Find the Number Occurring Odd Number of Times
Given an array of positive integers. All numbers occur an even number of times except one number which occurs an 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 : 5A Simple Solution is to run two nested loops. The outer loop picks all elements one by one and the inner loop counts the number of occurrences of the element picked by the outer loop. The 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
Time Complexity: O(n^2)
Auxiliary Space: O(1)
A Better Solution is to use Hashing. Use array elements as a key and their counts as values. Create an empty hash table. One by one traverse the given array elements and store counts. The 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
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,n)
{
let hmap = new Map();
// Putting all elements into the HashMap
for(let i = 0; i < n; i++)
{
if(hmap.has(arr[i]))
{
let val = hmap.get(arr[i]);
// If array element is already present then
// increase the count of that element.
hmap.set(arr[i], val + 1);
}
else
{
// if array element is not present then put
// element into the HashMap and initialize
// the count to one.
hmap.set(arr[i], 1);
}
}
// Checking for odd occurrence of each element present
// in the HashMap
for(let [key, value] of hmap.entries())
{
//document.write(hmap[a]+"<br>")
if(hmap.get(key) % 2 != 0)
return key;
}
return -1;
}
// driver code
let arr=[2, 3, 5, 4, 5, 2, 4, 3, 5, 2, 4, 4, 2];
let n = arr.length;
document.write(getOddOccurrence(arr, n));
// This code is contributed by unknown2108
</script>Output :
5
Time Complexity: O(n)
Auxiliary Space: O(n)
The Best Solution is to do bitwise XOR of all the elements. XOR of all elements gives us odd occurring elements. Please note that the XOR of two elements is 0 if both elements are the same and the 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)
Auxiliary Space: O(1)
Method 3:Using Built-in Python functions:
- Count the frequencies of every element using the Counter function
- Traverse in frequency dictionary
- Check which element has an odd frequency.
- Print that element and break the loop
Below is the implementation:
Python3
# importing counter from collections
from collections import Counter
# Python3 implementation to find
# odd frequency element
def oddElement(arr, n):
# Calculating frequencies using 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)
Auxiliary Space: O(1)
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