Open In App

Find the word from a given sentence having given word as prefix

Improve
Improve
Like Article
Like
Save
Share
Report

Given a string S representing a sentence and another string word, the task is to find the word from S which has the string word as a prefix. If no such word is present in the string, print -1.

Examples:

Input: S = “Welcome to Geeksforgeeks”, word=”Gee”
Output: Geeksforgeeks
Explanation:
The word “Geeksforgeeks” in the sentence has the prefix “Gee”.

Input: s=”Competitive Programming”, word=”kdflk”
Output: -1
Explanation:
No word in the string has “kdflk” as its prefix.

 

Approach: Follow the steps below to find which word has the given prefix:

  1. Extract the words from the sentence using the stringstream and store them in a vector of strings.
  2. Now, traverse the array and check which word contains the given word as its own prefix.
  3. If found to be true for any word, then print that word. Otherwise, if no such word is found, print -1.

Below is the implementation of the above approach:

C++




// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the position
// of the string having word as prefix
string isPrefixOfWord(string sentence,
                      string Word)
{
    stringstream ss(sentence);
 
    // Initialize a vector
    vector<string> v;
    string temp;
 
    // Extract words from the sentence
    while (ss >> temp) {
        v.push_back(temp);
    }
 
    // Traverse each word
    for (int i = 0; i < v.size(); i++) {
 
        // Traverse the characters of word
        for (int j = 0; j < v[i].size(); j++) {
 
            // If prefix does not match
            if (v[i][j] != Word[j])
                break;
 
            // Otherwise
            else if (j == Word.size() - 1)
 
                // Return  the word
                return v[i];
        }
    }
 
    // Return -1 if not present
    return "-1";
}
 
// Driver code
int main()
{
    string s = "Welcome to Geeksforgeeks";
    string word = "Gee";
 
    cout << isPrefixOfWord(s, word) << endl;
    return 0;
}


Java




// Java program for the above approach
import java.util.*;
import java.io.*;
import java.lang.Math;
 
class GFG{
      
// Function to find the position
// of the string having word as prefix
static String isPrefixOfWord(String sentence,
                             String Word)
{
    String a[] = sentence.split(" ");
   
    // Initialize an ArrayList
    ArrayList<String> v = new ArrayList<>();
     
    // Extract words from the sentence
    for(String e : a)
        v.add(e);
     
    // Traverse each word
    for(int i = 0; i < v.size(); i++)
    {
         
        // Traverse the characters of word
        for(int j = 0; j < v.get(i).length(); j++)
        {
             
            // If prefix does not match
            if (v.get(i).charAt(j) != Word.charAt(j))
                break;
                 
            // Otherwise
            else if (j == Word.length() - 1)
             
                // Return  the word
                return v.get(i);
        }
    }
     
    // Return -1 if not present
    return "-1";
}
 
// Driver code 
public static void main(final String[] args)
{
    String s = "Welcome to Geeksforgeeks";
    String word = "Gee";
   
    System.out.println(isPrefixOfWord(s, word));
}
}
 
// This code is contributed by bikram2001jha


Python3




# Python3 program for the
# above approach
 
# Function to find the
# position of the string
# having word as prefix
def isPrefixOfWord(sentence, word):
   
    a=sentence.split(" ")
 
    # Initialize an List
    v = []
 
    # Extract words from
    # the sentence
    for i in a:
        v.append(i)
 
    # Traverse each word
    for i in range(len(v)):
 
        # Traverse the characters of word
        for j in range(len(v[i])):
 
            # If prefix does not match
            if(v[i][j] != word[j]):
                break
 
            # Otherwise
            elif(j == len(word) - 1):
 
                # Return  the word
                return v[i]
 
    # Return -1 if not present
    return -1
 
# Driver code
s = "Welcome to Geeksforgeeks"
word = "Gee"
print(isPrefixOfWord(s, word))
 
# This code is contributed by avanitrachhadiya2155


C#




// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
      
// Function to find the position
// of the string having word as prefix
static String isPrefixOfWord(String sentence,
                             String Word)
{
  String []a = sentence.Split(' ');
 
  // Initialize an List
  List<String> v = new List<String>();
 
  // Extract words from the sentence
  foreach(String e in a)
    v.Add(e);
 
  // Traverse each word
  for(int i = 0; i < v.Count; i++)
  {
    // Traverse the characters of word
    for(int j = 0; j < v[i].Length; j++)
    {
      // If prefix does not match
      if (v[i][j] != Word[j])
        break;
 
      // Otherwise
      else if (j == Word.Length - 1)
 
        // Return  the word
        return v[i];
    }
  }
 
  // Return -1 if not present
  return "-1";
}
 
// Driver code 
public static void Main(String[] args)
{
  String s = "Welcome to Geeksforgeeks";
  String word = "Gee";
  Console.WriteLine(isPrefixOfWord(s, word));
}
}
 
// This code is contributed by Rajput-Ji


Javascript




<script>
//Javascript  program for the above approach
 
// Function to find the position
// of the string having word as prefix
function isPrefixOfWord(sentence, Word)
{
    var a = sentence.split(" ");
   
    // Initialize an ArrayList
    var v = [];
     
    // