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 :
C++
// 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 codeint 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
// 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 codeint 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
// Java program to find the maximum stolen valuepublic 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
# python3 code for the above approach# Function to find and Replace in Stringdef 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 codeif __name__ == "__main__" : arr = [ 2, 3, 3, 4, 4, 5 ] n = len(arr) #Function call printOdds(arr,n)# This code is contributed by adityapatil12 |
C#
// Include namespace systemusing System;// C# program to find the maximum stolen valuepublic 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
// 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 codelet arr = [ 2, 3, 3, 4, 4, 5 ];let n = arr.length;// function callprintOdds(arr, n);// This code is contributed by phasing17 |
Output
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:
C++
// 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 codeint 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 |
Java
/*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. |
Python3
# Python code to find two odd occurring elements# in an array where all other elements appear# even number of times.from collections import defaultdictdef 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 codearr = [2, 3, 3, 4, 4, 5]n = len(arr)# function callprintOdds(arr, n)# this code is contributed by akashish__ |
C#
// 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. |
Javascript
// 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. |
Output
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:
C++
// 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 codeint main(){ int arr[] = { 2, 3, 3, 4, 4, 5 }; int n = sizeof(arr) / sizeof(arr[0]); printOdds(arr, n); return 0;} |
Java
// 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 codepublic 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 |
Python3
# 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 codearr = [2, 3, 3, 4, 4, 5 ]n = len(arr) printOdds(arr, n)# This code is contributed# by Smitha |
C#
// 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 codepublic 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// 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?> |
Javascript
<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 codelet arr = [ 2, 3, 3, 4, 4, 5 ];let n = arr.length;printOdds(arr, n);// This code is contributed by subhammahato348</script> |
Output
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++
// 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 codeint 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 |
Python3
# Function to find element that occurs# odd times in the arraydef 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 codearr = [2, 3, 3, 4, 4, 5]n = len(arr)# Function callprintOdds(arr, n) |
Javascript
// Function to find element that occurs odd times in the arrayfunction 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 codelet arr = [2, 3, 3, 4, 4, 5];let n = arr.length;// Function callprintOdds(arr, n); |
C#
// C# Equivalentusing 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); }} |
Java
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 }} |
Output
2 5
Time Complexity: O(n*log2n), take log2n time for binary search function
Auxiliary Space: O(1)
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