Hacer una string a partir de otra borrando y reorganizando caracteres

Dadas dos strings, encuentre si podemos hacer la primera string a partir de la segunda eliminando algunos caracteres de la segunda y reorganizando los caracteres restantes.

Ejemplos: 

Input : s1 = ABHISHEKsinGH
      : s2 = gfhfBHkooIHnfndSHEKsiAnG
Output : Possible

Input : s1 = Hello
      : s2 = dnaKfhelddf
Output : Not Possible

Input : s1 = GeeksforGeeks
      : s2 = rteksfoGrdsskGeggehes
Output : Possible

Básicamente, necesitamos encontrar si una string contiene caracteres que son un subconjunto de caracteres en la segunda string. Primero contamos las ocurrencias de todos los caracteres en la segunda string. Luego recorremos la primera string y reducimos el conteo de cada carácter que está presente en la primera. Si en algún momento, la cuenta se vuelve menor que 0, devolvemos falso. Si todos los recuentos siguen siendo mayores o iguales a 0, devolvemos verdadero. 

Implementación:

C++

// CPP program to find if it is possible to
// make a string from characters present in
// other string.
#include <bits/stdc++.h>
using namespace std;
 
const int MAX_CHAR = 256;
 
// Returns true if it is possible to make
// s1 from characters present in s2.
bool isPossible(string& s1, string& s2)
{
    // Count occurrences of all characters
    // present in s2..
    int count[MAX_CHAR] = { 0 };
    for (int i = 0; i < s2.length(); i++)
        count[s2[i]]++;
 
    // For every character, present in s1,
    // reduce its count if count is more
    // than 0.
    for (int i = 0; i < s1.length(); i++) {
        if (count[s1[i]] == 0)
            return false;
        count[s1[i]]--;
    }
 
    return true;
}
 
// Driver code
int main()
{
    string s1 = "GeeksforGeeks",
           s2 = "rteksfoGrdsskGeggehes";
    if (isPossible(s1, s2))
        cout << "Possible";
    else
        cout << "Not Possible\n";
    return 0;
}

Java

// Java program to find if it is possible to
// make a string from characters present in
// other string.
class StringfromCharacters
{
    static final int MAX_CHAR = 256;
 
    // Returns true if it is possible to make
    // s1 from characters present in s2.
    static boolean isPossible(String s1, String s2)
    {
        // Count occurrences of all characters
        // present in s2..
        int count[] = new int[MAX_CHAR];
        for (int i = 0; i < s2.length(); i++)
            count[(int)s2.charAt(i)]++;
 
        // For every character, present in s1,
        // reduce its count if count is more
        // than 0.
        for (int i = 0; i < s1.length(); i++) {
            if (count[(int)s1.charAt(i)] == 0)
                return false;
            count[(int)s1.charAt(i)]--;
        }
     
        return true;
    }
     
    // Driver code
    public static void main(String args[])
    {
        String s1 = "GeeksforGeeks",
            s2 = "rteksfoGrdsskGeggehes";
        if (isPossible(s1, s2))
            System.out.println("Possible");
        else
            System.out.println("Not Possible");
    }
}
 
/* This code is contributed by Danish Kaleem */

Python 3

# Python 3 program to find if it is possible
# to make a string from characters present
# in other string.
MAX_CHAR = 256
 
# Returns true if it is possible to make
# s1 from characters present in s2.
def isPossible(s1, s2):
 
    # Count occurrences of all characters
    # present in s2..
    count = [0] * MAX_CHAR
    for i in range(len(s2)):
        count[ord(s2[i])] += 1
 
    # For every character, present in s1,
    # reduce its count if count is more
    # than 0.
    for i in range(len(s1)):
        if (count[ord(s1[i])] == 0):
            return False
        count[ord(s1[i])] -= 1
 
    return True
 
# Driver code
if __name__ == "__main__":
     
    s1 = "GeeksforGeeks"
    s2 = "rteksfoGrdsskGeggehes"
    if (isPossible(s1, s2)):
        print("Possible")
    else:
        print("Not Possible")
 
# This code is contributed by ita_c

C#

// C# program to find if it is possible to
// make a string from characters present
// in other string.
using System;
 
class GFG {
     
    static int MAX_CHAR = 256;
 
    // Returns true if it is possible to
    // make s1 from characters present
    // in s2.
    static bool isPossible(String s1, String s2)
    {
         
        // Count occurrences of all characters
        // present in s2..
        int []count = new int[MAX_CHAR];
         
        for (int i = 0; i < s2.Length; i++)
            count[(int)s2[i]]++;
 
        // For every character, present in s1,
        // reduce its count if count is more
        // than 0.
        for (int i = 0; i < s1.Length; i++)
        {
            if (count[(int)s1[i]] == 0)
                return false;
                 
            count[(int)s1[i]]--;
        }
     
        return true;
    }
     
    // Driver code
    public static void Main()
    {
        string s1 = "GeeksforGeeks",
               s2 = "rteksfoGrdsskGeggehes";
                
        if (isPossible(s1, s2))
            Console.WriteLine("Possible");
        else
            Console.WriteLine("Not Possible");
    }
}
 
// This code is contributed by vt_m.

Javascript

<script>
// Javascript program to find if it is possible to
// make a string from characters present in
// other string.
     
    let MAX_CHAR = 256;
     
    // Returns true if it is possible to make
    // s1 from characters present in s2.
    function isPossible(s1, s2)
    {
        // Count occurrences of all characters
        // present in s2..
        let count = new Array(MAX_CHAR);
        for(let i = 0; i < count.length; i++)
        {
            count[i] = 0;
        }
        for (let i = 0; i < s2.length; i++)
            count[s2[i].charCodeAt(0)]++;
   
        // For every character, present in s1,
        // reduce its count if count is more
        // than 0.
        for (let i = 0; i < s1.length; i++) {
            if (count[s1[i].charCodeAt(0)] == 0)
                return false;
            count[s1[i].charCodeAt(0)]--;
        }
       
        return true;
    }
     
    // Driver code
    let s1 = "GeeksforGeeks",
            s2 = "rteksfoGrdsskGeggehes";
    if (isPossible(s1, s2))
        document.write("Possible");
    else
        document.write("Not Possible");
         
    // This code is contributed by avanitrachhadiya2155
</script>
Producción

Possible

La complejidad del tiempo es: O(n)

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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

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