Comprobar si una palabra está presente en una oración

Dada una oración como una string str y una palabra palabra , la tarea es verificar si la palabra está presente en str o no. Una oración es una string compuesta de varias palabras y cada palabra está separada por espacios.
Ejemplos: 

Entrada: str = «Geeks for Geeks», word = «Geeks» 
Salida: Word está presente en la oración 

Entrada: str = «Geeks for Geeks», word = «eeks» 
Salida: Word no está presente en la oración 

Enfoque: en este algoritmo, se usa stringstream para dividir la oración en palabras y luego comparar cada palabra individual de la oración con la palabra dada. Si se encuentra la palabra, la función devuelve verdadero. 
Tenga en cuenta que esta implementación no busca una subsecuencia o substring, solo busca una sola palabra completa en una oración.
A continuación se muestra la implementación del enfoque de búsqueda que distingue entre mayúsculas y minúsculas: 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if the word is found
bool isWordPresent(string sentence, string word)
{
    // To break the sentence in words
    stringstream s(sentence);
 
    // To temporarily store each individual word
    string temp;
 
    while (s >> temp) {
 
        // Comparing the current word
        // with the word to be searched
        if (temp.compare(word) == 0) {
            return true;
        }
    }
    return false;
}
 
// Driver code
int main()
{
    string s = "Geeks for Geeks";
    string word = "Geeks";
 
    if (isWordPresent(s, word))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

// Java implementation of the approach
class GFG
{
 
// Function that returns true if the word is found
static boolean isWordPresent(String sentence, String word)
{
    // To break the sentence in words
    String []s = sentence.split(" ");
 
    // To temporarily store each individual word
    for ( String temp :s)
    {
 
        // Comparing the current word
        // with the word to be searched
        if (temp.compareTo(word) == 0)
        {
            return true;
        }
    }
    return false;
}
 
// Driver code
public static void main(String[] args)
{
    String s = "Geeks for Geeks";
    String word = "Geeks";
 
    if (isWordPresent(s, word))
        System.out.print("Yes");
    else
        System.out.print("No");
 
}
}
 
// This code is contributed by PrinciRaj1992

# Python3 implementation of the approach
 
# Function that returns true if the word is found
def isWordPresent(sentence, word):
     
    # To break the sentence in words
    s = sentence.split(" ")
 
    for i in s:
 
        # Comparing the current word
        # with the word to be searched
        if (i == word):
            return True
    return False
 
# Driver code
s = "Geeks for Geeks"
word = "Geeks"
 
if (isWordPresent(s, word)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by mohit kumar 29

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function that returns true if the word is found
static bool isWordPresent(String sentence, String word)
{
    // To break the sentence in words
    String []s = sentence.Split(' ');
 
    // To temporarily store each individual word
    foreach(String temp in s)
    {
 
        // Comparing the current word
        // with the word to be searched
        if (temp.CompareTo(word) == 0)
        {
            return true;
        }
    }
    return false;
}
 
// Driver code
public static void Main(String[] args)
{
    String s = "Geeks for Geeks";
    String word = "Geeks";
 
    if (isWordPresent(s, word))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by 29AjayKumar

<script>
 
// JavaScript implementation of the approach
 
// Function that returns true if the word is found
function isWordPresent(sentence, word)
{
     // To break the sentence in words
    let s = sentence.split(" ");
  
    // To temporarily store each individual word
    for ( let temp=0;temp<s.length;temp++)
    {
  
        // Comparing the current word
        // with the word to be searched
        if (s[temp] == (word) )
        {
            return true;
        }
    }
    return false;
}
 
// Driver code
let s = "Geeks for Geeks";
let  word = "Geeks";
 
if (isWordPresent(s, word))
        document.write("Yes");
    else
        document.write("No");
 
 
// This code is contributed by patel2127
 
</script>

Yes

Complejidad temporal: O(n) donde n es la longitud de la oración.

Espacio auxiliar: O(n) donde n es la longitud de la string.

A continuación se muestra la implementación del enfoque de búsqueda que no distingue entre mayúsculas y minúsculas: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if the word is found
bool isWordPresent(string sentence, string word)
{
    // To convert the word in uppercase
    transform(word.begin(),
              word.end(), word.begin(), ::toupper);
 
    // To convert the complete sentence in uppercase
    transform(sentence.begin(), sentence.end(),
              sentence.begin(), ::toupper);
 
    // Both strings are converted to the same case,
    // so that the search is not case-sensitive
 
    // To break the sentence in words
    stringstream s(sentence);
 
    // To store the individual words of the sentence
    string temp;
 
    while (s >> temp) {
 
        // Compare the current word
        // with the word to be searched
        if (temp.compare(word) == 0) {
            return true;
        }
    }
    return false;
}
 
// Driver code
int main()
{
    string s = "Geeks for Geeks";
    string word = "geeks";
 
    if (isWordPresent(s, word))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function that returns true if the word is found
static boolean isWordPresent(String sentence,
                            String word)
{
    // To convert the word in uppercase
    word = transform(word);
 
    // To convert the complete sentence in uppercase
    sentence = transform(sentence);
 
    // Both Strings are converted to the same case,
    // so that the search is not case-sensitive
 
    // To break the sentence in words
    String []s = sentence.split(" ");
 
    // To store the individual words of the sentence
    for ( String temp :s)
    {
 
        // Comparing the current word
        // with the word to be searched
        if (temp.compareTo(word) == 0)
        {
            return true;
        }
    }
    return false;
}
 
static String transform(String word)
{
    return word.toUpperCase();
}
 
// Driver code
public static void main(String[] args)
{
    String s = "Geeks for Geeks";
    String word = "geeks";
 
    if (isWordPresent(s, word))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}
 
// This code is contributed by PrinciRaj1992

# Python3 implementation of the approach
 
# Function that returns true if the word is found
def isWordPresent(sentence, word) :
     
    # To convert the word in uppercase
    word = word.upper()
 
    # To convert the complete sentence in uppercase
    sentence = sentence.upper()
 
    # Both strings are converted to the same case,
    # so that the search is not case-sensitive
 
    # To break the sentence in words
    s = sentence.split();
 
    for temp in s :
 
        # Compare the current word
        # with the word to be searched
        if (temp == word) :
            return True;
 
    return False;
 
# Driver code
if __name__ == "__main__" :
 
    s = "Geeks for Geeks";
    word = "geeks";
 
    if (isWordPresent(s, word)) :
        print("Yes");
    else :
        print("No");
 
# This code is contributed by AnkitRai01

// C# implementation of the approach
using System;
 
class GFG
{
 
// Function that returns true if the word is found
static bool isWordPresent(String sentence,
                            String word)
{
    // To convert the word in uppercase
    word = transform(word);
 
    // To convert the complete sentence in uppercase
    sentence = transform(sentence);
 
    // Both Strings are converted to the same case,
    // so that the search is not case-sensitive
 
    // To break the sentence in words
    String []s = sentence.Split(' ');
 
    // To store the individual words of the sentence
    foreach ( String temp in s)
    {
 
        // Comparing the current word
        // with the word to be searched
        if (temp.CompareTo(word) == 0)
        {
            return true;
        }
    }
    return false;
}
 
static String transform(String word)
{
    return word.ToUpper();
}
 
// Driver code
public static void Main(String[] args)
{
    String s = "Geeks for Geeks";
    String word = "geeks";
 
    if (isWordPresent(s, word))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}
 
// This code is contributed by 29AjayKumar

<script>
// Javascript implementation of the approach
 
// Function that returns true if the word is found
function isWordPresent(sentence,word)
{
    // To convert the word in uppercase
    word = transform(word);
  
    // To convert the complete sentence in uppercase
    sentence = transform(sentence);
  
    // Both Strings are converted to the same case,
    // so that the search is not case-sensitive
  
    // To break the sentence in words
    let s = sentence.split(" ");
  
    // To store the individual words of the sentence
    for ( let temp=0;temp<s.length;temp++)
    {
  
        // Comparing the current word
        // with the word to be searched
        if (s[temp] == (word))
        {
            return true;
        }
    }
    return false;
}
 
function transform(word)
{
    return word.toUpperCase();
}
 
// Driver code
let s = "Geeks for Geeks";
let word = "geeks";
if (isWordPresent(s, word))
    document.write("Yes");
else
    document.write("No");
 
 
 
// This code is contributed by unknown2108
</script>

Yes

Complejidad temporal: O(longitud(es))

Espacio auxiliar: O(1)

Método n.º 3: Uso de las funciones integradas de Python:

• Como todas las palabras en una oración están separadas por espacios.
• Tenemos que dividir la oración por espacios usando split().
• Dividimos todas las palabras por espacios y las almacenamos en una lista.
• Usamos la función count() para verificar si la palabra está en la array
• Si el valor de la cuenta es mayor que 0, la palabra está presente en la string

A continuación se muestra la implementación:

# Python3 implementation of the approach
 
# Function that returns true
# if the word is found
def isWordPresent(sentence, word):
 
    # To convert the word in uppercase
    word = word.upper()
 
    # To convert the complete
    # sentence in uppercase
    sentence = sentence.upper()
 
    # splitting the sentence to list
    lis = sentence.split()
    # checking if word is present
    if(lis.count(word) > 0):
        return True
    else:
        return False
 
 
# Driver code
s = "Geeks for Geeks"
word = "geeks"
if (isWordPresent(s, word)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by vikkycirus

<script>
 
// JavaScript implementation of the approach
 
// Function that returns true
// if the word is found
function isWordPresent(sentence, word){
 
    // To convert the word in uppercase
    word = word.toUpperCase()
 
    // To convert the complete
    // sentence in uppercase
    sentence = sentence.toUpperCase()
 
    // splitting the sentence to list
    let lis = sentence.split(' ')
    // checking if word is present
    if(lis.indexOf(word) != -1)
        return true
    else
        return false
}
 
// Driver code
let s = "Geeks for Geeks"
let word = "geeks"
if (isWordPresent(s, word))
    document.write("Yes","</br>")
else
    document.write("No","</br>")
 
// This code is contributed by shinjanpatra
 
</script>

Producción:

Yes

Complejidad temporal: O(longitud(es))

Espacio auxiliar: O(1)