Find the two numbers with odd occurrences in an unsorted array

Given an unsorted array that contains even number of occurrences for all numbers except two numbers. Find the two numbers which have odd occurrences in O(n) time complexity and O(1) extra space.

Examples:

Input: {12, 23, 34, 12, 12, 23, 12, 45}
Output: 34 and 45

Input: {4, 4, 100, 5000, 4, 4, 4, 4, 100, 100}
Output: 100 and 5000

Input: {10, 20}
Output: 10 and 20

A naive method to solve this problem is to run two nested loops. The outer loop picks an element and the inner loop counts the number of occurrences of the picked element. If the count of occurrences is odd then print the number. The time complexity of this method is O(n^2).



We can use sorting to get the odd occurring numbers in O(nLogn) time. First sort the numbers using an O(nLogn) sorting algorithm like Merge Sort, Heap Sort.. etc. Once the array is sorted, all we need to do is a linear scan of the array and print the odd occurring number.

We can also use hashing. Create an empty hash table which will have elements and their counts. Pick all elements of input array one by one. Look for the picked element in hash table. If the element is found in hash table, increment its count in table. If the element is not found, then enter it in hash table with count as 1. After all elements are entered in hash table, scan the hash table and print elements with odd count. This approach may take O(n) time on average, but it requires O(n) extra space.


A O(n) time and O(1) extra space solution:
The idea is similar to method 2 of this post. Let the two odd occurring numbers be x and y. We use bitwise XOR to get x and y. The first step is to do XOR of all elements present in array. XOR of all elements gives us XOR of x and y because of the following properties of XOR operation.
1) XOR of any number n with itself gives us 0, i.e., n ^ n = 0
2) XOR of any number n with 0 gives us n, i.e., n ^ 0 = n
3) XOR is cumulative and associative.

So we have XOR of x and y after the first step. Let the value of XOR be xor2. Every set bit in xor2 indicates that the corresponding bits in x and y have values different from each other. For example, if x = 6 (0110) and y is 15 (1111), then xor2 will be (1001), the two set bits in xor2 indicate that the corresponding bits in x and y are different. In the second step, we pick a set bit of xor2 and divide array elements in two groups. Both x and y will go to different groups. In the following code, the rightmost set bit of xor2 is picked as it is easy to get rightmost set bit of a number. If we do XOR of all those elements of array which have the corresponding bit set (or 1), then we get the first odd number. And if we do XOR of all those elements which have the corresponding bit 0, then we get the other odd occurring number. This step works because of the same properties of XOR. All the occurrences of a number will go in same set. XOR of all occurrences of a number which occur even number number of times will result in 0 in its set. And the xor of a set will be one of the odd occurring elements.

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// C++ Program to find the two odd occurring elements 
#include <bits/stdc++.h>
using namespace std;
  
/* Prints two numbers that occur odd number of times. The 
function assumes that the array size is at least 2 and 
there are exactly two numbers occurring odd number of times. */
void printTwoOdd(int arr[], int size) 
    int xor2 = arr[0]; /* Will hold XOR of two odd occurring elements */
    int set_bit_no; /* Will have only single set bit of xor2 */
    int i; 
    int n = size - 2; 
    int x = 0, y = 0; 
      
    /* Get the xor of all elements in arr[]. The xor will basically 
        be xor of two odd occurring elements */
    for(i = 1; i < size; i++) 
        xor2 = xor2 ^ arr[i]; 
      
    /* Get one set bit in the xor2. We get rightmost set bit 
        in the following line as it is easy to get */
    set_bit_no = xor2 & ~(xor2-1); 
      
    /* Now divide elements in two sets: 
        1) The elements having the corresponding bit as 1. 
        2) The elements having the corresponding bit as 0. */
    for(i = 0; i < size; i++) 
    
        /* XOR of first set is finally going to hold one odd 
        occurring number x */
        if(arr[i] & set_bit_no) 
        x = x ^ arr[i]; 
      
        /* XOR of second set is finally going to hold the other 
        odd occurring number y */
        else
        y = y ^ arr[i]; 
    
  
    cout << "The two ODD elements are " << x << " & " << y; 
  
/* Driver code */
int main() 
    int arr[] = {4, 2, 4, 5, 2, 3, 3, 1}; 
    int arr_size = sizeof(arr)/sizeof(arr[0]); 
    printTwoOdd(arr, arr_size); 
    return 0; 
  
// This is code is contributed by rathbhupendra
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// Program to find the two odd occurring elements
#include<stdio.h>
  
/* Prints two numbers that occur odd number of times. The
   function assumes that the array size is at least 2 and
   there are exactly two numbers occurring odd number of times. */
void printTwoOdd(int arr[], int size)
{
  int xor2 = arr[0]; /* Will hold XOR of two odd occurring elements */
  int set_bit_no;  /* Will have only single set bit of xor2 */
  int i;
  int n = size - 2;
  int x = 0, y = 0;
  
  /* Get the xor of all elements in arr[]. The xor will basically
     be xor of two odd occurring elements */
  for(i = 1; i < size; i++)
    xor2 = xor2 ^ arr[i];
  
  /* Get one set bit in the xor2. We get rightmost set bit
     in the following line as it is easy to get */
  set_bit_no = xor2 & ~(xor2-1);
  
  /* Now divide elements in two sets: 
    1) The elements having the corresponding bit as 1. 
    2) The elements having the corresponding bit as 0.  */
  for(i = 0; i < size; i++)
  {
     /* XOR of first set is finally going to hold one odd 
       occurring number x */ 
    if(arr[i] & set_bit_no)
      x = x ^ arr[i];
  
     /* XOR of second set is finally going to hold the other 
       odd occurring number y */ 
    else
      y = y ^ arr[i]; 
  }
  
  printf("\n The two ODD elements are %d & %d ", x, y);
}
  
/* Driver program to test above function */
int main()
{
  int arr[] = {4, 2, 4, 5, 2, 3, 3, 1};
  int arr_size = sizeof(arr)/sizeof(arr[0]);
  printTwoOdd(arr, arr_size);
  getchar();
  return 0;
}
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// Java program to find two odd
// occurring elements
  
import java.util.*;
  
class Main
{   
       
    /* Prints two numbers that occur odd 
       number of times. The function assumes 
       that the array size is at least 2 and
       there are exactly two numbers occurring 
       odd number of times. */
    static void printTwoOdd(int arr[], int size)
    {
      /* Will hold XOR of two odd occurring elements */    
      int xor2 = arr[0]; 
        
      /* Will have only single set bit of xor2 */
      int set_bit_no;  
      int i;
      int n = size - 2;
      int x = 0, y = 0;
       
      /* Get the xor of all elements in arr[]. 
         The xor will basically be xor of two 
         odd occurring elements */
      for(i = 1; i < size; i++)
        xor2 = xor2 ^ arr[i];
       
      /* Get one set bit in the xor2. We get 
         rightmost set bit in the following 
         line as it is easy to get */
      set_bit_no = xor2 & ~(xor2-1);
       
      /* Now divide elements in two sets: 
            1) The elements having the 
               corresponding bit as 1. 
            2) The elements having the 
               corresponding bit as 0.  */
      for(i = 0; i < size; i++)
      {
         /* XOR of first set is finally going 
            to hold one odd occurring number x */
        if((arr[i] & set_bit_no)>0)
          x = x ^ arr[i];
       
         /* XOR of second set is finally going 
            to hold the other odd occurring number y */
        else
          y = y ^ arr[i]; 
      }
       
      System.out.println("The two ODD elements are "+
                                        x + " & " + y);
    }
      
    // main function
    public static void main (String[] args) 
    {
        int arr[] = {4, 2, 4, 5, 2, 3, 3, 1};
        int arr_size = arr.length;
        printTwoOdd(arr, arr_size);
    }
}
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# Python3 program to find the
# two odd occurring elements
  
# Prints two numbers that occur odd
# number of times. The function assumes
# that the array size is at least 2 and
# there are exactly two numbers occurring
# odd number of times.
def printTwoOdd(arr, size):
      
    # Will hold XOR of two odd occurring elements 
    xor2 = arr[0
      
    # Will have only single set bit of xor2
    set_bit_no = 0  
    n = size - 2
    x, y = 0, 0
  
    # Get the xor of all elements in arr[]. 
    # The xor will basically be xor of two
    # odd occurring elements 
    for i in range(1, size):
        xor2 = xor2 ^ arr[i]
  
    # Get one set bit in the xor2. We get 
    # rightmost set bit in the following 
    # line as it is easy to get 
    set_bit_no = xor2 & ~(xor2 - 1)
  
    # Now divide elements in two sets: 
    # 1) The elements having the corresponding bit as 1. 
    # 2) The elements having the corresponding bit as 0. 
    for i in range(size):
      
        # XOR of first set is finally going to  
        # hold one odd  occurring number x 
        if(arr[i] & set_bit_no):
            x = x ^ arr[i]
  
        # XOR of second set is finally going 
        # to hold the other odd occurring number y 
        else:
            y = y ^ arr[i] 
  
    print("The two ODD elements are", x, "&", y)
  
# Driver Code
arr = [4, 2, 4, 5, 2, 3, 3, 1]
arr_size = len(arr)
printTwoOdd(arr, arr_size)
  
# This code is contributed by Anant Agarwal.
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// C# program to find two odd
// occurring elements
using System;
  
class main
{   
        
    // Prints two numbers that occur
    // odd number of times. Function
    // assumes that array size is at
    // least 2 and there are exactly
    // two numbers occurring odd number
    // of times.
    static void printTwoOdd(int []arr, int size) {
     
      // Will hold XOR of two odd
      //occurring elements   
      int xor2 = arr[0]; 
         
      // Will have only single set
      // bit of xor2
      int set_bit_no;  
      int i;
        
      //int n = size - 2;
      int x = 0, y = 0;
        
      // Get the xor of all the elements
      // in arr[].The xor will basically 
      // be xor of two odd occurring
      // elements
      for(i = 1; i < size; i++)
        xor2 = xor2 ^ arr[i];
        
      // Get one set bit in the xor2.
      // We get rightmost set bit in
      // the following line as it is
      // to get.
         set_bit_no = xor2 & ~(xor2-1);
        
      // divide elements in two sets: 
      // 1) The elements having the 
      // corresponding bit as 1. 
      // 2) The elements having the 
      // corresponding bit as 0.
      for(i = 0; i < size; i++)
      {
         // XOR of first set is finally 
         // going to hold one odd 
         // occurring number x
            if((arr[i] & set_bit_no)>0)
            x = x ^ arr[i];
        
            // XOR of second set is finally
            // going to hold the other 
            // odd occurring number y
            else
            y = y ^ arr[i]; 
      }
        
      Console.WriteLine("The two ODD elements are "+
                                        x + " & " + y);
    }
       
    // main function
    public static void Main() 
    {
        int []arr = {4, 2, 4, 5, 2, 3, 3, 1};
        int arr_size = arr.Length;
        printTwoOdd(arr, arr_size);
    }
}
  
//This code is contributed by Anant Agarwal.
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<?php
// PHP program to find the 
// two odd occurring elements
  
/* Prints two numbers that occur
   odd number of times. The
   function assumes that the 
   array size is at least 2 and
   there are exactly two numbers 
   occurring odd number of times. */
function printTwoOdd($arr, $size)
{
      
     // Will hold XOR of two 
     // odd occurring elements
     $xor2 = $arr[0];
       
     // Will have only single 
     // set bit of xor2 
     $set_bit_no;
       
    $i;
    $n = $size - 2;
    $x = 0;$y = 0;
  
     // Get the xor of all elements
     // in arr[]. The xor will basically
     // be xor of two odd occurring 
     // elements 
    for($i = 1; $i < $size; $i++)
        $xor2 = $xor2 ^ $arr[$i];
      
    // Get one set bit in the xor2. 
    // We get rightmost set bit
    // in the following line as 
    // it is easy to get 
    $set_bit_no = $xor2 & ~($xor2-1);
      
    /* Now divide elements in two sets: 
        1) The elements having the
           corresponding bit as 1. 
        2) The elements having the 
           corresponding bit as 0. */
    for($i = 0; $i < $size; $i++)
    {
          
        /* XOR of first set is finally 
           going to hold one odd 
           occurring number x */
        if($arr[$i] & $set_bit_no)
        $x = $x ^ $arr[$i];
      
        /* XOR of second set is finally
           going to hold the other 
           odd occurring number y */
        else
        $y = $y ^ $arr[$i]; 
    }
      
    echo "The two ODD elements are ", $x, " & ", $y;
    }
  
// Driver Code
$arr = array(4, 2, 4, 5, 2, 3, 3, 1);
$arr_size = sizeof($arr);
printTwoOdd($arr, $arr_size);
  
// This code is Contributed by Ajit
?>
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Output:
The two ODD elements are 5 & 1

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

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.




Improved By : jit_t, rathbhupendra



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