Escriba una función para verificar si dos strings dadas son anagramas entre sí o no. Un anagrama de una string es otra string que contiene los mismos caracteres, solo el orden de los caracteres puede ser diferente. Por ejemplo, «abcd» y «dabc» son un anagrama el uno del otro.
Le recomendamos encarecidamente que haga clic aquí y lo practique antes de pasar a la solución.
Método 1 (Usar clasificación)
- Ordenar ambas strings
- Comparar las strings ordenadas
A continuación se muestra la implementación de la idea anterior:
C++
// C++ program to check whether two strings are anagrams
// of each other
#include <bits/stdc++.h>
using namespace std;
/* function to check whether two strings are anagram of
each other */
bool areAnagram(string str1, string str2)
{
// Get lengths of both strings
int n1 = str1.length();
int n2 = str2.length();
// If length of both strings is not same, then they
// cannot be anagram
if (n1 != n2)
return false;
// Sort both the strings
sort(str1.begin(), str1.end());
sort(str2.begin(), str2.end());
// Compare sorted strings
for (int i = 0; i < n1; i++)
if (str1[i] != str2[i])
return false;
return true;
}
// Driver code
int main()
{
string str1 = "test";
string str2 = "ttew";
// Function Call
if (areAnagram(str1, str2))
cout << "The two strings are anagram of each other";
else
cout << "The two strings are not anagram of each "
"other";
return 0;
}
Java
// JAVA program to check whether two strings
// are anagrams of each other
import java.io.*;
import java.util.Arrays;
import java.util.Collections;
class GFG {
/* function to check whether two strings are
anagram of each other */
static boolean areAnagram(char[] str1, char[] str2)
{
// Get lengths of both strings
int n1 = str1.length;
int n2 = str2.length;
// If length of both strings is not same,
// then they cannot be anagram
if (n1 != n2)
return false;
// Sort both strings
Arrays.sort(str1);
Arrays.sort(str2);
// Compare sorted strings
for (int i = 0; i < n1; i++)
if (str1[i] != str2[i])
return false;
return true;
}
/* Driver Code*/
public static void main(String args[])
{
char str1[] = { 't', 'e', 's', 't' };
char str2[] = { 't', 't', 'e', 'w' };
// Function Call
if (areAnagram(str1, str2))
System.out.println("The two strings are"
+ " anagram of each other");
else
System.out.println("The two strings are not"
+ " anagram of each other");
}
}
// This code is contributed by Nikita Tiwari.
Python
class Solution:
# Function is to check whether two strings are anagram of each other or not.
def isAnagram(self, a, b):
if sorted(a) == sorted(b):
return True
else:
return False
# {
# Driver Code Starts
if __name__ == '__main__':
t = int(input())
for i in range(t):
a, b = map(str, input().strip().split())
if(Solution().isAnagram(a, b)):
print("The two strings are anagram of each other")
else:
print("The two strings are not anagram of each other")
# } Driver Code Ends
C#
// C# program to check whether two
// strings are anagrams of each other
using System;
using System.Collections;
class GFG {
/* function to check whether two
strings are anagram of each other */
public static bool areAnagram(ArrayList str1,
ArrayList str2)
{
// Get lengths of both strings
int n1 = str1.Count;
int n2 = str2.Count;
// If length of both strings is not
// same, then they cannot be anagram
if (n1 != n2) {
return false;
}
// Sort both strings
str1.Sort();
str2.Sort();
// Compare sorted strings
for (int i = 0; i < n1; i++) {
if (str1[i] != str2[i]) {
return false;
}
}
return true;
}
// Driver Code
public static void Main(string[] args)
{
// create and initialize new ArrayList
ArrayList str1 = new ArrayList();
str1.Add('t');
str1.Add('e');
str1.Add('s');
str1.Add('t');
// create and initialize new ArrayList
ArrayList str2 = new ArrayList();
str2.Add('t');
str2.Add('t');
str2.Add('e');
str2.Add('w');
// Function call
if (areAnagram(str1, str2)) {
Console.WriteLine("The two strings are"
+ " anagram of each other");
}
else {
Console.WriteLine("The two strings are not"
+ " anagram of each other");
}
}
}
// This code is contributed by Shrikant13
Javascript
<script>
// JavaScript program to check whether two strings
// are anagrams of each other
/* function to check whether two strings are
anagram of each other */
function areAnagram(str1,str2)
{
// Get lengths of both strings
let n1 = str1.length;
let n2 = str2.length;
// If length of both strings is not same,
// then they cannot be anagram
if (n1 != n2)
return false;
// Sort both strings
str1.sort();
str2.sort()
// Compare sorted strings
for (let i = 0; i < n1; i++)
if (str1[i] != str2[i])
return false;
return true;
}
/* Driver Code*/
let str1=['t', 'e', 's', 't' ];
let str2=['t', 't', 'e', 'w' ];
// Function Call
if (areAnagram(str1, str2))
document.write("The two strings are"
+ " anagram of each other<br>");
else
document.write("The two strings are not"
+ " anagram of each other<br>");
// This code is contributed by rag2127
</script>
Producción
The two strings are not anagram of each other
Complejidad de tiempo: O (nLogn)
Método 2 (Contar caracteres)
Este método asume que el conjunto de posibles caracteres en ambas strings es pequeño. En la siguiente implementación, se supone que los caracteres se almacenan utilizando 8 bits y puede haber 256 caracteres posibles.
- Cree arrays de conteo de tamaño 256 para ambas strings. Inicialice todos los valores en arrays de conteo como 0.
- Repita cada carácter de ambas strings e incremente el recuento de caracteres en las arrays de recuento correspondientes.
- Compara arrays de conteo. Si ambas arrays de conteo son iguales, devuelva verdadero.
A continuación se muestra la implementación de la idea anterior:
C++
// C++ program to check if two strings
// are anagrams of each other
#include <bits/stdc++.h>
using namespace std;
#define NO_OF_CHARS 256
/* function to check whether two strings are anagram of
each other */
bool areAnagram(char* str1, char* str2)
{
// Create 2 count arrays and initialize all values as 0
int count1[NO_OF_CHARS] = { 0 };
int count2[NO_OF_CHARS] = { 0 };
int i;
// For each character in input strings, increment count
// in the corresponding count array
for (i = 0; str1[i] && str2[i]; i++) {
count1[str1[i]]++;
count2[str2[i]]++;
}
// If both strings are of different length. Removing
// this condition will make the program fail for strings
// like "aaca" and "aca"
if (str1[i] || str2[i])
return false;
// Compare count arrays
for (i = 0; i < NO_OF_CHARS; i++)
if (count1[i] != count2[i])
return false;
return true;
}
/* Driver code*/
int main()
{
char str1[] = "geeksforgeeks";
char str2[] = "forgeeksgeeks";
// Function Call
if (areAnagram(str1, str2))
cout << "The two strings are anagram of each other";
else
cout << "The two strings are not anagram of each "
"other";
return 0;
}
// This is code is contributed by rathbhupendra
C
// C program to check if two strings
// are anagrams of each other
#include <stdio.h>
#define NO_OF_CHARS 256
/* function to check whether two strings are anagram of
each other */
bool areAnagram(char* str1, char* str2)
{
// Create 2 count arrays and initialize all values as 0
int count1[NO_OF_CHARS] = { 0 };
int count2[NO_OF_CHARS] = { 0 };
int i;
// For each character in input strings, increment count
// in the corresponding count array
for (i = 0; str1[i] && str2[i]; i++) {
count1[str1[i]]++;
count2[str2[i]]++;
}
// If both strings are of different length. Removing
// this condition will make the program fail for strings
// like "aaca" and "aca"
if (str1[i] || str2[i])
return false;
// Compare count arrays
for (i = 0; i < NO_OF_CHARS; i++)
if (count1[i] != count2[i])
return false;
return true;
}
/* Driver code*/
int main()
{
char str1[] = "geeksforgeeks";
char str2[] = "forgeeksgeeks";
// Function Call
if (areAnagram(str1, str2))
printf("The two strings are anagram of each other");
else
printf("The two strings are not anagram of each "
"other");
return 0;
}
Java
// JAVA program to check if two strings
// are anagrams of each other
import java.io.*;
import java.util.*;
class GFG {
static int NO_OF_CHARS = 256;
/* function to check whether two strings
are anagram of each other */
static boolean areAnagram(char str1[], char str2[])
{
// Create 2 count arrays and initialize
// all values as 0
int count1[] = new int[NO_OF_CHARS];
Arrays.fill(count1, 0);
int count2[] = new int[NO_OF_CHARS];
Arrays.fill(count2, 0);
int i;
// For each character in input strings,
// increment count in the corresponding
// count array
for (i = 0; i < str1.length && i < str2.length;
i++) {
count1[str1[i]]++;
count2[str2[i]]++;
}
// If both strings are of different length.
// Removing this condition will make the program
// fail for strings like "aaca" and "aca"
if (str1.length != str2.length)
return false;
// Compare count arrays
for (i = 0; i < NO_OF_CHARS; i++)
if (count1[i] != count2[i])
return false;
return true;
}
/* Driver code*/
public static void main(String args[])
{
char str1[] = ("geeksforgeeks").toCharArray();
char str2[] = ("forgeeksgeeks").toCharArray();
// Function call
if (areAnagram(str1, str2))
System.out.println("The two strings are"
+ "anagram of each other");
else
System.out.println("The two strings are not"
+ " anagram of each other");
}
}
// This code is contributed by Nikita Tiwari.
Python
# Python program to check if two strings are anagrams of # each other NO_OF_CHARS = 256 # Function to check whether two strings are anagram of # each other def areAnagram(str1, str2): # Create two count arrays and initialize all values as 0 count1 = [0] * NO_OF_CHARS count2 = [0] * NO_OF_CHARS # For each character in input strings, increment count # in the corresponding count array for i in str1: count1[ord(i)] += 1 for i in str2: count2[ord(i)] += 1 # If both strings are of different length. Removing this # condition will make the program fail for strings like # "aaca" and "aca" if len(str1) != len(str2): return 0 # Compare count arrays for i in xrange(NO_OF_CHARS): if count1[i] != count2[i]: return 0 return 1 # Driver code str1 = "geeksforgeeks" str2 = "forgeeksgeeks" # Function call if areAnagram(str1, str2): print "The two strings are anagram of each other" else: print "The two strings are not anagram of each other" # This code is contributed by Bhavya Jain
C#
// C# program to check if two strings
// are anagrams of each other
using System;
public class GFG {
static int NO_OF_CHARS = 256;
/* function to check whether two strings
are anagram of each other */
static bool areAnagram(char[] str1, char[] str2)
{
// Create 2 count arrays and initialize
// all values as 0
int[] count1 = new int[NO_OF_CHARS];
int[] count2 = new int[NO_OF_CHARS];
int i;
// For each character in input strings,
// increment count in the corresponding
// count array
for (i = 0; i < str1.Length && i < str2.Length;
i++) {
count1[str1[i]]++;
count2[str2[i]]++;
}
// If both strings are of different length.
// Removing this condition will make the program
// fail for strings like "aaca" and "aca"
if (str1.Length != str2.Length)
return false;
// Compare count arrays
for (i = 0; i < NO_OF_CHARS; i++)
if (count1[i] != count2[i])
return false;
return true;
}
/* Driver code*/
public static void Main()
{
char[] str1 = ("geeksforgeeks").ToCharArray();
char[] str2 = ("forgeeksgeeks").ToCharArray();
// Function Call
if (areAnagram(str1, str2))
Console.WriteLine("The two strings are"
+ "anagram of each other");
else
Console.WriteLine("The two strings are not"
+ " anagram of each other");
}
}
// This code contributed by 29AjayKumar
Javascript
<script>
// JAVAscript program to check if two strings
// are anagrams of each other
let NO_OF_CHARS = 256;
/* function to check whether two strings
are anagram of each other */
function areAnagram(str1, str2)
{
// Create 2 count arrays and initialize
// all values as 0
let count1 = new Array(NO_OF_CHARS);
let count2 = new Array(NO_OF_CHARS);
for(let i = 0; i < NO_OF_CHARS; i++)
{
count1[i] = 0;
count2[i] = 0;
}
let i;
// For each character in input strings,
// increment count in the corresponding
// count array
for (i = 0; i < str1.length && i < str2.length;
i++) {
count1[str1[i].charCodeAt(0)]++;
count2[str1[i].charCodeAt(0)]++;
}
// If both strings are of different length.
// Removing this condition will make the program
// fail for strings like "aaca" and "aca"
if (str1.length != str2.length)
return false;
// Compare count arrays
for (i = 0; i < NO_OF_CHARS; i++)
if (count1[i] != count2[i])
return false;
return true;
}
/* Driver code*/
let str1 = ("geeksforgeeks").split("");
let str2 = ("forgeeksgeeks").split("");
// Function call
if (areAnagram(str1, str2))
document.write("The two strings are"
+ "anagram of each other<br>");
else
document.write("The two strings are not"
+ " anagram of each other<br>");
// This code is contributed by avanitrachhadiya2155
</script>
Producción
The two strings are anagram of each other
Complejidad de tiempo : O(n)
Complejidad espacial: O(NO_OF_CHAR) = O(256) = O(1)(uso constante del espacio)
Método 3 (contar caracteres usando una array)
La implementación anterior puede ser más avanzada para usar solo una array de conteo en lugar de dos. Podemos incrementar el valor en la array de conteo para caracteres en str1 y disminuir para caracteres en str2. Finalmente, si todos los valores de conteo son 0, entonces las dos strings son anagramas entre sí. Gracias a Ace por sugerir esta optimización.
C++
// C++ program to check if two strings
// are anagrams of each other
#include <bits/stdc++.h>
using namespace std;
#define NO_OF_CHARS 256
bool areAnagram(char* str1, char* str2)
{
// Create a count array and initialize all values as 0
int count[NO_OF_CHARS] = { 0 };
int i;
// For each character in input strings, increment count
// in the corresponding count array
for (i = 0; str1[i] && str2[i]; i++) {
count[str1[i]]++;
count[str2[i]]--;
}
// If both strings are of different length. Removing
// this condition will make the program fail for strings
// like "aaca" and "aca"
if (str1[i] || str2[i])
return false;
// See if there is any non-zero value in count array
for (i = 0; i < NO_OF_CHARS; i++)
if (count[i])
return false;
return true;
}
// Driver code
int main()
{
char str1[] = "geeksforgeeks";
char str2[] = "forgeeksgeeks";
// Function call
if (areAnagram(str1, str2))
cout << "The two strings are anagram of each other";
else
cout << "The two strings are not anagram of each "
"other";
return 0;
}
Java
// Java program to check if two strings
// are anagrams of each other
class GFG{
static int NO_OF_CHARS = 256;
// function to check if two strings
// are anagrams of each other
static boolean areAnagram(char[] str1,
char[] str2)
{
// Create a count array and initialize
// all values as 0
int[] count = new int[NO_OF_CHARS];
int i;
// If both strings are of different
// length. Removing this condition
// will make the program fail for
// strings like "aaca" and "aca"
if (str1.length != str2.length)
return false;
// For each character in input strings,
// increment count in the corresponding
// count array
for(i = 0; i < str1.length; i++)
{
count[str1[i]]++;
count[str2[i]]--;
}
// See if there is any non-zero
// value in count array
for(i = 0; i < NO_OF_CHARS; i++)
if (count[i] != 0)
{
return false;
}
return true;
}
// Driver code
public static void main(String[] args)
{
char str1[] = "geeksforgeeks".toCharArray();
char str2[] = "forgeeksgeeks".toCharArray();
// Function call
if (areAnagram(str1, str2))
System.out.print("The two strings are " +
"anagram of each other");
else
System.out.print("The two strings are " +
"not anagram of each other");
}
}
// This code is contributed by mark_85
Python3
# Python program to check if two strings
# are anagrams of each other
NO_OF_CHARS = 256
# function to check if two strings
# are anagrams of each other
def areAnagram(str1,str2):
# If both strings are of different
# length. Removing this condition
# will make the program fail for
# strings like "aaca" and "aca"
if(len(str1) != len(str2)):
return False;
# Create a count array and initialize
# all values as 0
count=[0 for i in range(NO_OF_CHARS)]
i=0
# For each character in input strings,
# increment count in the corresponding
# count array
for i in range(len(str1)):
count[ord(str1[i]) - ord('a')] += 1;
count[ord(str2[i]) - ord('a')] -= 1;
# See if there is any non-zero
# value in count array
for i in range(NO_OF_CHARS):
if (count[i] != 0):
return False
return True
# Driver code
str1="geeksforgeeks"
str2="forgeeksgeeks"
# Function call
if (areAnagram(str1, str2)):
print("The two strings are anagram of each other")
else:
print("The two strings are not anagram of each other")
# This code is contributed by patel2127
C#
// C# program to check if two strings
// are anagrams of each other
using System;
class GFG{
static int NO_OF_CHARS = 256;
// function to check if two strings
// are anagrams of each other
static bool areAnagram(char[] str1,
char[] str2)
{
// If both strings are of different
// Length. Removing this condition
// will make the program fail for
// strings like "aaca" and "aca"
if (str1.Length != str2.Length)
return false;
// Create a count array and initialize
// all values as 0
int[] count = new int[NO_OF_CHARS];
int i;
// For each character in input strings,
// increment count in the corresponding
// count array
for(i = 0; i < str1.Length; i++)
{
count[str1[i] - 'a']++;
count[str2[i] - 'a']--;
}
// See if there is any non-zero
// value in count array
for(i = 0; i < NO_OF_CHARS; i++)
if (count[i] != 0)
{
return false;
}
return true;
}
// Driver code
public static void Main(String []args)
{
char []str1 = "geeksforgeeks".ToCharArray();
char []str2 = "forgeeksgeeks".ToCharArray();
// Function call
if (areAnagram(str1, str2))
Console.Write("The two strings are " +
"anagram of each other");
else
Console.Write("The two strings are " +
"not anagram of each other");
}
}
// This code is contributed by shivanisinghss2110
Javascript
<script>
// Javascript program to check if two strings
// are anagrams of each other
let NO_OF_CHARS = 256;
// function to check if two strings
// are anagrams of each other
function areAnagram(str1, str2)
{
// If both strings are of different
// length. Removing this condition
// will make the program fail for
// strings like "aaca" and "aca"
if (str1.length != str2.length)
return false;
// Create a count array and initialize
// all values as 0
let count = new Array(NO_OF_CHARS);
for(let i = 0; i < NO_OF_CHARS; i++)
{
count[i] = 0;
}
let i;
// For each character in input strings,
// increment count in the corresponding
// count array
for(i = 0; i < str1.length; i++)
{
count[str1[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
count[str2[i].charCodeAt(0) - 'a'.charCodeAt(0)]--;
}
// See if there is any non-zero
// value in count array
for(i = 0; i < NO_OF_CHARS; i++)
if (count[i] != 0)
{
return false;
}
return true;
}
// Driver code
let str1 = "geeksforgeeks".split("");
let str2 = "forgeeksgeeks".split("");
// Function call
if (areAnagram(str1, str2))
document.write("The two strings are " +
"anagram of each other");
else
document.write("The two strings are " +
"not anagram of each other");
// This code is contributed by unknown2108.
</script>
Producción
The two strings are anagram of each other
Complejidad de tiempo: O(n)
Complejidad de tiempo: O(NO_OF_CHAR) = O(256) = O(1) (uso de espacio constante)
Método 4 (Poner todos los caracteres en HashMap)
En la implementación anterior, estamos usando espacio adicional ya que estamos creando una array de 256 caracteres, pero podemos optimizarla usando HashMap donde podemos almacenar caracteres y contar caracteres en HashMap. La idea es poner todos los caracteres de una string en HashMap y reducirlos a medida que los encontramos mientras recorremos otra string.
C++
// C++ implementation of the approach
// Function that returns true if a and b
// are anagarams of each other
#include <bits/stdc++.h>
using namespace std;
bool isAnagram(string a,string b)
{
// Check if length of both strings is same or not
if (a.length() != b.length()) {
return false;
}
// Create a HashMap containing Character as Key and
// Integer as Value. We will be storing character as
// Key and count of character as Value.
unordered_map<char,int> Map;
// Loop over all character of String a and put in
// HashMap.
for (int i = 0; i < a.length(); i++) {
Map[a[i]]++;
}
// Now loop over String b
for (int i = 0; i < b.length(); i++) {
// Check if current character already exists in
// HashMap/map
if (Map.find(b[i]) != Map.end()) {
// If contains reduce count of that
// character by 1 to indicate that current
// character has been already counted as
// idea here is to check if in last count of
// all characters in last is zero which
// means all characters in String a are
// present in String b.
Map[b[i]] -= 1;
}
}
// Loop over all keys and check if all keys are 0.
// If so it means it is anagram.
for (auto items : Map) {
if (items.second != 0) {
return false;
}
}
// Returning True as all keys are zero
return true;
}
// Driver code
int main()
{
string str1 = "geeksforgeeks";
string str2 = "forgeeksgeeks";
// Function call
if (isAnagram(str1, str2))
cout<<"The two strings are anagram of each other"<<endl;
else
cout<<"The two strings are not anagram of each other"<<endl;
}
// This code is contributed by shinjanpatra
Java
import java.io.*;
import java.util.*;
class GFG {
public static boolean isAnagram(String a, String b)
{
// Check if length of both strings is same or not
if (a.length() != b.length()) {
return false;
}
// Create a HashMap containing Character as Key and
// Integer as Value. We will be storing character as
// Key and count of character as Value.
HashMap<Character, Integer> map = new HashMap<>();
// Loop over all character of String a and put in
// HashMap.
for (int i = 0; i < a.length(); i++) {
// Check if HashMap already contain current
// character or not
if (map.containsKey(a.charAt(i))) {
// If contains increase count by 1 for that
// character
map.put(a.charAt(i),
map.get(a.charAt(i)) + 1);
}
else {
// else put that character in map and set
// count to 1 as character is encountered
// first time
map.put(a.charAt(i), 1);
}
}
// Now loop over String b
for (int i = 0; i < b.length(); i++) {
// Check if current character already exists in
// HashMap/map
if (map.containsKey(b.charAt(i))) {
// If contains reduce count of that
// character by 1 to indicate that current
// character has been already counted as
// idea here is to check if in last count of
// all characters in last is zero which
// means all characters in String a are
// present in String b.
map.put(b.charAt(i),
map.get(b.charAt(i)) - 1);
}
}
// Extract all keys of HashMap/map
Set<Character> keys = map.keySet();
// Loop over all keys and check if all keys are 0.
// If so it means it is anagram.
for (Character key : keys) {
if (map.get(key) != 0) {
return false;
}
}
// Returning True as all keys are zero
return true;
}
public static void main(String[] args)
{
String str1 = "geeksforgeeks";
String str2 = "forgeeksgeeks";
// Function call
if (isAnagram(str1, str2))
System.out.print("The two strings are "
+ "anagram of each other");
else
System.out.print("The two strings are "
+ "not anagram of each other");
}
}
Python3
# Python3 implementation of the approach
# Function that returns True if a and b
# are anagarams of each other
def isAnagram(a, b):
# Check if length of both strings is same or not
if (len(a) != len(b)):
return False
# Create a HashMap containing Character as Key and
# Integer as Value. We will be storing character as
# Key and count of character as Value.
map = {}
# Loop over all character of String a and put in
# HashMap.
for i in range(len(a)):
# Check if HashMap already contain current
# character or not
if (a[i] in map):
# If contains increase count by 1 for that
# character
map[a[i]] += 1
else:
# else set that character in map and set
# count to 1 as character is encountered
# first time
map[a[i]] = 1
# Now loop over String b
for i in range(len(b)):
# Check if current character already exists in
# HashMap/map
if (b[i] in map):
# If contains reduce count of that
# character by 1 to indicate that current
# character has been already counted as
# idea here is to check if in last count of
# all characters in last is zero which
# means all characters in String a are
# present in String b.
map[b[i]] -= 1
# Extract all keys of HashMap/map
keys = map.keys()
# Loop over all keys and check if all keys are 0.
# If so it means it is anagram.
for key in keys:
if (map[key] != 0):
return False
# Returning True as all keys are zero
return True
# Driver code
str1 = "geeksforgeeks"
str2 = "forgeeksgeeks"
# Function call
if (isAnagram(str1, str2)):
print("The two strings are anagram of each other")
else:
print("The two strings are not anagram of each other")
# This code is contributed by shinjanpatra
C#
using System;
using System.Collections.Generic;
public class GFG {
public static bool isAnagram(String a, String b)
{
// Check if length of both strings is same or not
if (a.Length != b.Length) {
return false;
}
// Create a Dictionary containing char as Key and
// int as Value. We will be storing character as
// Key and count of character as Value.
Dictionary<char, int> map = new Dictionary<char, int>();
// Loop over all character of String a and put in
// Dictionary.
for (int i = 0; i < a.Length; i++)
{
// Check if Dictionary already contain current
// character or not
if (map.ContainsKey(a[i]))
{
// If contains increase count by 1 for that
// character
map[a[i]] =
map[a[i]] + 1;
}
else
{
// else put that character in map and set
// count to 1 as character is encountered
// first time
map.Add(a[i], 1);
}
}
// Now loop over String b
for (int i = 0; i < b.Length; i++)
{
// Check if current character already exists in
// Dictionary/map
if (map.ContainsKey(b[i]))
{
// If contains reduce count of that
// character by 1 to indicate that current
// character has been already counted as
// idea here is to check if in last count of
// all characters in last is zero which
// means all characters in String a are
// present in String b.
map[b[i]]=
map[b[i]] - 1;
}
}
// Extract all keys of Dictionary/map
var keys = map.Keys;
// Loop over all keys and check if all keys are 0.
// If so it means it is anagram.
foreach (char key in keys) {
if (map[key] != 0) {
return false;
}
}
// Returning True as all keys are zero
return true;
}
// Driver code
public static void Main(String[] args)
{
String str1 = "geeksforgeeks";
String str2 = "forgeeksgeeks";
// Function call
if (isAnagram(str1, str2))
Console.Write("The two strings are "
+ "anagram of each other");
else
Console.Write("The two strings are "
+ "not anagram of each other");
}
}
// This code is contributed by shikhasingrajput
Javascript
<script>
// JavaScript implementation of the approach
// Function that returns true if a and b
// are anagarams of each other
function isAnagram(a, b)
{
// Check if length of both strings is same or not
if (a.length != b.length) {
return false;
}
// Create a HashMap containing Character as Key and
// Integer as Value. We will be storing character as
// Key and count of character as Value.
let map = new Map();
// Loop over all character of String a and put in
// HashMap.
for (let i = 0; i < a.length; i++) {
// Check if HashMap already contain current
// character or not
if (map.has(a[i])) {
// If contains increase count by 1 for that
// character
map.set(a[i],
map.get(a[i]) + 1);
}
else {
// else set that character in map and set
// count to 1 as character is encountered
// first time
map.set(a[i], 1);
}
}
// Now loop over String b
for (let i = 0; i < b.length; i++) {
// Check if current character already exists in
// HashMap/map
if (map.has(b[i])) {
// If contains reduce count of that
// character by 1 to indicate that current
// character has been already counted as
// idea here is to check if in last count of
// all characters in last is zero which
// means all characters in String a are
// present in String b.
map.set(b[i],
map.get(b[i]) - 1);
}
}
// Extract all keys of HashMap/map
let keys = map.keys();
// Loop over all keys and check if all keys are 0.
// If so it means it is anagram.
for (let key of keys) {
if (map.get(key) != 0) {
return false;
}
}
// Returning True as all keys are zero
return true;
}
// Driver code
let str1 = "geeksforgeeks";
let str2 = "forgeeksgeeks";
// Function call
if (isAnagram(str1, str2))
document.write("The two strings are anagram of each other");
else
document.write("The two strings are not anagram of each other");
// This code is contributed by shinjanpatra
</script>
Producción
The two strings are anagram of each other
Complejidad de tiempo : O(n)
Complejidad espacial: O(1) (uso constante del espacio)
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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