Dada una array de n enteros. La tarea es imprimir los duplicados en la array dada. Si no hay duplicados, imprima -1.
Ejemplos:
Input: {2, 10,10, 100, 2, 10, 11,2,11,2}
Output: 2 10 11
Input: {5, 40, 1, 40, 100000, 1, 5, 1}
Output: 5 40 1
Nota: Los elementos duplicados se pueden imprimir en cualquier orden.
Enfoque simple : la idea es usar un bucle anidado y, para cada elemento, verificar si el elemento está presente en la array más de una vez o no. Si está presente, guárdelo en un Hash-map. De lo contrario, continúe revisando otros elementos.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the
// above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the Duplicates,
// if duplicate occurs 2 times or
// more than 2 times in array so,
// it will print duplicate value
// only once at output
void findDuplicates(int arr[], int len)
{
// Initialize ifPresent as false
bool ifPresent = false;
// ArrayList to store the output
vector<int> al;
for(int i = 0; i < len - 1; i++)
{
for(int j = i + 1; j < len; j++)
{
if (arr[i] == arr[j])
{
// Checking if element is
// present in the ArrayList
// or not if present then break
auto it = std::find(al.begin(),
al.end(), arr[i]);
if (it != al.end())
{
break;
}
// If element is not present in the
// ArrayList then add it to ArrayList
// and make ifPresent at true
else
{
al.push_back(arr[i]);
ifPresent = true;
}
}
}
}
// If duplicates is present
// then print ArrayList
if (ifPresent == true)
{
cout << "[" << al[0] << ", ";
for(int i = 1; i < al.size() - 1; i++)
{
cout << al[i] << ", ";
}
cout << al[al.size() - 1] << "]";
}
else
{
cout << "No duplicates present in arrays";
}
}
// Driver code
int main()
{
int arr[] = { 12, 11, 40, 12,
5, 6, 5, 12, 11 };
int n = sizeof(arr) / sizeof(arr[0]);
findDuplicates(arr, n);
return 0;
}
// This code is contributed by divyeshrabadiya07
Java
// Java implementation of the
// above approach
import java.util.ArrayList;
public class GFG {
// Function to find the Duplicates,
// if duplicate occurs 2 times or
// more than 2 times in
// array so, it will print duplicate
// value only once at output
static void findDuplicates(
int arr[], int len)
{
// initialize ifPresent as false
boolean ifPresent = false;
// ArrayList to store the output
ArrayList<Integer> al = new ArrayList<Integer>();
for (int i = 0; i < len - 1; i++) {
for (int j = i + 1; j < len; j++) {
if (arr[i] == arr[j]) {
// checking if element is
// present in the ArrayList
// or not if present then break
if (al.contains(arr[i])) {
break;
}
// if element is not present in the
// ArrayList then add it to ArrayList
// and make ifPresent at true
else {
al.add(arr[i]);
ifPresent = true;
}
}
}
}
// if duplicates is present
// then print ArrayList
if (ifPresent == true) {
System.out.print(al + " ");
}
else {
System.out.print(
"No duplicates present in arrays");
}
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 12, 11, 40, 12, 5, 6, 5, 12, 11 };
int n = arr.length;
findDuplicates(arr, n);
}
}
Python3
# Python3 implementation of the
# above approach
# Function to find the Duplicates,
# if duplicate occurs 2 times or
# more than 2 times in array so,
# it will print duplicate value
# only once at output
def findDuplicates(arr, Len):
# Initialize ifPresent as false
ifPresent = False
# ArrayList to store the output
a1 = []
for i in range(Len - 1):
for j in range(i + 1, Len):
# Checking if element is
# present in the ArrayList
# or not if present then break
if (arr[i] == arr[j]):
if arr[i] in a1:
break
# If element is not present in the
# ArrayList then add it to ArrayList
# and make ifPresent at true
else:
a1.append(arr[i])
ifPresent = True
# If duplicates is present
# then print ArrayList
if (ifPresent):
print(a1, end = " ")
else:
print("No duplicates present in arrays")
# Driver Code
arr = [ 12, 11, 40, 12, 5, 6, 5, 12, 11 ]
n = len(arr)
findDuplicates(arr, n)
# This code is contributed by rag2127
C#
// C# implementation of the
// above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the Duplicates,
// if duplicate occurs 2 times or
// more than 2 times in array so,
// it will print duplicate value
// only once at output
static void findDuplicates(int[] arr, int len)
{
// Initialize ifPresent as false
bool ifPresent = false;
// ArrayList to store the output
List<int> al = new List<int>();
for(int i = 0; i < len - 1; i++)
{
for(int j = i + 1; j < len; j++)
{
if (arr[i] == arr[j])
{
// Checking if element is
// present in the ArrayList
// or not if present then break
if (al.Contains(arr[i]))
{
break;
}
// If element is not present in the
// ArrayList then add it to ArrayList
// and make ifPresent at true
else
{
al.Add(arr[i]);
ifPresent = true;
}
}
}
}
// If duplicates is present
// then print ArrayList
if (ifPresent == true)
{
Console.Write("[" + al[0] + ", ");
for(int i = 1; i < al.Count - 1; i++)
{
Console.Write(al[i] + ", ");
}
Console.Write(al[al.Count - 1] + "]");
}
else
{
Console.Write("No duplicates present in arrays");
}
}
// Driver code
static void Main()
{
int[] arr = { 12, 11, 40, 12,
5, 6, 5, 12, 11 };
int n = arr.Length;
findDuplicates(arr, n);
}
}
// This code is contributed by divyesh072019
Javascript
<script>
// JavaScript implementation of the
// above approach
// Function to find the Duplicates,
// if duplicate occurs 2 times or
// more than 2 times in
// array so, it will print duplicate
// value only once at output
function findDuplicates(arr, len) {
// initialize ifPresent as false
let ifPresent = false;
// ArrayList to store the output
let al = new Array();
for (let i = 0; i < len - 1; i++) {
for (let j = i + 1; j < len; j++) {
if (arr[i] == arr[j]) {
// checking if element is
// present in the ArrayList
// or not if present then break
if (al.includes(arr[i])) {
break;
}
// if element is not present in the
// ArrayList then add it to ArrayList
// and make ifPresent at true
else {
al.push(arr[i]);
ifPresent = true;
}
}
}
}
// if duplicates is present
// then print ArrayList
if (ifPresent == true) {
document.write(`[${al}]`);
}
else {
document.write("No duplicates present in arrays");
}
}
// Driver Code
let arr = [12, 11, 40, 12, 5, 6, 5, 12, 11];
let n = arr.length;
findDuplicates(arr, n);
</script>
Producción
[12, 11, 5]
Tiempo Complejidad: O(N 2 )
Espacio Auxiliar: O(N)
Enfoque eficiente : use unordered_map para hash. Cuente la frecuencia de aparición de cada elemento y se imprimen los elementos con una frecuencia superior a 1. se utiliza unordered_map ya que no se conoce el rango de enteros. Para Python, use el diccionario para almacenar el número como clave y su frecuencia como valor. Se puede usar el diccionario ya que no se conoce el rango de números enteros.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find
// duplicates in the given array
#include <bits/stdc++.h>
using namespace std;
// function to find and print duplicates
void printDuplicates(int arr[], int n)
{
// unordered_map to store frequencies
unordered_map<int, int> freq;
for (int i=0; i<n; i++)
freq[arr[i]]++;
bool dup = false;
unordered_map<int, int>:: iterator itr;
for (itr=freq.begin(); itr!=freq.end(); itr++)
{
// if frequency is more than 1
// print the element
if (itr->second > 1)
{
cout << itr->first << " ";
dup = true;
}
}
// no duplicates present
if (dup == false)
cout << "-1";
}
// Driver program to test above
int main()
{
int arr[] = {12, 11, 40, 12, 5, 6, 5, 12, 11};
int n = sizeof(arr) / sizeof(arr[0]);
printDuplicates(arr, n);
return 0;
}
Java
// Java program to find
// duplicates in the given array
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
public class FindDuplicatedInArray
{
// Driver program
public static void main(String[] args)
{
int arr[] = {12, 11, 40, 12, 5, 6, 5, 12, 11};
int n = arr.length;
printDuplicates(arr, n);
}
// function to find and print duplicates
private static void printDuplicates(int[] arr, int n)
{
Map<Integer,Integer> map = new HashMap<>();
int count = 0;
boolean dup = false;
for(int i = 0; i < n; i++){
if(map.containsKey(arr[i])){
count = map.get(arr[i]);
map.put(arr[i], count + 1);
}
else{
map.put(arr[i], 1);
}
}
for(Entry<Integer,Integer> entry : map.entrySet())
{
// if frequency is more than 1
// print the element
if(entry.getValue() > 1){
System.out.print(entry.getKey()+ " ");
dup = true;
}
}
// no duplicates present
if(!dup){
System.out.println("-1");
}
}
}
Python3
# Python3 program to find duplicates
# using dictionary approach.
def printDuplicates(arr):
dict = {}
for ele in arr:
try:
dict[ele] += 1
except:
dict[ele] = 1
for item in dict:
# if frequency is more than 1
# print the element
if(dict[item] > 1):
print(item, end=" ")
print("\n")
# Driver Code
if __name__ == "__main__":
list = [12, 11, 40, 12,
5, 6, 5, 12, 11]
printDuplicates(list)
# This code is contributed
# by Sushil Bhile
C#
// C# program to find
// duplicates in the given array
using System;
using System.Collections.Generic;
class GFG
{
// function to find and print duplicates
static void printDuplicates(int[] arr, int n)
{
Dictionary<int,
int> map = new Dictionary<int,
int>();
int count = 0;
bool dup = false;
for (int i = 0; i < n; i++)
{
if (map.ContainsKey(arr[i]))
{
count = map[arr[i]];
map[arr[i]]++;
}
else
map.Add(arr[i], 1);
}
foreach (KeyValuePair<int,
int> entry in map)
{
// if frequency is more than 1
// print the element
if (entry.Value > 1)
Console.Write(entry.Key + " ");
dup = true;
}
// no duplicates present
if (!dup)
Console.WriteLine("-1");
}
// Driver Code
public static void Main(String[] args)
{
int[] arr = { 12, 11, 40, 12,
5, 6, 5, 12, 11 };
int n = arr.Length;
printDuplicates(arr, n);
}
}
// This code is contributed by
// sanjeev2552
Javascript
<script>
// JavaScript program to find
// duplicates in the given array
// Function to find and print duplicates
function printDuplicates(arr, n)
{
// unordered_map to store frequencies
var freq = new Map();
for(let i = 0; i < n; i++)
{
if (freq.has(arr[i]))
{
freq.set(arr[i], freq.get(arr[i]) + 1);
}
else
{
freq.set(arr[i], 1);
}
}
var dup = false;
for(let [key, value] of freq)
{
// If frequency is more than 1
// print the element
if (value > 1)
{
document.write(key + " ");
dup = true;
}
}
// No duplicates present
if (dup == false)
document.write("-1");
}
// Driver code
var arr = [ 12, 11, 40, 12,
5, 6, 5, 12, 11 ];
var n = arr.length;
printDuplicates(arr, n);
// This code is contributed by lokeshpotta20
</script>
Producción
5 12 11
Complejidad temporal: O(N)
Espacio auxiliar: O(N)
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA