Given an array where all elements appear even number of times except two, print the two odd occurring elements. It may be assumed that the size of array is at-least two.
Examples:
Input : arr[] = {2, 3, 8, 4, 4, 3, 7, 8} Output : 2 7 Input : arr[] = {15, 10, 10, 50 7, 5, 5, 50, 50, 50, 50, 50} Output : 7 15
Simple solution :
Approach :
A simple solution is to use two nested loops. The outer loop traverses through all elements. The inner loop counts occurrences of the current element. We print the elements whose counts of occurrences are odd.
Below is the code of the given approach :
// CPP code to find two odd occurring elements // in an array where all other elements appear // even number of times. #include <bits/stdc++.h> using namespace std;
void printOdds( int arr[], int n)
{ for ( int i = 0; i < n; i++) {
int count = 0;
for ( int j = 0; j < n; j++) {
if (arr[i] == arr[j]) {
count += 1;
}
}
if (count % 2 != 0) {
cout << arr[i]
<< " " ; // Print the elements that occur
// odd number of times in an array
}
}
cout << endl;
} // Driver code int main()
{ int arr[] = { 2, 3, 3, 4, 4, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
// function call
printOdds(arr, n);
return 0;
} // This code is contributed by Suruchi Kumari |
// C code to find two odd occurring elements // in an array where all other elements appear // even number of times. #include <stdio.h> void printOdds( int arr[], int n)
{ for ( int i = 0; i < n; i++) {
int count = 0;
for ( int j = 0; j < n; j++) {
if (arr[i] == arr[j]) {
count += 1;
}
}
if (count % 2 != 0) {
printf (
"%d " ,
arr[i]); // Print the elements that occur
// odd number of times in an array
}
}
printf ( "\n" );
} // Driver code int main()
{ int arr[] = { 2, 3, 3, 4, 4, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
// function call
printOdds(arr, n);
return 0;
} // This code is contributed by Suruchi Kumari |
// Java program to find the maximum stolen value public class GFG {
// calculate the maximum stolen value
static void printOdds( int [] arr, int n)
{
for ( int i = 0 ; i < n; i++) {
int count = 0 ;
for ( int j = 0 ; j < n; j++) {
if (arr[i] == arr[j]) {
count += 1 ;
}
}
if (count % 2 != 0 ) {
System.out.print(arr[i]+ " " ); // Print the elements that occur
// odd number of times in an array
}
}
System.out.print( "\n" );
}
// Driver Code
public static void main (String[] args)
{
int arr[] = { 2 , 3 , 3 , 4 , 4 , 5 };
int n = arr.length;
// function call
printOdds(arr, n);
}
} // This code is contributed by adityapatil12 |
# python3 code for the above approach # Function to find and Replace in String def printOdds(arr, n) :
for i in range ( 0 ,n) :
count = 0
for j in range ( 0 ,n) :
if arr[i] = = arr[j] :
count + = 1
if count % 2 ! = 0 :
print (arr[i],end = ' ' ) # Print the elements that occur
# odd number of times in an array
# Driver code if __name__ = = "__main__" :
arr = [ 2 , 3 , 3 , 4 , 4 , 5 ]
n = len (arr)
#Function call
printOdds(arr,n)
# This code is contributed by adityapatil12 |
// Include namespace system using System;
// C# program to find the maximum stolen value public class GFG
{ // calculate the maximum stolen value
public static void printOdds( int [] arr, int n)
{
for ( int i = 0; i < n; i++)
{
var count = 0;
for ( int j = 0; j < n; j++)
{
if (arr[i] == arr[j])
{
count += 1;
}
}
if (count % 2 != 0)
{
Console.Write(arr[i].ToString() + " " );
}
}
Console.Write( "\n" );
}
// Driver Code
public static void Main(String[] args)
{
int [] arr = {2, 3, 3, 4, 4, 5};
var n = arr.Length;
// function call
GFG.printOdds(arr, n);
}
} // This code is contributed by aadityaburujwale. |
// JavaScript code to find two odd occurring elements // in an array where all other elements appear // even number of times. function printOdds(arr, n)
{ for ( var i = 0; i < n; i++) {
let count = 0;
for ( var j = 0; j < n; j++) {
if (arr[i] == arr[j]) {
count += 1;
}
}
if (count % 2 != 0) {
process.stdout.write(arr[i] + " " );
// Print the elements that occur
// odd number of times in an array
}
}
process.stdout.write( "\n" );
} // Driver code let arr = [ 2, 3, 3, 4, 4, 5 ]; let n = arr.length; // function call printOdds(arr, n); // This code is contributed by phasing17 |
2 5
Complexity Analysis:
- Time complexity : O(n^2)
- Auxiliary space : O(n)
A better solution is to use hashing. Time complexity of this solution is O(n) but it requires extra space.
We can construct a frequency hashmap, then iterate over all of its key – value pairs, and print all keys whose values (frequency) are odd.
Implementation:
// CPP code to find two odd occurring elements // in an array where all other elements appear // even number of times. #include <bits/stdc++.h> using namespace std;
void printOdds( int arr[], int n)
{ //declaring an unordered map
unordered_map< int , int > freq;
//building the frequency table
for ( int i = 0; i < n; i++)
freq[arr[i]]++;
//iterating over the map
for ( auto & it: freq) {
//if the frequency is odd
//print the element
if (it.second % 2)
cout << it.first << " " ;
}
} // Driver code int main()
{ int arr[] = { 2, 3, 3, 4, 4, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
//function call
printOdds(arr, n);
return 0;
} //this code is contributed by phasing17 |
/*package whatever //do not write package name here */ import java.util.*;
class GFG {
static void printOdds( int arr[], int n)
{
// declaring an unordered map
TreeMap<Integer, Integer> freq = new TreeMap<>(Collections.reverseOrder());
// building the frequency table
for ( int i = 0 ; i < n; i++) {
freq.put(arr[i],
freq.getOrDefault(arr[i], 0 ) + 1 );
}
// iterating over the map
for ( int i : freq.keySet()) {
// if the frequency is odd
// print the element
if (freq.get(i) % 2 == 1 )
System.out.print(i + " " );
}
}
public static void main(String[] args)
{
int arr[] = { 2 , 3 , 3 , 4 , 4 , 5 };
int n = arr.length;
// function call
printOdds(arr, n);
}
} // This code is contributed by aadityaburujwale. |
# Python code to find two odd occurring elements # in an array where all other elements appear # even number of times. from collections import defaultdict
def printOdds(arr, n):
freq = defaultdict( int )
# building the frequency table
for i in range (n):
freq[arr[i]] + = 1
# iterating over the map
for key, value in freq.items():
# if the frequency is odd
# print the element
if value % 2 :
print (key,end = " " )
# Driver code arr = [ 2 , 3 , 3 , 4 , 4 , 5 ]
n = len (arr)
# function call printOdds(arr, n) # this code is contributed by akashish__ |
// C# code to find two odd occurring elements // in an array where all other elements appear // even number of times. using System;
using System.Collections.Generic;
public class GFG {
static void printOdds( int [] arr, int n)
{
// declaring an unordered map
Dictionary< int , int > freq
= new Dictionary< int , int >();
// building the frequency table
for ( int i = 0; i < n; i++) {
if (freq.ContainsKey(arr[i])) {
int val = freq[arr[i]];
freq.Remove(arr[i]);
freq.Add(arr[i], val + 1);
}
else {
freq.Add(arr[i], 1);
}
}
// iterating over the map
foreach (KeyValuePair< int , int > entry in freq)
{
// if the frequency is odd
// print the element
if (entry.Value % 2 != 0)
Console.Write(entry.Key + " " );
}
}
// Driver code
static public void Main()
{
int [] arr = { 2, 3, 3, 4, 4, 5 };
int n = 6;
// function call
printOdds(arr, n);
}
} // This code is contributed by garg28harsh. |
// function code to find two odd occurring elements // in an array where all other elements appear // even number of times. function printOdds(arr, n)
{ // declaring an unordered map
var mp = new Map();
// building the frequency table
for ( var i = 0; i < n; i++)
{
if (mp.has(arr[i]))
mp.set(arr[i], mp.get(arr[i])+1);
else
mp.set(arr[i], 1);
}
let el=[];
mp.forEach(( key,value) => {
if (key%2!=0)
{
el.push(value);
}
});
console.log(el);
} // Driver code let arr = [2, 3, 3, 4, 4, 5 ];
let n = arr.length;
// function call
printOdds(arr, n);
// This code is contributed by garg28harsh. |
5 2
Complexity Analysis:
- Time Complexity: O(n)
- Space Complexity: O(n)
An efficient solution is to use bitwise operators. The idea is based on approach used in two missing elements and two repeating elements.
Implementation:
// CPP code to find two odd occurring elements // in an array where all other elements appear // even number of times. #include <bits/stdc++.h> using namespace std;
void printOdds( int arr[], int n)
{ // Find XOR of all numbers
int res = 0;
for ( int i = 0; i < n; i++)
res = res ^ arr[i];
// Find a set bit in the XOR (We find
// rightmost set bit here)
int set_bit = res & (~(res - 1));
// Traverse through all numbers and
// divide them in two groups
// (i) Having set bit set at same
// position as the only set bit
// in set_bit
// (ii) Having 0 bit at same position
// as the only set bit in set_bit
int x = 0, y = 0;
for ( int i = 0; i < n; i++) {
if (arr[i] & set_bit)
x = x ^ arr[i];
else
y = y ^ arr[i];
}
// XOR of two different sets are our
// required numbers.
cout << x << " " << y;
} // Driver code int main()
{ int arr[] = { 2, 3, 3, 4, 4, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
printOdds(arr, n);
return 0;
} |
// Java code to find two // odd occurring elements // in an array where all // other elements appear // even number of times. class GFG
{ static void printOdds( int arr[],
int n)
{ // Find XOR of
// all numbers
int res = 0 ;
for ( int i = 0 ; i < n; i++)
res = res ^ arr[i];
// Find a set bit in the
// XOR (We find rightmost
// set bit here)
int set_bit = res &
(~(res - 1 ));
// Traverse through all
// numbers and divide them
// in two groups (i) Having
// set bit set at same position
// as the only set bit in
// set_bit (ii) Having 0 bit at
// same position as the only
// set bit in set_bit
int x = 0 , y = 0 ;
for ( int i = 0 ; i < n; i++)
{
if ((arr[i] & set_bit) != 0 )
x = x ^ arr[i];
else
y = y ^ arr[i];
}
// XOR of two different
// sets are our required
// numbers.
System.out.println( x + " " + y);
} // Driver code public static void main(String [] args)
{ int arr[] = { 2 , 3 , 3 ,
4 , 4 , 5 };
int n = arr.length;
printOdds(arr, n);
} } // This code is contributed by // Smitha Dinesh Semwal |
# Python 3 code to find two # odd occurring elements in # an array where all other # elements appear even number # of times. def printOdds(arr, n):
# Find XOR of all numbers
res = 0
for i in range ( 0 , n):
res = res ^ arr[i]
# Find a set bit in
# the XOR (We find
# rightmost set bit here)
set_bit = res & (~(res - 1 ))
# Traverse through all numbers
# and divide them in two groups
# (i) Having set bit set at
# same position as the only set
# bit in set_bit
# (ii) Having 0 bit at same
# position as the only set
# bit in set_bit
x = 0
y = 0
for i in range ( 0 , n):
if (arr[i] & set_bit):
x = x ^ arr[i]
else :
y = y ^ arr[i]
# XOR of two different
# sets are our
# required numbers.
print (x , y, end = "")
# Driver code arr = [ 2 , 3 , 3 , 4 , 4 , 5 ]
n = len (arr)
printOdds(arr, n) # This code is contributed # by Smitha |
// C# code to find two // odd occurring elements // in an array where all // other elements appear // even number of times. using System;
class GFG
{ static void printOdds( int []arr,
int n)
{ // Find XOR of
// all numbers
int res = 0;
for ( int i = 0; i < n; i++)
res = res ^ arr[i];
// Find a set bit in the
// XOR (We find rightmost
// set bit here)
int set_bit = res &
(~(res - 1));
// Traverse through all
// numbers and divide them
// in two groups (i) Having
// set bit set at same position
// as the only set bit in
// set_bit (ii) Having 0 bit at
// same position as the only
// set bit in set_bit
int x = 0, y = 0;
for ( int i = 0; i < n; i++)
{
if ((arr[i] & set_bit) != 0)
x = x ^ arr[i];
else
y = y ^ arr[i];
}
// XOR of two different
// sets are our required
// numbers.
Console.WriteLine(x + " " + y);
} // Driver code public static void Main()
{ int []arr = { 2, 3, 3,
4, 4, 5 };
int n = arr.Length;
printOdds(arr, n);
} } // This code is contributed by // Akanksha Rai(Abby_akku) |
<?php // PHP code to find two odd // occurring elements in an // array where all other elements // appear even number of times. function printOdds( $arr , $n )
{ // Find XOR of all numbers
$res = 0;
for ( $i = 0; $i < $n ; $i ++)
$res = $res ^ $arr [ $i ];
// Find a set bit in the
// XOR (We find rightmost
// set bit here)
$set_bit = $res & (~( $res - 1));
// Traverse through all numbers
// and divide them in two groups
// (i) Having set bit set at same
// position as the only set bit
// in set_bit
// (ii) Having 0 bit at same position
// as the only set bit in set_bit
$x = 0;
$y = 0;
for ( $i = 0; $i < $n ; $i ++)
{
if ( $arr [ $i ] & $set_bit )
$x = $x ^ $arr [ $i ];
else
$y = $y ^ $arr [ $i ];
}
// XOR of two different sets
// are our required numbers.
echo ( $x . " " . $y );
} // Driver code $arr = array ( 2, 3, 3, 4, 4, 5 );
$n = sizeof( $arr );
printOdds( $arr , $n );
// This code is contributed by Smitha ?> |
<script> // Javascript code to find two odd // occurring elements in an array // where all other elements appear // even number of times. function printOdds(arr, n)
{ // Find XOR of all numbers
let res = 0;
for (let i = 0; i < n; i++)
res = res ^ arr[i];
// Find a set bit in the XOR (We find
// rightmost set bit here)
let set_bit = res & (~(res - 1));
// Traverse through all numbers and
// divide them in two groups
// (i) Having set bit set at same
// position as the only set bit
// in set_bit
// (ii) Having 0 bit at same position
// as the only set bit in set_bit
let x = 0, y = 0;
for (let i = 0; i < n; i++)
{
if (arr[i] & set_bit)
x = x ^ arr[i];
else
y = y ^ arr[i];
}
// XOR of two different sets are our
// required numbers.
document.write(x + " " + y);
} // Driver code let arr = [ 2, 3, 3, 4, 4, 5 ]; let n = arr.length; printOdds(arr, n); // This code is contributed by subhammahato348 </script> |
5 2
Complexity Analysis:
- Time Complexity : O(n)
- Auxiliary Space : O(1)
Another Efficient Solution (Using binary search) : Sort the array for binary search . Then we can find frequency of all array elements using binary search function . Then we can check if frequency of array element is odd or not , If frequency is odd , then print that element .
Below is the implementation of above approach:
// C++ implementation of the above approach #include <bits/stdc++.h> using namespace std;
// Function to find element that occurs // odd times in the array void printOdds( int arr[], int n)
{ int count = 0;
sort(arr,arr+n); //sort array for binary search
for ( int i = 0 ; i < n ;i++)
{
//index of first and last occ of arr[i]
int first_index = lower_bound(arr,arr+n,arr[i])- arr;
int last_index = upper_bound(arr,arr+n,arr[i])- arr-1;
i = last_index; // assign i to last_index to avoid printing
// same element multiple time
int fre = last_index-first_index+1; //finding frequency
//( occurences of arr[i] in the array )
if (fre % 2 != 0)
{ // if element occurs odd times then print that number
cout << arr[i]<< " " ;
}
}
} // Drive code int main()
{ int arr[] = { 2, 3, 3, 4, 4, 5 };
int n = sizeof (arr) / sizeof (arr[0]);
// Function call
printOdds(arr, n);
return 0;
} // This Code is contributed by nikhilsainiofficial546 |
# Function to find element that occurs # odd times in the array def printOdds(arr, n):
count = 0
arr.sort() # sort array for binary search
i = 0
while i < n:
# index of first and last occ of arr[i]
first_index = arr.index(arr[i])
last_index = len (arr) - arr[:: - 1 ].index(arr[i]) - 1
i = last_index # assign i to last_index to avoid printing
# same element multiple time
fre = last_index - first_index + 1 # finding frequency
# (occurrences of arr[i] in the array)
if fre % 2 ! = 0 :
# if element occurs odd times then print that number
print (arr[i], end = ' ' )
i + = 1
# Drive code arr = [ 2 , 3 , 3 , 4 , 4 , 5 ]
n = len (arr)
# Function call printOdds(arr, n) |
// Function to find element that occurs odd times in the array function printOdds(arr, n) {
let count = 0;
arr.sort(); // sort array for binary search
let i = 0; let ans= "" ;
while (i < n) {
// index of first and last occ of arr[i]
let first_index = arr.indexOf(arr[i]);
let last_index = arr.lastIndexOf(arr[i]);
i = last_index; // assign i to last_index to avoid printing same element multiple times
let fre = last_index - first_index + 1; // finding frequency (occurrences of arr[i] in the array)
if (fre % 2 !== 0) {
// if element occurs odd times then print that number
ans = ans + arr[i] + " " ;
}
i += 1;
}console.log(ans);
} // Drive code let arr = [2, 3, 3, 4, 4, 5]; let n = arr.length; // Function call printOdds(arr, n); |
// C# Equivalent using System;
public class GFG {
// Function to find element that occurs odd times in the
// array
static void printOdds( int [] arr, int n)
{
int count = 0;
Array.Sort(arr); // sort array for binary search
int i = 0;
string ans = "" ;
while (i < n) {
// index of first and last occ of arr[i]
int first_index = Array.IndexOf(arr, arr[i]);
int last_index = Array.LastIndexOf(arr, arr[i]);
i = last_index; // assign i to last_index to
// avoid printing same element
// multiple times
int fre = last_index - first_index
+ 1; // finding frequency (occurrences
// of arr[i] in the array)
if (fre % 2 != 0) {
// if element occurs odd times then print
// that number
ans = ans + arr[i] + " " ;
}
i += 1;
}
Console.WriteLine(ans);
}
public static void Main()
{
// Driver's code
int [] arr = new int [] { 2, 3, 3, 4, 4, 5 };
int n = arr.Length;
// Function call
printOdds(arr, n);
}
} |
import java.util.*;
class Main {
// Function to find element that occurs
// odd times in the array
static void printOdds( int [] arr, int n)
{
int count = 0 ;
Arrays.sort(arr); // sort array for binary search
for ( int i = 0 ; i < n; i++) {
// index of first and last occ of arr[i]
int first_index
= Arrays.binarySearch(arr, arr[i]);
int last_index = first_index;
// handling case where element is not found by
// binary search
if (first_index < 0 ) {
continue ;
}
// loop to find last index of arr[i] in the
// array
while (last_index + 1 < n
&& arr[last_index + 1 ] == arr[i]) {
last_index++;
}
i = last_index; // assign i to last_index to
// avoid printing same element
// multiple times
int fre = last_index - first_index
+ 1 ; // finding frequency
// (occurrences of arr[i] in the
// array)
if (fre % 2 != 0 ) {
// if element occurs odd times then print
// that number
System.out.print(arr[i] + " " );
}
}
}
// Drive code
public static void main(String[] args)
{
int [] arr = { 2 , 3 , 3 , 4 , 4 , 5 };
int n = arr.length;
// Function call
printOdds(arr, n); // Output: 2 5
}
} |
2 5
Time Complexity: O(n*log2n), take log2n time for binary search function
Auxiliary Space: O(1)