Check if a word is present in a sentence

Given a sentence as a string str and a word word, the task is to check if the word is present in str or not. A sentence is a string comprised of multiple words and each word is separated with spaces.
Examples:

Input: str = “Geeks for Geeks”, word = “Geeks” 
Output: Word is present in the sentence 

Input: str = “Geeks for Geeks”, word = “eeks” 
Output: Word is not present in the sentence 

Approach: In this algorithm, stringstream is used to break the sentence into words then compare each individual word of the sentence with the given word. If the word is found then the function returns true. 
Note that this implementation does not search for a sub-sequence or sub-string, it only searches for a complete single word in a sentence.
Below is the implementation for the case-sensitive search approach: 

CPP

// 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

// 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

Python

# 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#

// 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

Javascript

<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>

Output:

Yes

Time complexity: O(n) where n is the length of the sentence.

Auxiliary space: O(n) where n is the length of string.

Below is the implementation for the case-insensitive search approach: 

C++

// 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

// 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

# 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#

// 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

Javascript

<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>

Output:

Yes

Time complexity: O(length(s))

Auxiliary space: O(1)

Method #3:Using Built-in Python Functions:

• As all the words in a sentence are separated by spaces.
• We have to split the sentence by spaces using split().
• We split all the words by spaces and store them in a list.
• We use count() function to check whether the word is in array
• If the value of count is greater than 0 then word is present in string

Below is the implementation:

Python3

# 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

Javascript

<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>

Output:

Yes

Time complexity: O(length(s))

Auxiliary space: O(1)