Compruebe si la string dada es un subarreglo de prefijo de la array dada

Dada una string str y una array de palabras word[] , la tarea es encontrar si str es una string de prefijos de word[] .

Ejemplos:

Entrada: str = “indiaismycountry”,  
palabra[] = {“india”, “es”, “mi”, “país”, “y”, “yo”, “amor”, “india”}
Salida: verdadero
Explicación: La string str se puede hacer concatenando «india», «es», «mi» y «país» juntos.

Entrada: str = “indianismo”,  
palabra[] = {“india”, “es”, “mi”, “país”, “y”, “yo”, “amor”, “india”}
Salida: false
Explicación: Es imposible hacer str usando los prefijos de la array de palabras.

Enfoque: Este es un problema simple relacionado con la implementación. Siga los pasos que se mencionan a continuación:

• Tome una string vacía llamada ans .
• Repita la array de palabras y siga agregando cada elemento de la array de palabras a ans .
• Después de agregar a ans comparándolo con s , si ambos coinciden, simplemente devuelven verdadero; de lo contrario, continúa.
• Si la iteración finaliza y ans no coincide con la s , devuelve falso.

A continuación se muestra el programa C++ para implementar el enfoque anterior:

// C++ program to implement the
// given approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check whether string
// is prefix
bool isPrefixString(string s,
                    vector<string>& word)
{
    // ans is taken as an empty string
    string ans = "";
 
    // N is used to store
    // the size of word array
    int N = word.size();
 
    // Iterating over the word array
    for (int i = 0; i < N; i++) {
        // Adding element by element
        // of the array
        ans += word[i];
 
        // If ans and str are same
        // return true
        if (ans == s)
            return true;
    }
 
    // As iteration is ending which means
    // string is not prefix so return false.
    return false;
}
 
// Driver code
int main()
{
    string str = "indiaismycountry";
    vector<string> word
        = { "india", "is", "my",
            "country", "and", "i",
            "love", "india" };
    bool ans = isPrefixString(str, word);
    if (ans)
        cout << "True";
    else
        cout << "False";
    return 0;
}

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to check whether string
  // is prefix
  static Boolean isPrefixString(String s,
                                String word[])
  {
     
    // ans is taken as an empty string
    String ans = "";
 
    // N is used to store
    // the size of word array
    int N = word.length;
 
    // Iterating over the word array
    for (int i = 0; i < N; i++) {
      // Adding element by element
      // of the array
      ans += word[i];
      // If ans and str are same
      // return true
      if (ans.equals(s))
        return true;
    }
 
    // As iteration is ending which means
    // string is not prefix so return false.
    return false;
  }
 
  // Driver code
  public static void main (String[] args)
  {
    String str = "indiaismycountry";
    String word[]
      = { "india", "is", "my",
         "country", "and", "i",
         "love", "india" };
    Boolean ans = isPrefixString(str, word);
    if (ans)
      System.out.println("True");
    else
      System.out.println("False");
  }
}
 
// This code is contributed by hrithikgarg03188.

# Pyhton program to implement the
# given approach
 
# Function to check whether string
# is prefix
def isPrefixString(s, word):
 
    # ans is taken as an empty string
    ans = ""
 
    # N is used to store
    # the size of word array
    N = len(word)
 
    # Iterating over the word array
    for i in range(0, N):
 
        # Adding element by element
        # of the array
        ans = ans + (word[i])
 
        # If ans and str are same
        # return true
        if (ans == s):
            return True
 
    # As iteration is ending which means
    # string is not prefix so return false.
    return False
 
# Driver code
str = "indiaismycountry"
word = ["india", "is", "my", "country", "and", "i", "love", "india"]
 
ans = isPrefixString(str, word)
if (ans == True):
    print("True")
else:
    print("False")
 
# This code is contributed by Samim Hossain Mondal.

// C# program for the above approach
using System;
class GFG {
 
  // Function to check whether string
  // is prefix
  static bool isPrefixString(string s,
                             string []word)
  {
 
    // ans is taken as an empty string
    string ans = "";
 
    // N is used to store
    // the size of word array
    int N = word.Length;
 
    // Iterating over the word array
    for (int i = 0; i < N; i++)
    {
 
      // Adding element by element
      // of the array
      ans += word[i];
 
      // If ans and str are same
      // return true
      if (ans.Equals(s))
        return true;
    }
 
    // As iteration is ending which means
    // string is not prefix so return false.
    return false;
  }
 
  // Driver code
  public static void Main ()
  {
    string str = "indiaismycountry";
    string []word
      = { "india", "is", "my",
         "country", "and", "i",
         "love", "india" };
    bool ans = isPrefixString(str, word);
    if (ans)
      Console.WriteLine("True");
    else
      Console.WriteLine("False");
  }
}
 
// This code is contributed by Samim Hossain Mondal.

<script>
      // JavaScript code for the above approach
 
      // Function to check whether string
      // is prefix
      function isPrefixString(s, word)
      {
       
          // ans is taken as an empty string
          let ans = "";
 
          // N is used to store
          // the size of word array
          let N = word.length;
 
          // Iterating over the word array
          for (let i = 0; i < N; i++)
          {
           
              // Adding element by element
              // of the array
              ans += word[i];
 
              // If ans and str are same
              // return true
              if (ans == s)
                  return true;
          }
 
          // As iteration is ending which means
          // string is not prefix so return false.
          return false;
      }
 
      // Driver code
      let str = "indiaismycountry";
      let word
          = ["india", "is", "my",
              "country", "and", "i",
              "love", "india"];
      let ans = isPrefixString(str, word);
      if (ans)
          document.write("True");
      else
          document.write("False");
 
     // This code is contributed by Potta Lokesh
  </script>

True

Complejidad de tiempo: O(N), N es el tamaño de la array de palabras.
Complejidad espacial: O(M) donde M es la longitud de str