Given an array that contains odd number of occurrences for all numbers except for a few elements which are present even number of times. Find the elements which have even occurrences in the array in O(n) time complexity and O(1) extra space.
Assume array contain elements in the range 0 to 63.
Examples :
Input: [9, 12, 23, 10, 12, 12, 15, 23, 14, 12, 15] Output: 12, 23 and 15 Input: [1, 4, 7, 5, 9, 7, 3, 4, 6, 8, 3, 0, 3] Output: 4 and 7 Input: [4, 4, 10, 10, 4, 4, 4, 4, 10, 10] Output: 4 and 10
A simple method would be to traverse the array and store frequencies of its elements in a map. Later, print elements that have even count in the map. The solution takes O(n) time but requires extra space for storing frequencies. Below is an interesting method to solve this problem using bitwise operators.
This method assumes that long long integers are stored using 64 bits. The idea is to use XOR operator. We know that
1 XOR 1 = 0 1 XOR 0 = 1 0 XOR 1 = 1 0 XOR 0 = 0
Consider below input –
1, 4, 7, 5, 9, 7, 3, 4, 6, 8, 3, 0, 3
If we left-shift 1 by value of each element of the array and take XOR of all the items, we will get below binary integer –
1101101011
Each 1 in the i’th index from the right represents odd occurrence of element i. And each 0 in the i’th index from the right represents even or non-occurrence of element i in the array.
0 is present in 2nd, 4th and 7th position in above binary number. But 2 is not present in our array. So our answer is 4 and 7.
Below is the implementation of above idea
C++
// C++ Program to find the even occurring elements// in given array#include <iostream>using namespace std;// Function to find the even occurring elements// in given arrayvoid printRepeatingEven(int arr[], int n){ long long _xor = 0L; long long pos; // do for each element of array for( int i = 0; i < n; ++i) { // left-shift 1 by value of current element pos = 1 << arr[i]; // Toggle the bit everytime element gets repeated _xor ^= pos; } // Traverse array again and use _xor to find even // occurring elements for (int i = 0; i < n; ++i) { // left-shift 1 by value of current element pos = 1 << arr[i]; // Each 0 in _xor represents an even occurrence if (!(pos & _xor)) { // print the even occurring numbers cout << arr[i] << " "; // set bit as 1 to avoid printing duplicates _xor ^= pos; } }}// Driver codeint main(){ int arr[] = { 9, 12, 23, 10, 12, 12, 15, 23, 14, 12, 15 }; int n = sizeof(arr) / sizeof(arr[0]); printRepeatingEven(arr, n); return 0;} |
Java
// Java Program to find the even occurring// elements in given array class GFG {// Function to find the even occurring // elements in given array static void printRepeatingEven(int arr[], int n){ long _xor = 0L; long pos; // do for each element of array for (int i = 0; i < n; ++i) { // left-shift 1 by value of // current element pos = 1 << arr[i]; // Toggle the bit everytime // element gets repeated _xor ^= pos; } // Traverse array again and use _xor // to find even occurring elements for (int i = 0; i < n; ++i) { // left-shift 1 by value of // current element pos = 1 << arr[i]; // Each 0 in _xor represents // an even occurrence if (!((pos & _xor)!=0)) { // print the even occurring numbers System.out.print(arr[i] + " "); // set bit as 1 to avoid // printing duplicates _xor ^= pos; } }}// Driver code public static void main(String args[]) { int arr[] = {9, 12, 23, 10, 12, 12, 15, 23, 14, 12, 15}; int n = arr.length; printRepeatingEven(arr, n);}}// This code is contributed // by 29AjayKumar |
C#
// C# Program to find the even occurring// elements in given array using System;class GFG {// Function to find the even occurring // elements in given array static void printRepeatingEven(int[] arr, int n){ long _xor = 0L; long pos; // do for each element of array for (int i = 0; i < n; ++i) { // left-shift 1 by value of // current element pos = 1 << arr[i]; // Toggle the bit everytime // element gets repeated _xor ^= pos; } // Traverse array again and use _xor // to find even occurring elements for (int i = 0; i < n; ++i) { // left-shift 1 by value of // current element pos = 1 << arr[i]; // Each 0 in _xor represents // an even occurrence if (!((pos & _xor) != 0)) { // print the even occurring numbers Console.Write(arr[i] + " "); // set bit as 1 to avoid // printing duplicates _xor ^= pos; } }}// Driver code public static void Main() { int[] arr = {9, 12, 23, 10, 12, 12, 15, 23, 14, 12, 15}; int n = arr.Length; printRepeatingEven(arr, n);}}// This code is contributed // by Mukul singh |
PHP
<?php// PHP Program to find the even // occurring elements in given array// Function to find the even // occurring elements in given arrayfunction printRepeatingEven($arr, $n){ $_xor = 0; $pos; // do for each element of array for( $i = 0; $i < $n; ++$i) { // left-shift 1 by value // of current element $pos = 1 << $arr[$i]; // Toggle the bit everytime // element gets repeated $_xor ^= $pos; } // Traverse array again and // use _xor to find even // occurring elements for ($i = 0; $i < $n; ++$i) { // left-shift 1 by value // of current element $pos = 1 << $arr[$i]; // Each 0 in _xor represents // an even occurrence if (!($pos & $_xor)) { // print the even // occurring numbers echo $arr[$i], " "; // set bit as 1 to avoid // printing duplicates $_xor ^= $pos; } }}// Driver code$arr = array(9, 12, 23, 10, 12, 12, 15, 23, 14, 12, 15 );$n = sizeof($arr);printRepeatingEven($arr, $n);// This code is contributed by aj_36?> |
Python 3
# Python 3 program to find the even # occurring elements in given array# Function to find the even occurring # elements in given arraydef printRepeatingEven(arr, n): axor = 0; # do for each element of array for i in range(0, n): # left-shift 1 by value of # current element pos = 1 << arr[i]; # Toggle the bit every time # element gets repeated axor ^= pos; # Traverse array again and use _xor # to find even occurring elements for i in range(0, n - 1): # left-shift 1 by value of # current element pos = 1 << arr[i]; # Each 0 in _xor represents an # even occurrence if (not(pos & axor)): # print the even occurring numbers print(arr[i], end = " "); # set bit as 1 to avoid printing # duplicates axor ^= pos; # Driver codearr = [9, 12, 23, 10, 12, 12, 15, 23, 14, 12, 15 ];n = len(arr) ;printRepeatingEven(arr, n);# This code is contributed # by Shivi_Aggarwal |
Javascript
<script>// Javascript Program to find the even occurring// elements in given array // Function to find the even occurring // elements in given array function printRepeatingEven(arr, n){ let _xor = 0; let pos; // do for each element of array for (let i = 0; i < n; ++i) { // left-shift 1 by value of // current element pos = 1 << arr[i]; // Toggle the bit everytime // element gets repeated _xor ^= pos; } // Traverse array again and use _xor // to find even occurring elements for (let i = 0; i < n; ++i) { // left-shift 1 by value of // current element pos = 1 << arr[i]; // Each 0 in _xor represents // an even occurrence if (!((pos & _xor)!=0)) { // print the even occurring numbers document.write(arr[i] + " "); // set bit as 1 to avoid // printing duplicates _xor ^= pos; } }}// Driver Code let arr = [9, 12, 23, 10, 12, 12, 15, 23, 14, 12, 15]; let n = arr.length; printRepeatingEven(arr, n); </script> |
Output :
12 23 15
Time Complexity: O(n)
Auxiliary Space: O(1)
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