Dadas dos strings en minúsculas, la tarea es convertirlas en anagrama . La única operación permitida es eliminar un carácter de cualquier string. ¿Encontrar el número mínimo de caracteres que se eliminarán para hacer que ambas strings sean un anagrama?
Si dos strings contienen el mismo conjunto de datos en cualquier orden, las strings se denominan Anagramas .
Ejemplos:
C++
// C++ program to find minimum number of characters
// to be removed to make two strings anagram.
#include<bits/stdc++.h>
using namespace std;
const int CHARS = 26;
// function to calculate minimum numbers of characters
// to be removed to make two strings anagram
int remAnagram(string str1, string str2)
{
// make hash array for both string and calculate
// frequency of each character
int count1[CHARS] = {0}, count2[CHARS] = {0};
// count frequency of each character in first string
for (int i=0; str1[i]!='\0'; i++)
count1[str1[i]-'a']++;
// count frequency of each character in second string
for (int i=0; str2[i]!='\0'; i++)
count2[str2[i]-'a']++;
// traverse count arrays to find number of characters
// to be removed
int result = 0;
for (int i=0; i<26; i++)
result += abs(count1[i] - count2[i]);
return result;
}
// Driver program to run the case
int main()
{
string str1 = "bcadeh", str2 = "hea";
cout << remAnagram(str1, str2);
return 0;
}
Java
// Java program to find minimum number of
// characters to be removed to make two
// strings anagram.
import java.util.*;
class GFG {
// function to calculate minimum numbers
// of characters to be removed to make
// two strings anagram
static int remAnagram(String str1, String str2)
{
// make hash array for both string
// and calculate frequency of each
// character
int count1[] = new int[26];
int count2[] = new int[26];
// count frequency of each character
// in first string
for (int i = 0; i < str1.length() ; i++)
count1[str1.charAt(i) -'a']++;
// count frequency of each character
// in second string
for (int i = 0; i < str2.length() ; i++)
count2[str2.charAt(i) -'a']++;
// traverse count arrays to find
// number of characters to be removed
int result = 0;
for (int i = 0; i < 26; i++)
result += Math.abs(count1[i] -
count2[i]);
return result;
}
// Driver program to run the case
public static void main(String[] args)
{
String str1 = "bcadeh", str2 = "hea";
System.out.println(remAnagram(str1, str2));
}
}
// This code is contributed by Prerna Saini
Python3
# Python 3 program to find minimum
# number of characters
# to be removed to make two
# strings anagram.
CHARS = 26
# function to calculate minimum
# numbers of characters
# to be removed to make two
# strings anagram
def remAnagram(str1, str2):
# make hash array for both string
# and calculate
# frequency of each character
count1 = [0]*CHARS
count2 = [0]*CHARS
# count frequency of each character
# in first string
i = 0
while i < len(str1):
count1[ord(str1[i])-ord('a')] += 1
i += 1
# count frequency of each character
# in second string
i =0
while i < len(str2):
count2[ord(str2[i])-ord('a')] += 1
i += 1
# traverse count arrays to find
# number of characters
# to be removed
result = 0
for i in range(26):
result += abs(count1[i] - count2[i])
return result
# Driver program to run the case
if __name__ == "__main__":
str1 = "bcadeh"
str2 = "hea"
print(remAnagram(str1, str2))
# This code is contributed by
# ChitraNayal
C#
// C# program to find minimum
// number of characters to be
// removed to make two strings
// anagram.
using System;
class GFG
{
// function to calculate
// minimum numbers of
// characters to be removed
// to make two strings anagram
static int remAnagram(string str1,
string str2)
{
// make hash array for both
// string and calculate frequency
// of each character
int []count1 = new int[26];
int []count2 = new int[26];
// count frequency of each
// character in first string
for (int i = 0; i < str1.Length ; i++)
count1[str1[i] -'a']++;
// count frequency of each
// character in second string
for (int i = 0; i < str2.Length ; i++)
count2[str2[i] -'a']++;
// traverse count arrays to
// find number of characters
// to be removed
int result = 0;
for (int i = 0; i < 26; i++)
result += Math.Abs(count1[i] -
count2[i]);
return result;
}
// Driver Code
public static void Main()
{
string str1 = "bcadeh",
str2 = "hea";
Console.Write(remAnagram(str1, str2));
}
}
// This code is contributed
// by nitin mittal.
PHP
<?php
// PHP program to find minimum number of
// characters to be removed to make two
// strings anagram.
// function to calculate minimum numbers
// of characters to be removed to make
// two strings anagram
function remAnagram($str1, $str2)
{
// make hash array for both string
// and calculate frequency of each
// character
$count1 = array(26);
$count2 = array(26);
// count frequency of each character
// in first string
for ($i = 0; $i < strlen($str1) ; $i++)
$count1[$str1[$i] - 'a']++;
// count frequency of each character
// in second string
for ($i = 0; $i < strlen($str2) ; $i++)
$count2[$str2[$i] -'a']++;
// traverse count arrays to find
// number of characters to be removed
$result = 0;
for ($i = 0; $i < 26; $i++)
$result += abs($count1[$i] -
$count2[$i]);
return $result;
}
// Driver Code
{
$str1 = "bcadeh"; $str2 = "hea";
echo(remAnagram($str1, $str2));
}
// This code is contributed by Code_Mech
Javascript
<script>
// javascript program to find minimum number of
// characters to be removed to make two
// strings anagram.
// function to calculate minimum numbers
// of characters to be removed to make
// two strings anagram
function remAnagram(str1, str2)
{
// make hash array for both string
// and calculate frequency of each
// character
var count1 = Array.from({length: 26}, (_, i) => 0);
var count2 = Array.from({length: 26}, (_, i) => 0);
// count frequency of each character
// in first string
for (i = 0; i < str1.length ; i++)
count1[str1.charAt(i).charCodeAt(0) -'a'.charCodeAt(0)]++;
// count frequency of each character
// in second string
for (i = 0; i < str2.length ; i++)
count2[str2.charAt(i).charCodeAt(0) -'a'.charCodeAt(0)]++;
// traverse count arrays to find
// number of characters to be removed
var result = 0;
for (i = 0; i < 26; i++)
result += Math.abs(count1[i] -
count2[i]);
return result;
}
// Driver program to run the case
var str1 = "bcadeh", str2 = "hea";
document.write(remAnagram(str1, str2));
// This code is contributed by Amit Katiyar
</script>
C++
// C++ implementation to count number of deletions
// required from two strings to create an anagram
#include <bits/stdc++.h>
using namespace std;
const int CHARS = 26;
int countDeletions(string str1, string str2)
{
int arr[CHARS] = {0};
for (int i = 0; i < str1.length(); i++)
arr[str1[i] - 'a']++;
for (int i = 0; i < str2.length(); i++)
arr[str2[i] - 'a']--;
long long int ans = 0;
for(int i = 0; i < CHARS; i++)
ans +=abs(arr[i]);
return ans;
}
int main() {
string str1 = "bcadeh", str2 = "hea";
cout << countDeletions(str1, str2);
return 0;
}
Java
// Java implementation to count number of deletions
// required from two strings to create an anagram
class GFG {
final static int CHARS = 26;
static int countDeletions(String str1, String str2) {
int arr[] = new int[CHARS];
for (int i = 0; i < str1.length(); i++) {
arr[str1.charAt(i) - 'a']++;
}
for (int i = 0; i < str2.length(); i++) {
arr[str2.charAt(i) - 'a']--;
}
int ans = 0;
for (int i = 0; i < CHARS; i++) {
ans += Math.abs(arr[i]);
}
return ans;
}
static public void main(String[] args) {
String str1 = "bcadeh", str2 = "hea";
System.out.println(countDeletions(str1, str2));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python3 program to find minimum
# number of characters to be
# removed to make two strings
# anagram.
# function to calculate minimum
# numbers of characters to be
# removed to make two strings anagram
def makeAnagram(a, b):
buffer = [0] * 26
for char in a:
buffer[ord(char) - ord('a')] += 1
for char in b:
buffer[ord(char) - ord('a')] -= 1
return sum(map(abs, buffer))
# Driver Code
if __name__ == "__main__" :
str1 = "bcadeh"
str2 = "hea"
print(makeAnagram(str1, str2))
# This code is contributed
# by Raghib Ahsan
C#
// C# implementation to count number of deletions
// required from two strings to create an anagram
using System;
public class GFG {
readonly static int CHARS = 26;
static int countDeletions(String str1, String str2) {
int []arr = new int[CHARS];
for (int i = 0; i < str1.Length; i++) {
arr[str1[i]- 'a']++;
}
for (int i = 0; i < str2.Length; i++) {
arr[str2[i] - 'a']--;
}
int ans = 0;
for (int i = 0; i < CHARS; i++) {
ans += Math.Abs(arr[i]);
}
return ans;
}
static public void Main() {
String str1 = "bcadeh", str2 = "hea";
Console.WriteLine(countDeletions(str1, str2));
}
}
//This code is contributed by PrinciRaj1992
PHP
<?php
// PHP implementation to count number of deletions
// required from two strings to create an anagram
function countDeletions($str1, $str2)
{
$CHARS = 26;
$arr = array();
for ($i = 0; $i < strlen($str1); $i++)
{
$arr[ord($str1[$i]) - ord('a')]++;
}
for ($i = 0; $i < strlen($str2); $i++)
{
$arr[ord($str2[$i]) - ord('a')]--;
}
$ans = 0;
for ($i = 0; $i < $CHARS; $i++)
{
$ans += abs($arr[$i]);
}
return $ans;
}
// Driver Code
$str1 = "bcadeh"; $str2 = "hea";
echo(countDeletions($str1, $str2));
// This code is contributed by Code_Mech
?>
Javascript
<script>
// Javascript implementation to count
// number of deletions required from
// two strings to create an anagram
CHARS = 26;
function countDeletions(str1, str2)
{
var arr = Array.from({length: CHARS}, (_, i) => 0);
for(i = 0; i < str1.length; i++)
{
arr[str1.charAt(i).charCodeAt(0) -
'a'.charCodeAt(0)]++;
}
for(i = 0; i < str2.length; i++)
{
arr[str2.charAt(i).charCodeAt(0) -
'a'.charCodeAt(0)]--;
}
var ans = 0;
for(i = 0; i < CHARS; i++)
{
ans += Math.abs(arr[i]);
}
return ans;
}
// Driver code
str1 = "bcadeh", str2 = "hea";
document.write(countDeletions(str1, str2));
// This code is contributed by Rajput-Ji
</script>
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA