[evenInd]; arr[evenInd] = arr[oddInd]; arr[oddInd] = temp; } else break ; } } // function to print the array public static void printArray( int arr[], int n) { for ( int i = 0 ; i < n; i++) System.out.print(arr[i] + " " ); } // Driver function public static void main(String argc[]){ int arr[] = { 3 , 6 , 12 , 1 , 5 , 8 }; int n = 6 ; System.out.print( "Original Array: " ); printArray(arr, n); arrangeOddAndEven(arr, n); System.out.print( "\nModified Array: " ); printArray(arr, n); } } // This code is contributed by Sagar Shukla |
Python3
# Python3 implementation to # arrange odd and even numbers def arrangeOddAndEven(arr, n): oddInd = 1 evenInd = 0 while ( True ): while (evenInd < n and arr[evenInd] % 2 = = 0 ): evenInd + = 2 while (oddInd < n and arr[oddInd] % 2 = = 1 ): oddInd + = 2 if (evenInd < n and oddInd < n): temp = arr[evenInd] arr[evenInd] = arr[oddInd] arr[oddInd] = temp; else : break # function to print the array def printArray(arr, n): for i in range ( 0 ,n): print (arr[i] , " ",end=" ") # Driver function def main(): arr = [ 3 , 6 , 12 , 1 , 5 , 8 ] n = 6 print ( "Original Array: " ,end = "") printArray(arr, n) arrangeOddAndEven(arr, n) print ( "\nModified Array: " ,end = "") printArray(arr, n) if __name__ = = '__main__' : main() # This code is contributed by 29AjayKumar |
C#
// C# implementation to // arrange odd and even numbers using System; class GFG { // function to arrange // odd and even numbers public static void arrangeOddAndEven( int [] arr, int n) { int oddInd = 1; int evenInd = 0; while ( true ) { while (evenInd < n && arr[evenInd] % 2 == 0) evenInd += 2; while (oddInd < n && arr[oddInd] % 2 == 1) oddInd += 2; if (evenInd < n && oddInd < n) { int temp = arr[evenInd]; arr[evenInd] = arr[oddInd]; arr[oddInd] = temp; } else break ; } } // function to print the array public static void printArray( int [] arr, int n) { for ( int i = 0; i < n; i++) Console.Write(arr[i] + " " ); } // Driver function public static void Main() { int [] arr = { 3, 6, 12, 1, 5, 8 }; int n = 6; Console.Write( "Original Array: " ); printArray(arr, n); arrangeOddAndEven(arr, n); Console.Write( "\nModified Array: " ); printArray(arr, n); } } // This code is contributed by Sam007 |
Javascript
<script> // Javascript implementation to arrange // odd and even numbers // function to arrange odd and even numbers function arrangeOddAndEven(arr, n) { let oddInd = 1; let evenInd = 0; while ( true ) { while (evenInd < n && arr[evenInd] % 2 == 0) evenInd += 2; while (oddInd < n && arr[oddInd] % 2 == 1) oddInd += 2; if (evenInd < n && oddInd < n) { let temp; temp = arr[evenInd]; arr[evenInd] = arr[oddInd]; arr[oddInd] = temp; } else break ; } } // function to print the array function printArray(arr, n) { for (let i = 0; i < n; i++) document.write(arr[i] + " " ); } // Driver program to test above let arr = [ 3, 6, 12, 1, 5, 8 ]; let n = arr.length; document.write( "Original Array: " ); printArray(arr, n); arrangeOddAndEven(arr, n); document.write( "<br>" + "Modified Array: " ); printArray(arr, n); // This code is contributed by Mayank Tyagi </script> |
Output :
Original Array: 3 6 12 1 5 8 Modified Array: 6 3 12 1 8 5
Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(1), as we are not using any extra space.
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