Dada una array de solo lectura de tamaño ( n+1 ), encuentre uno de los múltiples elementos repetidos en la array donde la array contiene números enteros solo entre 1 y n.
Una array de solo lectura significa que el contenido de la array no se puede modificar.
Ejemplos:
Input : n = 5
arr[] = {1, 1, 2, 3, 5, 4}
Output : One of the numbers repeated in the array is: 1
Input : n = 10
arr[] = {10, 1, 2, 3, 5, 4, 9, 8, 5, 6, 4}
Output : One of the numbers repeated in the array is: 4 OR 5
Método 1: dado que el tamaño de la array es n+1 y los elementos varían de 1 a n, se confirma que habrá al menos un elemento repetido. Una solución simple es crear una array de conteo y almacenar conteos de todos los elementos. Tan pronto como encontramos un elemento con un recuento de más de 1, lo devolvemos.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to find one of the repeating
// elements in a read only array
#include <bits/stdc++.h>
using namespace std;
// Function to find one of the repeating
// elements
int findRepeatingNumber(const 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++;
}
if (count > 1)
return arr[i];
}
// If no repeating element exists
return -1;
}
// Driver Code
int main()
{
// Read only array
const int arr[] = { 1, 1, 2, 3, 5, 4 };
int n = 5;
cout << "One of the numbers repeated in"
" the array is: "
<< findRepeatingNumber(arr, n) << endl;
}
Java
public class GFG {
// Java program to find one of the repeating
// elements in a read only array
// Function to find one of the repeating
// elements
public static int findRepeatingNumber(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++;
}
}
if (count > 1) {
return arr[i];
}
}
// If no repeating element exists
return -1;
}
// Driver Code
public static void main(String[] args)
{
// Read only array
final int[] arr = { 1, 1, 2, 3, 5, 4 };
int n = 5;
System.out.print("One of the numbers repeated in"
+ " the array is: ");
System.out.print(findRepeatingNumber(arr, n));
System.out.print("\n");
}
}
// This code is contributed by Aarti_Rathi
Python3
# Python 3program to find one of the repeating
# elements in a read only array
from math import sqrt
# Function to find one of the repeating
# elements
def findRepeatingNumber(arr, n):
for i in arr:
count = 0;
for j in arr:
if i == j:
count=count+1
if(count>1):
return i
# return -1 if no repeating element exists
return -1
# Driver Code
if __name__ == '__main__':
# read only array, not to be modified
arr = [1, 1, 2, 3, 5, 4]
# array of size 6(n + 1) having
# elements between 1 and 5
n = 5
print("One of the numbers repeated in the array is:",
findRepeatingNumber(arr, n))
# This code is contributed by Arpit Jain
C#
// C# program to find one of the repeating
// elements in a read only array
using System;
public class GFG {
// Function to find one of the repeating
// elements
public static int findRepeatingNumber(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++;
}
}
if (count > 1) {
return arr[i];
}
}
// If no repeating element exists
return -1;
}
// Driver Code
public static void Main(String[] args)
{
// Read only array
int[] arr = { 1, 1, 2, 3, 5, 4 };
int n = 5;
Console.Write("One of the numbers repeated in"
+ " the array is: ");
Console.Write(findRepeatingNumber(arr, n));
Console.Write("\n");
}
}
// This code is contributed by Abhijeet Kumar(abhijeet19403)
Producción
One of the numbers repeated in the array is: 1
Tiempo Complejidad: O(n 2 )
Espacio Auxiliar: O(1)
Una solución optimizada para el espacio es dividir el rango dado (de 1 a n) en bloques de tamaño igual a sqrt(n). Mantenemos el recuento de elementos pertenecientes a cada bloque para cada bloque. Ahora, como el tamaño de una array es (n+1) y los bloques son de tamaño sqrt(n), entonces habrá uno de esos bloques cuyo tamaño será mayor que sqrt(n). Para el bloque cuyo recuento es mayor que sqrt(n), podemos usar hash para los elementos de este bloque para encontrar qué elemento aparece más de una vez.
Explicación :
el método descrito anteriormente funciona por los dos motivos siguientes:
- Siempre habrá un bloque que tenga un recuento mayor que sqrt(n) debido a un elemento adicional. Incluso cuando se ha agregado un elemento adicional, ocupará una posición en uno de los bloques solamente, haciendo que ese bloque sea seleccionado.
- El bloque seleccionado definitivamente tiene un elemento repetitivo. Considere que el i -ésimo bloque está seleccionado. El tamaño del bloque es mayor que sqrt(n) (Por lo tanto, se selecciona) Máximo de elementos distintos en este bloque = sqrt(n). Por lo tanto, el tamaño puede ser mayor que sqrt(n) solo si hay un elemento repetido en el rango ( i*sqrt(n), (i+1)*sqrt(n) ].
Nota: El último bloque formado puede o no tener un rango igual a sqrt(n). Por lo tanto, verificar si este bloque tiene un elemento repetido será diferente a otros bloques. Sin embargo, esta dificultad se puede superar desde el punto de vista de la implementación inicializando el bloque seleccionado con el último bloque. Esto es seguro porque al menos un bloque tiene que ser seleccionado.
A continuación se muestra el algoritmo paso a paso para resolver este problema :
- Divida la array en bloques de tamaño sqrt(n).
- Cree una array de conteo que almacene el conteo de elementos para cada bloque.
- Elija el bloque que tenga un recuento de más de sqrt(n), configurando el último bloque
como predeterminado. - Para los elementos pertenecientes al bloque seleccionado, utilice el método de hashing (explicado en el siguiente paso) para encontrar el elemento repetido en ese bloque.
- Podemos crear una array hash de pares clave-valor, donde la clave es el elemento en el bloque y el valor es el recuento de la cantidad de veces que aparece la clave dada. Esto se puede implementar fácilmente usando unordered_map en C++ STL.
A continuación se muestra la implementación de la idea anterior:
C++
// C++ program to find one of the repeating
// elements in a read only array
#include <bits/stdc++.h>
using namespace std;
// Function to find one of the repeating
// elements
int findRepeatingNumber(const int arr[], int n)
{
// Size of blocks except the
// last block is sq
int sq = sqrt(n);
// Number of blocks to incorporate 1 to
// n values blocks are numbered from 0
// to range-1 (both included)
int range = (n / sq) + 1;
// Count array maintains the count for
// all blocks
int count[range] = {0};
// Traversing the read only array and
// updating count
for (int i = 0; i <= n; i++)
{
// arr[i] belongs to block number
// (arr[i]-1)/sq i is considered
// to start from 0
count[(arr[i] - 1) / sq]++;
}
// The selected_block is set to last
// block by default. Rest of the blocks
// are checked
int selected_block = range - 1;
for (int i = 0; i < range - 1; i++)
{
if (count[i] > sq)
{
selected_block = i;
break;
}
}
// after finding block with size > sq
// method of hashing is used to find
// the element repeating in this block
unordered_map<int, int> m;
for (int i = 0; i <= n; i++)
{
// checks if the element belongs to the
// selected_block
if ( ((selected_block * sq) < arr[i]) &&
(arr[i] <= ((selected_block + 1) * sq)))
{
m[arr[i]]++;
// repeating element found
if (m[arr[i]] > 1)
return arr[i];
}
}
// return -1 if no repeating element exists
return -1;
}
// Driver Program
int main()
{
// read only array, not to be modified
const int arr[] = { 1, 1, 2, 3, 5, 4 };
// array of size 6(n + 1) having
// elements between 1 and 5
int n = 5;
cout << "One of the numbers repeated in"
" the array is: "
<< findRepeatingNumber(arr, n) << endl;
}
Java
// Java program to find one of the repeating
// elements in a read only array
import java.io.*;
import java.util.*;
class GFG
{
// Function to find one of the repeating
// elements
static int findRepeatingNumber(int[] arr, int n)
{
// Size of blocks except the
// last block is sq
int sq = (int) Math.sqrt(n);
// Number of blocks to incorporate 1 to
// n values blocks are numbered from 0
// to range-1 (both included)
int range = (n / sq) + 1;
// Count array maintains the count for
// all blocks
int[] count = new int[range];
// Traversing the read only array and
// updating count
for (int i = 0; i <= n; i++)
{
// arr[i] belongs to block number
// (arr[i]-1)/sq i is considered
// to start from 0
count[(arr[i] - 1) / sq]++;
}
// The selected_block is set to last
// block by default. Rest of the blocks
// are checked
int selected_block = range - 1;
for (int i = 0; i < range - 1; i++)
{
if (count[i] > sq)
{
selected_block = i;
break;
}
}
// after finding block with size > sq
// method of hashing is used to find
// the element repeating in this block
HashMap<Integer, Integer> m = new HashMap<>();
for (int i = 0; i <= n; i++)
{
// checks if the element belongs to the
// selected_block
if ( ((selected_block * sq) < arr[i]) &&
(arr[i] <= ((selected_block + 1) * sq)))
{
m.put(arr[i], 1);
// repeating element found
if (m.get(arr[i]) == 1)
return arr[i];
}
}
// return -1 if no repeating element exists
return -1;
}
// Driver code
public static void main(String args[])
{
// read only array, not to be modified
int[] arr = { 1, 1, 2, 3, 5, 4 };
// array of size 6(n + 1) having
// elements between 1 and 5
int n = 5;
System.out.println("One of the numbers repeated in the array is: " +
findRepeatingNumber(arr, n));
}
}
// This code is contributed by rachana soma
Python3
# Python 3program to find one of the repeating
# elements in a read only array
from math import sqrt
# Function to find one of the repeating
# elements
def findRepeatingNumber(arr, n):
# Size of blocks except the
# last block is sq
sq = sqrt(n)
# Number of blocks to incorporate 1 to
# n values blocks are numbered from 0
# to range-1 (both included)
range__= int((n / sq) + 1)
# Count array maintains the count for
# all blocks
count = [0 for i in range(range__)]
# Traversing the read only array and
# updating count
for i in range(0, n + 1, 1):
# arr[i] belongs to block number
# (arr[i]-1)/sq i is considered
# to start from 0
count[int((arr[i] - 1) / sq)] += 1
# The selected_block is set to last
# block by default. Rest of the blocks
# are checked
selected_block = range__ - 1
for i in range(0, range__ - 1, 1):
if (count[i] > sq):
selected_block = i
break
# after finding block with size > sq
# method of hashing is used to find
# the element repeating in this block
m = {i:0 for i in range(n)}
for i in range(0, n + 1, 1):
# checks if the element belongs
# to the selected_block
if (((selected_block * sq) < arr[i]) and
(arr[i] <= ((selected_block + 1) * sq))):
m[arr[i]] += 1
# repeating element found
if (m[arr[i]] > 1):
return arr[i]
# return -1 if no repeating element exists
return -1
# Driver Code
if __name__ == '__main__':
# read only array, not to be modified
arr = [1, 1, 2, 3, 5, 4]
# array of size 6(n + 1) having
# elements between 1 and 5
n = 5
print("One of the numbers repeated in the array is:",
findRepeatingNumber(arr, n))
# This code is contributed by
# Sahil_shelangia
C#
// C# program to find one of the repeating
// elements in a read only array
using System;
using System.Collections.Generic;
class GFG
{
// Function to find one of the repeating
// elements
static int findRepeatingNumber(int[] arr, int n)
{
// Size of blocks except the
// last block is sq
int sq = (int) Math.Sqrt(n);
// Number of blocks to incorporate 1 to
// n values blocks are numbered from 0
// to range-1 (both included)
int range = (n / sq) + 1;
// Count array maintains the count for
// all blocks
int[] count = new int[range];
// Traversing the read only array and
// updating count
for (int i = 0; i <= n; i++)
{
// arr[i] belongs to block number
// (arr[i]-1)/sq i is considered
// to start from 0
count[(arr[i] - 1) / sq]++;
}
// The selected_block is set to last
// block by default. Rest of the blocks
// are checked
int selected_block = range - 1;
for (int i = 0; i < range - 1; i++)
{
if (count[i] > sq)
{
selected_block = i;
break;
}
}
// after finding block with size > sq
// method of hashing is used to find
// the element repeating in this block
Dictionary<int,int> m = new Dictionary<int,int>();
for (int i = 0; i <= n; i++)
{
// checks if the element belongs to the
// selected_block
if ( ((selected_block * sq) < arr[i]) &&
(arr[i] <= ((selected_block + 1) * sq)))
{
m.Add(arr[i], 1);
// repeating element found
if (m[arr[i]] == 1)
return arr[i];
}
}
// return -1 if no repeating element exists
return -1;
}
// Driver code
public static void Main(String []args)
{
// read only array, not to be modified
int[] arr = { 1, 1, 2, 3, 5, 4 };
// array of size 6(n + 1) having
// elements between 1 and 5
int n = 5;
Console.WriteLine("One of the numbers repeated in the array is: " +
findRepeatingNumber(arr, n));
}
}
// This code contributed by Rajput-Ji
Javascript
<script>
// JavaScript program to find one of the repeating
// elements in a read only array
// Function to find one of the repeating
// elements
function findRepeatingNumber(arr, n) {
// Size of blocks except the
// last block is sq
let sq = Math.sqrt(n);
// Number of blocks to incorporate 1 to
// n values blocks are numbered from 0
// to range-1 (both included)
let range = Math.floor(n / sq) + 1;
// Count array maintains the count for
// all blocks
let count = new Array(range).fill(0);
// Traversing the read only array and
// updating count
for (let i = 0; i <= n; i++) {
// arr[i] belongs to block number
// (arr[i]-1)/sq i is considered
// to start from 0
count[Math.floor((arr[i] - 1) / sq)]++;
}
// The selected_block is set to last
// block by default. Rest of the blocks
// are checked
let selected_block = range - 1;
for (let i = 0; i < range - 1; i++) {
if (count[i] > sq) {
selected_block = i;
break;
}
}
// after finding block with size > sq
// method of hashing is used to find
// the element repeating in this block
let m = new Map();
for (let i = 0; i <= n; i++) {
// checks if the element belongs to the
// selected_block
if (((selected_block * sq) < arr[i]) &&
(arr[i] <= ((selected_block + 1) * sq))) {
m[arr[i]]++;
if (m.has(arr[i])) {
m.set(arr[i], m.get(arr[i]) + 1)
} else {
m.set(arr[i], 1)
}
// repeating element found
if (m.get(arr[i]) > 1)
return arr[i];
}
}
// return -1 if no repeating element exists
return -1;
}
// Driver Program
// read only array, not to be modified
const arr = [1, 1, 2, 3, 5, 4];
// array of size 6(n + 1) having
// elements between 1 and 5
let n = 5;
document.write("One of the numbers repeated in"
+ " the array is: "
+ findRepeatingNumber(arr, n) + "<br>");
</script>
Producción
One of the numbers repeated in the array is: 1
Complejidad de tiempo : O(N)
Espacio auxiliar : sqrt(N)
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA