Given an array of positive integers. All numbers occur even number of times except one number which occurs odd number of times. Find the number in O(n) time & constant space.
Examples :
Input : arr = {1, 2, 3, 2, 3, 1, 3} Output : 3 Input : arr = {5, 7, 2, 7, 5, 2, 5} Output : 5
A Simple Solution is to run two nested loops. The outer loop picks all elements one by one and inner loop counts number of occurrences of the element picked by outer loop. Time complexity of this solution is O(n2).
Below is the implementation of the brute force approach :
C++
// C++ program to find the element // occurring odd number of times #include<bits/stdc++.h> using namespace std; // Function to find the element // occurring odd number of times int getOddOccurrence( int arr[], int arr_size) { for ( int i = 0; i < arr_size; i++) { int count = 0; for ( int j = 0; j < arr_size; j++) { if (arr[i] == arr[j]) count++; } if (count % 2 != 0) return arr[i]; } return -1; } // driver code int main() { int arr[] = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 }; int n = sizeof (arr) / sizeof (arr[0]); // Function calling cout << getOddOccurrence(arr, n); return 0; } |
Java
// Java program to find the element occurring // odd number of times class OddOccurrence { // function to find the element occurring odd // number of times static int getOddOccurrence( int arr[], int arr_size) { int i; for (i = 0 ; i < arr_size; i++) { int count = 0 ; for ( int j = 0 ; j < arr_size; j++) { if (arr[i] == arr[j]) count++; } if (count % 2 != 0 ) return arr[i]; } return - 1 ; } // driver code public static void main(String[] args) { int arr[] = new int []{ 2 , 3 , 5 , 4 , 5 , 2 , 4 , 3 , 5 , 2 , 4 , 4 , 2 }; int n = arr.length; System.out.println(getOddOccurrence(arr, n)); } } // This code has been contributed by Kamal Rawal |
Python3
# Python program to find the element occurring # odd number of times # function to find the element occurring odd # number of times def getOddOccurrence(arr, arr_size): for i in range ( 0 ,arr_size): count = 0 for j in range ( 0 , arr_size): if arr[i] = = arr[j]: count + = 1 if (count % 2 ! = 0 ): return arr[i] return - 1 # driver code arr = [ 2 , 3 , 5 , 4 , 5 , 2 , 4 , 3 , 5 , 2 , 4 , 4 , 2 ] n = len (arr) print (getOddOccurrence(arr, n)) # This code has been contributed by # Smitha Dinesh Semwal |
C#
// C# program to find the element // occurring odd number of times using System; class GFG { // Function to find the element // occurring odd number of times static int getOddOccurrence( int []arr, int arr_size) { for ( int i = 0; i < arr_size; i++) { int count = 0; for ( int j = 0; j < arr_size; j++) { if (arr[i] == arr[j]) count++; } if (count % 2 != 0) return arr[i]; } return -1; } // Driver code public static void Main() { int []arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 }; int n = arr.Length; Console.Write(getOddOccurrence(arr, n)); } } // This code is contributed by Sam007 |
PHP
<?php // PHP program to find the // element occurring odd // number of times // Function to find the element // occurring odd number of times function getOddOccurrence(& $arr , $arr_size ) { $count = 0; for ( $i = 0; $i < $arr_size ; $i ++) { for ( $j = 0; $j < $arr_size ; $j ++) { if ( $arr [ $i ] == $arr [ $j ]) $count ++; } if ( $count % 2 != 0) return $arr [ $i ]; } return -1; } // Driver code $arr = array (2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2); $n = sizeof( $arr ); // Function calling echo (getOddOccurrence( $arr , $n )); // This code is contributed // by Shivi_Aggarwal ?> |
Javascript
<script> // Javascript program to find the element // occurring odd number of times // Function to find the element // occurring odd number of times function getOddOccurrence(arr, arr_size) { for (let i = 0; i < arr_size; i++) { let count = 0; for (let j = 0; j < arr_size; j++) { if (arr[i] == arr[j]) count++; } if (count % 2 != 0) return arr[i]; } return -1; } // driver code let arr = [ 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 ]; let n = arr.length; // Function calling document.write(getOddOccurrence(arr, n)); // This code is contributed by Mayank Tyagi </script> |
Output :
5
A Better Solution is to use Hashing. Use array elements as key and their counts as value. Create an empty hash table. One by one traverse the given array elements and store counts. Time complexity of this solution is O(n). But it requires extra space for hashing.
