Dada una array de enteros positivos. Todos los números ocurren un número par de veces excepto un número que ocurre un número impar de veces. Encuentra el número en tiempo O(n) y espacio constante.
Ejemplos:
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>
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>
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>
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
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA