Dada una array de strings arr[] que consta solo de caracteres en minúsculas y mayúsculas, la tarea es modificar la array eliminando los caracteres de las strings que se repiten en la misma string o en cualquier otra string. Imprime la array modificada.
Ejemplos:
Entrada: arr[] = {“Geeks”, “For”, “Geeks”}
Salida: {“Geks”, “For”}
Explicación:
En arr[0[, ‘e’ aparece dos veces en la string. Quitar una sola ‘e’ de la primera string modifica «Geeks» a «Geks».
En arr[1], todos los caracteres no se repiten. Por lo tanto, la string permanece sin cambios.
En arr[2], la string es la misma que arr[0]. Por lo tanto, se requiere eliminar la string completa.Entrada: arr[] = {“Geeks”, “Para”, “Geeks”, “Publicar”}
Salida: {“Geks”, “Para”, “Pt”}
Enfoque: Siga los pasos para resolver el problema:
- Inicialice un conjunto desordenado para almacenar los caracteres de la string mientras recorre la array .
- Recorra la array y para cada string, realice las siguientes operaciones:
- Iterar sobre los caracteres de la string.
- Si el carácter actual ya está presente en el Conjunto , sáltelo. De lo contrario, agréguelo a la string de salida.
- Inserte la string recién generada en la lista inicializada para almacenar la salida.
- Imprime la lista de strings obtenidas como respuesta.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to remove duplicate
// characters across the strings
void removeDuplicateCharacters(vector<string> arr)
{
// Stores distinct characters
unordered_set<char> cset;
// Size of the array
int n = arr.size();
// Stores the list of
// modified strings
vector<string> out;
// Traverse the array
for (auto str : arr) {
// Stores the modified string
string out_curr = "";
// Iterate over the characters
// of the modified string
for (auto ch : str) {
// If character is already present
if (cset.find(ch) != cset.end())
continue;
out_curr += ch;
// Insert character into the Set
cset.insert(ch);
}
if (out_curr.size())
out.push_back(out_curr);
}
// Print the list of modified strings
for (int i = 0; i < out.size(); i++) {
// Print each string
cout << out[i] << " ";
}
}
// Driver Code
int main()
{
// Given array of strings
vector<string> arr
= { "Geeks", "For", "Geeks", "Post" };
// Function Call to modify the
// given array of strings
removeDuplicateCharacters(arr);
}
Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to remove duplicate
// characters across the strings
static void removeDuplicateCharacters(String arr[])
{
// Stores distinct characters
HashSet<Character> cset = new HashSet<>();
// Size of the array
int n = arr.length;
// Stores the list of
// modified strings
ArrayList<String> out = new ArrayList<>();
// Traverse the array
for(String str : arr)
{
// Stores the modified string
String out_curr = "";
// Iterate over the characters
// of the modified string
for(char ch : str.toCharArray())
{
// If character is already present
if (cset.contains(ch))
continue;
out_curr += ch;
// Insert character into the Set
cset.add(ch);
}
if (out_curr.length() != 0)
out.add(out_curr);
}
// Print the list of modified strings
for(int i = 0; i < out.size(); i++)
{
// Print each string
System.out.print(out.get(i) + " ");
}
}
// Driver Code
public static void main(String[] args)
{
// Given array of strings
String arr[] = { "Geeks", "For", "Geeks", "Post" };
// Function Call to modify the
// given array of strings
removeDuplicateCharacters(arr);
}
}
// This code is contributed by Kingash
Python3
# Python 3 program for the above approach # Function to remove duplicate # characters across the strings def removeDuplicateCharacters(arr): # Stores distinct characters cset = set([]) # Size of the array n = len(arr) # Stores the list of # modified strings out = [] # Traverse the array for st in arr: # Stores the modified string out_curr = "" # Iterate over the characters # of the modified string for ch in st: # If character is already present if (ch in cset): continue out_curr += ch # Insert character into the Set cset.add(ch) if (len(out_curr)): out.append(out_curr) # Print the list of modified strings for i in range(len(out)): # Print each string print(out[i], end = " ") # Driver Code if __name__ == "__main__": # Given array of strings arr = ["Geeks", "For", "Geeks", "Post"] # Function Call to modify the # given array of strings removeDuplicateCharacters(arr) # This code is contributed by ukasp.
C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to remove duplicate
// characters across the strings
static void removeDuplicateCharacters(string[] arr)
{
// Stores distinct characters
HashSet<int> cset = new HashSet<int>();
// Size of the array
int n = arr.Length;
// Stores the list of
// modified strings
List<string> Out = new List<string>();
// Traverse the array
foreach(string str in arr)
{
// Stores the modified string
string out_curr = "";
// Iterate over the characters
// of the modified string
foreach(char ch in str.ToCharArray())
{
// If character is already present
if (cset.Contains(ch))
continue;
out_curr += ch;
// Insert character into the Set
cset.Add(ch);
}
if (out_curr.Length != 0)
Out.Add(out_curr);
}
// Print the list of modified strings
for(int i = 0; i < Out.Count; i++)
{
// Print each string
Console.Write(Out[i] + " ");
}
}
static public void Main (){
// Given array of strings
string[] arr = { "Geeks", "For",
"Geeks", "Post" };
// Function Call to modify the
// given array of strings
removeDuplicateCharacters(arr);
}
}
// This code is contributed by avanitrachhadiya2155
Javascript
<script>
// JavaScript program for the above approach
// Function to remove duplicate
// characters across the strings
function removeDuplicateCharacters(arr)
{
// Stores distinct characters
var cset = new Set();
// Size of the array
var n = arr.length;
// Stores the list of
// modified strings
var out = [];
// Traverse the array
arr.forEach(str => {
// Stores the modified string
var out_curr = "";
// Iterate over the characters
// of the modified string
str.split('').forEach(ch => {
// If character is already present
if (!cset.has(ch))
{
out_curr += ch;
// Insert character into the Set
cset.add(ch);
}
});
if (out_curr.size!=0)
out.push(out_curr);
});
// Print the list of modified strings
for (var i = 0; i < out.length; i++) {
// Print each string
document.write( out[i] + " ");
}
}
// Driver Code
// Given array of strings
var arr
= ["Geeks", "For", "Geeks", "Post"];
// Function Call to modify the
// given array of strings
removeDuplicateCharacters(arr);
</script>
Producción:
Geks For Pt
Complejidad de tiempo: O(N * M) donde M es la longitud de la string más larga de la array.
Espacio Auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por DevanshuAgarwal y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA