Dados dos números enteros A y B , la tarea es verificar si los números dados son anagramas entre sí o no. Al igual que las strings, se dice que un número es un anagrama de otro número si se puede hacer igual al otro número simplemente mezclando los dígitos en él.
Ejemplos:
Entrada: A = 204, B = 240
Salida: SíEntrada: A = 23, B = 959
Salida: No
Enfoque: Cree dos arrays freqA[] y freqB[] donde freqA[i] y freqB[i] almacenarán la frecuencia del dígito i en a y b respectivamente. Ahora recorra las arrays de frecuencia y para cualquier dígito i si freqA[i] != freqB[i] entonces los números no son anagramas entre sí, sino que lo son.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
const int TEN = 10;
// Function to update the frequency array
// such that freq[i] stores the
// frequency of digit i in n
void updateFreq(int n, int freq[])
{
// While there are digits
// left to process
while (n) {
int digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n /= TEN;
}
}
// Function that returns true if a and b
// are anagarams of each other
bool areAnagrams(int a, int b)
{
// To store the frequencies of
// the digits in a and b
int freqA[TEN] = { 0 };
int freqB[TEN] = { 0 };
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for (int i = 0; i < TEN; i++) {
// If frequency differs for any digit
// then the numbers are not
// anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Driver code
int main()
{
int a = 240, b = 204;
if (areAnagrams(a, b))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// Java implementation of the approach
class GFG
{
static final int TEN = 10;
// Function to update the frequency array
// such that freq[i] stores the
// frequency of digit i in n
static void updateFreq(int n, int [] freq)
{
// While there are digits
// left to process
while (n > 0)
{
int digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n /= TEN;
}
}
// Function that returns true if a and b
// are anagarams of each other
static boolean areAnagrams(int a, int b)
{
// To store the frequencies of
// the digits in a and b
int [] freqA = new int[TEN];
int [] freqB = new int[TEN];
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for (int i = 0; i < TEN; i++)
{
// If frequency differs for any digit
// then the numbers are not
// anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Driver code
public static void main (String[] args)
{
int a = 204, b = 240;
if (areAnagrams(a, b))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by ihirtik
Python3
# Python3 implementation of the approach
TEN = 10
# Function to update the frequency array
# such that freq[i] stores the
# frequency of digit i in n
def updateFreq(n, freq) :
# While there are digits
# left to process
while (n) :
digit = n % TEN
# Update the frequency of
# the current digit
freq[digit] += 1
# Remove the last digit
n //= TEN
# Function that returns true if a and b
# are anagarams of each other
def areAnagrams(a, b):
# To store the frequencies of
# the digits in a and b
freqA = [ 0 ] * TEN
freqB = [ 0 ] * TEN
# Update the frequency of
# the digits in a
updateFreq(a, freqA)
# Update the frequency of
# the digits in b
updateFreq(b, freqB)
# Match the frequencies of
# the common digits
for i in range(TEN):
# If frequency differs for any digit
# then the numbers are not
# anagrams of each other
if (freqA[i] != freqB[i]):
return False
return True
# Driver code
a = 240
b = 204
if (areAnagrams(a, b)):
print("Yes")
else:
print("No")
# This code is contributed by
# divyamohan123
C#
// C# implementation of the approach
using System;
class GFG
{
static int TEN = 10;
// Function to update the frequency array
// such that freq[i] stores the
// frequency of digit i in n
static void updateFreq(int n, int [] freq)
{
// While there are digits
// left to process
while (n > 0)
{
int digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n /= TEN;
}
}
// Function that returns true if a and b
// are anagarams of each other
static bool areAnagrams(int a, int b)
{
// To store the frequencies of
// the digits in a and b
int [] freqA = new int[TEN];
int [] freqB = new int[TEN];;
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for (int i = 0; i < TEN; i++)
{
// If frequency differs for any digit
// then the numbers are not
// anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Driver code
public static void Main ()
{
int a = 204, b = 240;
if (areAnagrams(a, b))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by ihirtik
Javascript
<script>
// Javascript implementation of the approach
const TEN = 10;
// Function to update the frequency array
// such that freq[i] stores the
// frequency of digit i in n
function updateFreq(n, freq)
{
// While there are digits
// left to process
while (n) {
let digit = n % TEN;
// Update the frequency of
// the current digit
freq[digit]++;
// Remove the last digit
n = parseInt(n / TEN);
}
}
// Function that returns true if a and b
// are anagarams of each other
function areAnagrams(a, b)
{
// To store the frequencies of
// the digits in a and b
let freqA = new Array(TEN).fill(0);
let freqB = new Array(TEN).fill(0);
// Update the frequency of
// the digits in a
updateFreq(a, freqA);
// Update the frequency of
// the digits in b
updateFreq(b, freqB);
// Match the frequencies of
// the common digits
for (let i = 0; i < TEN; i++) {
// If frequency differs for any digit
// then the numbers are not
// anagrams of each other
if (freqA[i] != freqB[i])
return false;
}
return true;
}
// Driver code
let a = 240, b = 204;
if (areAnagrams(a, b))
document.write("Yes");
else
document.write("Yes");
</script>
Producción:
Yes
Método #2: Usando la clasificación():
- Convierte dos enteros en strings.
- Ordena las strings y comprueba si son iguales.
- Escriba Sí si son iguales.
- De lo contrario no.
A continuación se muestra la implementación:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function that returns true if a and b
// are anagarams of each other
bool areAnagrams(int a, int b)
{
// Converting numbers to strings
string c = to_string(a);
string d = to_string(b);
// Checking if the sorting values
// of two strings are equal
sort(c.begin(), c.end());
sort(d.begin(), d.end());
if (c == d)
return true;
else
return false;
}
// Driver code
int main()
{
int a = 240;
int b = 204;
if (areAnagrams(a, b))
cout << "Yes";
else
cout << "No";
}
// This code is contributed by ukasp.
Python3
# Python3 implementation of the approach
# Function that returns true if a and b
# are anagarams of each other
def areAnagrams(a, b):
# Converting numbers to strings
a = str(a)
b = str(b)
# Checking if the sorting values
# of two strings are equal
if(sorted(a) == sorted(b)):
return True
else:
return False
# Driver code
a = 240
b = 204
if (areAnagrams(a, b)):
print("Yes")
else:
print("No")
# This code is contributed by vikkycirus
Javascript
<script>
// JavaScript implementation of the approach
// Function that returns true if a and b
// are anagarams of each other
function areAnagrams(a, b){
// Converting numbers to strings
a = a.toString().split("").sort().join("")
b = b.toString().split("").sort().join("")
// Checking if the sorting values
// of two strings are equal
if(a == b){
return true
}
else
return false
}
// Driver code
let a = 240
let b = 204
if (areAnagrams(a, b))
document.write("Yes")
else
document.write("No")
// This code is contributed by shinjanpatra
</script>
Producción:
Yes