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 numbersvoid 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 arrayvoid printArray(int arr[], int n){ for (int i = 0; i < n; i++) cout << arr[i] << " ";}// Driver program to test aboveint 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 numbersimport java.util.*;import java.lang.*;class GfG {// function to arrange// odd and even numberspublic 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 arraypublic 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 numbersdef 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 arraydef 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 numbersusing 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.
Approach: Two-Pointer Swap
The steps to solve the problem using this approach are as follows:
- Initialize two pointers i and j to 0 and 1, respectively. These pointers will help us to keep track of the current indices of the elements in the array that we are processing.
- Loop through the array until i or j reaches the end of the array.
- If the element at index i is odd and the element at index j is even, swap them. This will ensure that the even number is moved to an even index and odd number is moved to an odd index.
- Increment both i and j by 2 to move them to the next odd and even index respectively.
- If the element at index i is already even, increment i by 2.
- Similarly, if the element at index j is already odd, increment j by 2.
- Once we have looped through the entire array, the even numbers will be at even indices and odd numbers will be at odd indices.
- Return the modified array.
C++
#include <iostream>#include <vector>using namespace std;vector<int> arrange_array(vector<int> arr) { int i = 0; int j = 1; int n = arr.size(); while (i < n && j < n) { if (arr[i] % 2 != 0 && arr[j] % 2 == 0) { swap(arr[i], arr[j]); i += 2; j += 2; } else { if (arr[i] % 2 == 0) { i += 2; } if (arr[j] % 2 != 0) { j += 2; } } } return arr;}int main() { // Example 1 vector<int> arr1 = {3, 6, 12, 1, 5, 8}; vector<int> arranged_arr1 = arrange_array(arr1); for (auto i : arranged_arr1) { cout << i << " "; } cout << endl; // Example 2 vector<int> arr2 = {10, 9, 7, 18, 13, 19, 4, 20, 21, 14}; vector<int> arranged_arr2 = arrange_array(arr2); for (auto i : arranged_arr2) { cout << i << " "; } cout << endl; return 0;} |
Java
import java.util.Arrays;public class Main { public static void main(String[] args) { int[] arr1 = { 3, 6, 12, 1, 5, 8 }; int[] arrangedArr1 = arrangeArray(arr1); System.out.println(Arrays.toString(arrangedArr1)); int[] arr2 = { 10, 9, 7, 18, 13, 19, 4, 20, 21, 14 }; int[] arrangedArr2 = arrangeArray(arr2); System.out.println(Arrays.toString(arrangedArr2)); } // Function to arrange the array such that odd numbers // are at odd indices and even numbers are at even // indices public static int[] arrangeArray(int[] arr) { int i = 0; // starting index for odd numbers int j = 1; // starting index for even numbers int n = arr.length; // length of the array while (i < n && j < n) { // if the number at i is odd and the number at j // is even, swap them and move to next pair of // indices if (arr[i] % 2 != 0 && arr[j] % 2 == 0) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; i += 2; // move to next odd index j += 2; // move to next even index } else { // if the number at i is even, move to next // odd index if (arr[i] % 2 == 0) { i += 2; } // if the number at j is odd, move to next // even index if (arr[j] % 2 != 0) { j += 2; } } } return arr; // return the arranged array }} |
Python3
def arrange_array(arr): i = 0 j = 1 n = len(arr) while i < n and j < n: if arr[i] % 2 != 0 and arr[j] % 2 == 0: arr[i], arr[j] = arr[j], arr[i] i += 2 j += 2 else: if arr[i] % 2 == 0: i += 2 if arr[j] % 2 != 0: j += 2 return arr# Example 1arr = [3, 6, 12, 1, 5, 8]arranged_arr = arrange_array(arr)print(arranged_arr) # Output: [6, 3, 12, 1, 8, 5]#Examplearr = [10, 9, 7, 18, 13, 19, 4, 20, 21, 14]arranged_arr = arrange_array(arr)print(arranged_arr) # Output: [10, 9, 18, 7, 20, 19, 4, 13, 14, 21] |
Javascript
function arrangeArray(arr) { let i = 0; // starting index for odd numbers let j = 1; // starting index for even numbers let n = arr.length; // length of the array while (i < n && j < n) { // if the number at i is odd and the number at j is even, swap them and move to next pair of indices if (arr[i] % 2 !== 0 && arr[j] % 2 === 0) { [arr[i], arr[j]] = [arr[j], arr[i]]; // swapping values i += 2; // move to next odd index j += 2; // move to next even index } else { // if the number at i is even, move to next odd index if (arr[i] % 2 === 0) { i += 2; } // if the number at j is odd, move to next even index if (arr[j] % 2 !== 0) { j += 2; } } } return arr; // return the arranged array}// Example 1let arr1 = [3, 6, 12, 1, 5, 8];let arrangedArr1 = arrangeArray(arr1);console.log(arrangedArr1.join(" "));// Example 2let arr2 = [10, 9, 7, 18, 13, 19, 4, 20, 21, 14];let arrangedArr2 = arrangeArray(arr2);console.log(arrangedArr2.join(" ")); |
C#
using System;using System.Collections.Generic;class Program{ static List<int> arrangeArray(List<int> arr) { int i = 0; int j = 1; int n = arr.Count; while (i < n && j < n) { if (arr[i] % 2 != 0 && arr[j] % 2 == 0) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; i += 2; j += 2; } else { if (arr[i] % 2 == 0) { i += 2; } if (arr[j] % 2 != 0) { j += 2; } } } return arr; } static void Main(string[] args) { // Example 1 List<int> arr1 = new List<int> { 3, 6, 12, 1, 5, 8 }; List<int> arranged_arr1 = arrangeArray(arr1); foreach (int i in arranged_arr1) { Console.Write(i + " "); } Console.WriteLine(); // Example 2 List<int> arr2 = new List<int> { 10, 9, 7, 18, 13, 19, 4, 20, 21, 14 }; List<int> arranged_arr2 = arrangeArray(arr2); foreach (int i in arranged_arr2) { Console.Write(i + " "); } Console.WriteLine(); Console.ReadLine(); }} |
Output
[6, 3, 12, 1, 8, 5] [10, 9, 18, 7, 20, 19, 4, 13, 14, 21]
This algorithm runs in O(n) time complexity and O(1) space complexity
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