Dadas dos strings ‘s’ y ‘q’, verifique si todos los caracteres de q están presentes en ‘s’.
Ejemplos:
Example:
Input: s = "abctd"
q = "cat"
Output: Yes
Explanation:
All characters of "cat" are
present in "abctd"
Input: s = dog
hod
Output: No
Explanation:
Given query 'hod' hod has the letter
'h' which is not available in set 'dog',
hence the output is no.
Una solución simple es probar todos los personajes uno por uno. Encuentre su número de ocurrencias en ‘q’, luego en ‘s’. El número de ocurrencias en ‘q’ debe ser menor o igual al mismo en ‘s’. Si todos los caracteres cumplen esta condición, devuelve verdadero. De lo contrario, devuelve falso.
Una solución eficiente es crear una array de frecuencias de longitud 256 (Número de caracteres posibles) e inicializarla en 0. Luego calculamos la frecuencia de cada elemento presente en ‘s’. Después de contar los caracteres en ‘s’, recorremos ‘q’ y verificamos si la frecuencia de cada carácter es menor que su frecuencia en ‘s’, reduciendo su frecuencia en la array de frecuencia construida para ‘s’.
A continuación se muestra la implementación del enfoque anterior.
C++
// CPP program to check if a query string
// is present is given set.
#include <bits/stdc++.h>
using namespace std;
const int MAX_CHAR = 256;
bool isPresent(string s, string q)
{
// Count occurrences of all characters
// in s.
int freq[MAX_CHAR] = { 0 };
for (int i = 0; i < s.length(); i++)
freq[s[i]]++;
// Check if number of occurrences of
// every character in q is less than
// or equal to that in s.
for (int i = 0; i < q.length(); i++) {
freq[q[i]]--;
if (freq[q[i]] < 0)
return false;
}
return true;
}
// driver program
int main()
{
string s = "abctd";
string q = "cat";
if (isPresent(s, q))
cout << "Yes";
else
cout << "No";
return 0;
}
Java
// java program to check if a query
// string is present is given set.
import java.io.*;
public class GFG {
static int MAX_CHAR = 256;
static boolean isPresent(String s, String q)
{
// Count occurrences of all
// characters in s.
int []freq = new int[MAX_CHAR];
for (int i = 0; i < s.length(); i++)
freq[s.charAt(i)]++;
// Check if number of occurrences of
// every character in q is less than
// or equal to that in s.
for (int i = 0; i < q.length(); i++)
{
freq[q.charAt(i)]--;
if (freq[q.charAt(i)] < 0)
return false;
}
return true;
}
// driver program
static public void main (String[] args)
{
String s = "abctd";
String q = "cat";
if (isPresent(s, q))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by vt_m.
Python 3
# Python 3 program to check if a query
# string is present is given set.
MAX_CHAR = 256
def isPresent(s, q):
# Count occurrences of all characters
# in s.
freq = [0] * MAX_CHAR
for i in range(0 , len(s)):
freq[ord(s[i])] += 1
# Check if number of occurrences of
# every character in q is less than
# or equal to that in s.
for i in range(0, len(q)):
freq[ord(q[i])] -= 1
if (freq[ord(q[i])] < 0):
return False
return True
# driver program
s = "abctd"
q = "cat"
if (isPresent(s, q)):
print("Yes")
else:
print("No")
# This code is contributed by Smitha
C#
// C# program to check if a query
// string is present is given set.
using System;
public class GFG {
static int MAX_CHAR = 256;
static bool isPresent(string s, string q)
{
// Count occurrences of all
// characters in s.
int []freq = new int[MAX_CHAR];
for (int i = 0; i < s.Length; i++)
freq[s[i]]++;
// Check if number of occurrences of
// every character in q is less than
// or equal to that in s.
for (int i = 0; i < q.Length; i++)
{
freq[q[i]]--;
if (freq[q[i]] < 0)
return false;
}
return true;
}
// driver program
static public void Main ()
{
string s = "abctd";
string q = "cat";
if (isPresent(s, q))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by vt_m.
PHP
<?php
// PHP program to check if a query string
// is present is given set.
function isPresent($s, $q)
{
// Count occurrences of
// all characters in s.
$freq = array(256);
for ($i = 0; $i < 256; $i++)
$freq[$i] = 0;
for ($i = 0; $i < strlen($s); $i++)
$freq[ ord($s[$i]) - ord('a') ]++ ;
// Check if number of occurrences of
// every character in q is less than
// or equal to that in s.
for ($i = 0; $i < strlen($q); $i++)
{
$freq[ord($q[$i]) - ord('a')]--;
if ($freq[ord($q[$i]) - ord('a')] < 0)
return false;
}
return true;
}
// Driver Code
$s = "abctd";
$q = "cat";
if (isPresent($s, $q))
echo "Yes";
else
echo "No";
// This code is contributed by Sam007
?>
Javascript
<script>
// Javascript program to check if a query
// string is present is given set.
let MAX_CHAR = 256;
function isPresent(s, q)
{
// Count occurrences of all
// characters in s.
let freq = new Array(MAX_CHAR);
freq.fill(0);
for (let i = 0; i < s.length; i++)
freq[s[i]]++;
// Check if number of occurrences of
// every character in q is less than
// or equal to that in s.
for (let i = 0; i < q.length; i++)
{
freq[q[i]]--;
if (freq[q[i]] < 0)
return false;
}
return true;
}
let s = "abctd";
let q = "cat";
if (isPresent(s, q))
document.write("Yes");
else
document.write("No");
</script>
Yes
Complejidad temporal: O(n)
Espacio auxiliar: O(256)
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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