Check if a word is present in a sentence
Last Updated :
12 Jul, 2025
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 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>
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 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>
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:
C++
#include <iostream>
#include <algorithm>
#include <vector>
#include <sstream>
using namespace std;
bool isWordPresent(string sentence, string word)
{
// Convert the word to uppercase
transform(word.begin(), word.end(), word.begin(), ::toupper);
// Convert the sentence to uppercase
transform(sentence.begin(), sentence.end(), sentence.begin(), ::toupper);
// Split the sentence into words
stringstream ss(sentence);
vector<string> words;
string w;
while (ss >> w)
words.push_back(w);
// Check if the word is present
for (const auto& w : words) {
if (w == word) {
return true;
}
}
return false;
}
int main()
{
string s = "Geeks for Geeks";
string word = "geeks";
if (isWordPresent(s, word))
cout << "Yes\n";
else
cout << "No\n";
return 0;
}
// Java implementation of the approach
import java.io.*;
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 = word.toUpperCase();
// To convert the complete
// sentence in uppercase
sentence = sentence.toUpperCase();
// Splitting the sentence to an array
String[] words = sentence.split(" ");
// Checking if word is present
for (String w : words) {
if (w.equals(word)) {
return true;
}
}
return false;
}
public static void main(String[] args)
{
String s = "Geeks for Geeks";
String word = "geeks";
if (isWordPresent(s, word))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by karthik
# 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
// C# implementation of the approach
using System;
public 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 = word.ToUpper();
// To convert the complete sentence in uppercase
sentence = sentence.ToUpper();
// Splitting the sentence to an array
string[] words = sentence.Split(' ');
// Checking if word is present
foreach(string w in words)
{
if (w.Equals(word)) {
return true;
}
}
return false;
}
static public void Main()
{
// Code
string s = "Geeks for Geeks";
string word = "geeks";
if (isWordPresent(s, word))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// This code is contributed by sankar
<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)
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