Program :
C++
// C++ program to find the element // occurring odd number of times #include <bits/stdc++.h> using namespace std; // function to find the element // occurring odd number of times int getOddOccurrence( int arr[], int size) { // Defining HashMap in C++ unordered_map< int , int > hash; // Putting all elements into the HashMap for ( int i = 0; i < size; i++) { hash[arr[i]]++; } // Iterate through HashMap to check an element // occurring odd number of times and return it for ( auto i : hash) { if (i.second % 2 != 0) { return i.first; } } return -1; } // Driver code int main() { int arr[] = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 }; int n = sizeof (arr) / sizeof (arr[0]); // Function calling cout << getOddOccurrence(arr, n); return 0; } // This code is contributed by codeMan_d. |
Java
// Java program to find the element occurring odd // number of times import java.io.*; import java.util.HashMap; class OddOccurrence { // function to find the element occurring odd // number of times static int getOddOccurrence( int arr[], int n) { HashMap<Integer,Integer> hmap = new HashMap<>(); // Putting all elements into the HashMap for ( int i = 0 ; i < n; i++) { if (hmap.containsKey(arr[i])) { int val = hmap.get(arr[i]); // If array element is already present then // increase the count of that element. hmap.put(arr[i], val + 1 ); } else // if array element is not present then put // element into the HashMap and initialize // the count to one. hmap.put(arr[i], 1 ); } // Checking for odd occurrence of each element present // in the HashMap for (Integer a:hmap.keySet()) { if (hmap.get(a) % 2 != 0 ) return a; } return - 1 ; } // driver code public static void main(String[] args) { int arr[] = new int []{ 2 , 3 , 5 , 4 , 5 , 2 , 4 , 3 , 5 , 2 , 4 , 4 , 2 }; int n = arr.length; System.out.println(getOddOccurrence(arr, n)); } } // This code is contributed by Kamal Rawal |
Python3
# Python3 program to find the element # occurring odd number of times # function to find the element # occurring odd number of times def getOddOccurrence(arr,size): # Defining HashMap in C++ Hash = dict () # Putting all elements into the HashMap for i in range (size): Hash [arr[i]] = Hash .get(arr[i], 0 ) + 1 ; # Iterate through HashMap to check an element # occurring odd number of times and return it for i in Hash : if ( Hash [i] % 2 ! = 0 ): return i return - 1 # Driver code arr = [ 2 , 3 , 5 , 4 , 5 , 2 , 4 , 3 , 5 , 2 , 4 , 4 , 2 ] n = len (arr) # Function calling print (getOddOccurrence(arr, n)) # This code is contributed by mohit kumar |
C#
// C# program to find the element occurring odd // number of times using System; using System.Collections.Generic; public class OddOccurrence { // function to find the element occurring odd // number of times static int getOddOccurrence( int []arr, int n) { Dictionary< int , int > hmap = new Dictionary< int , int >(); // Putting all elements into the HashMap for ( int i = 0; i < n; i++) { if (hmap.ContainsKey(arr[i])) { int val = hmap[arr[i]]; // If array element is already present then // increase the count of that element. hmap.Remove(arr[i]); hmap.Add(arr[i], val + 1); } else // if array element is not present then put // element into the HashMap and initialize // the count to one. hmap.Add(arr[i], 1); } // Checking for odd occurrence of each element present // in the HashMap foreach (KeyValuePair< int , int > entry in hmap) { if (entry.Value % 2 != 0) { return entry.Key; } } return -1; } // Driver code public static void Main(String[] args) { int []arr = new int []{2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2}; int n = arr.Length; Console.WriteLine(getOddOccurrence(arr, n)); } } // This code is contributed by Princi Singh |
Output :
5
The Best Solution is to do bitwise XOR of all the elements. XOR of all elements gives us odd occurring element. Please note that XOR of two elements is 0 if both elements are same and XOR of a number x with 0 is x.
Below is the implementation of the above approach.
C++
// C++ program to find the element // occurring odd number of times #include <bits/stdc++.h> using namespace std; // Function to find element occurring // odd number of times int getOddOccurrence( int ar[], int ar_size) { int res = 0; for ( int i = 0; i < ar_size; i++) res = res ^ ar[i]; return res; } /* Driver function to test above function */ int main() { int ar[] = {2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2}; int n = sizeof (ar)/ sizeof (ar[0]); // Function calling cout << getOddOccurrence(ar, n); return 0; } |
C
// C program to find the element // occurring odd number of times #include <stdio.h> // Function to find element occurring // odd number of times int getOddOccurrence( int ar[], int ar_size) { int res = 0; for ( int i = 0; i < ar_size; i++) res = res ^ ar[i]; return res; } /* Driver function to test above function */ int main() { int ar[] = {2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2}; int n = sizeof (ar) / sizeof (ar[0]); // Function calling printf ( "%d" , getOddOccurrence(ar, n)); return 0; } |
Java
//Java program to find the element occurring odd number of times class OddOccurance { int getOddOccurrence( int ar[], int ar_size) { int i; int res = 0 ; for (i = 0 ; i < ar_size; i++) { res = res ^ ar[i]; } return res; } public static void main(String[] args) { OddOccurance occur = new OddOccurance(); int ar[] = new int []{ 2 , 3 , 5 , 4 , 5 , 2 , 4 , 3 , 5 , 2 , 4 , 4 , 2 }; int n = ar.length; System.out.println(occur.getOddOccurrence(ar, n)); } } // This code has been contributed by Mayank Jaiswal |
Python3
# Python program to find the element occurring odd number of times def getOddOccurrence(arr): # Initialize result res = 0 # Traverse the array for element in arr: # XOR with the result res = res ^ element return res # Test array arr = [ 2 , 3 , 5 , 4 , 5 , 2 , 4 , 3 , 5 , 2 , 4 , 4 , 2 ] print ( "%d" % getOddOccurrence(arr)) |
C#
// C# program to find the element // occurring odd number of times using System; class GFG { // Function to find the element // occurring odd number of times static int getOddOccurrence( int []arr, int arr_size) { int res = 0; for ( int i = 0; i < arr_size; i++) { res = res ^ arr[i]; } return res; } // Driver code public static void Main() { int []arr = { 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 }; int n = arr.Length; Console.Write(getOddOccurrence(arr, n)); } } // This code is contributed by Sam007 |
PHP
<?php // PHP program to find the // element occurring odd // number of times // Function to find element // occurring odd number of times function getOddOccurrence(& $ar , $ar_size ) { $res = 0; for ( $i = 0; $i < $ar_size ; $i ++) $res = $res ^ $ar [ $i ]; return $res ; } // Driver Code $ar = array (2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2); $n = sizeof( $ar ); // Function calling echo (getOddOccurrence( $ar , $n )); // This code is contributed // by Shivi_Aggarwal ?> |
Javascript
<script> // JavaScript program to find the element // occurring odd number of times // Function to find the element // occurring odd number of times function getOddOccurrence( ar, ar_size) { let res = 0; for (let i = 0; i < ar_size; i++) res = res ^ ar[i]; return res; } // driver code let arr = [ 2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2 ]; let n = arr.length; // Function calling document.write(getOddOccurrence(arr, n)); </script> |
Output :
5
Time Complexity: O(n)
Method 3:Using Built in python functions:
- Count the frequencies of every element using Counter function
- Traverse in frequency dictionary
- Check which element has a odd frequency.
- Print that element and break the loop
Below is the implementation:
Python3
# importing counter from colleections from collections import Counter # Python3 implementation to find # odd frequeency element def oddElement(arr, n): # Calculating freequencies usinf Counter count_map = Counter(arr) for i in range ( 0 , n): # If count of element is odd we return if (count_map[arr[i]] % 2 ! = 0 ): return arr[i] # Driver Code if __name__ = = "__main__" : arr = [ 1 , 1 , 3 , 3 , 5 , 6 , 6 ] n = len (arr) print (oddElement(arr, n)) # This code is contributed by vikkycirus |
Output:
5
Time Complexity: O(N)
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