The basic idea of this approach is to check each word of the given sentence ‘S’ if it matches with the given word ‘W’.

Consider the following steps:

1. Start iterating through each character of sentence string ‘S’ using a variable ‘i’ such that 0 <= ‘i’ < |S|
   • Create a string “temp” which stores the current word.
   • Add all the subsequent characters of the sentence till space is detected or if all end of the string is reached.
   • Check if the word ‘W’ matches with “temp”. Return true if it matches.
2. After the loop ends return false because the word is not present in the sentence.

O(|S|), where |S| is the length of the string.

Since we are using a variable to store the words of the sentence string and in the worst case, the whole string can be a single word. So the overall space complexity will be O(|S|).

O(|S|), where |S| is the length of the string.

Since we are iterating through the input sentence string ‘S’ once and checking if the word matches with the given word ‘W’. So the overall time complexity will be O(|S|). Please refer here for more details link

/*

        Time Complexity: O(N)

        Space Complexity: O(1)



       where N is the length input sentence S.

*/



public class Solution 

{



	public static boolean findWord(String S, String W) 

	{



		int n = S.length();



		for (int i = 0; i < n; i++) {



			StringBuilder temp = new StringBuilder();



			while (i < n && S.charAt(i) != ' ') 

			{

				temp.append(S.charAt(i));

				i++;

			}



			if (temp.toString().compareTo(W) == 0) 

			{

				return true;

			}

		}

		return false;

	}

}

'''

    Time Complexity: O(|S|)

    Space Complexity: O(|S|)



    where |S| is the length of the input sentence.

'''



def findWord(s, w):

    n = len(s)



    i = 0



    while i < n:

        #  To store the current word.

        temp = ""



        #  Processing the string to find individual words.

        while (i < n and s[i] != ' '):

            temp += s[i]

            i += 1



        # Checking if the current word is equal to W or not.

        if (temp == w):

            return True

        i += 1

        

    # If no word matches, then return false.

    return False

/*
    Time Complexity: O(|S|)
    Space Complexity: O(|S|)

    where |S| is the length of the input sentence.
*/

bool findWord(string &s, string w)
{
	int n = s.size();

	for (int i = 0; i < n; i++)
	{

		//	To store the current word.
		string temp = "";

		//	Processing the string to find individual words.
		while (i < n && s[i] != ' ')
		{
			temp += s[i];
			i++;
		}

		// Checking if the current word is equal to W or not.
		if (temp == w)
		{
			return true;
		}
	}

	// If no word matches, then return false.
	return false;
}

In the last approach, we were explicitly storing the words of the sentence in a set, but there are some in-built libraries (like stringstream in C++) that can do this ...read more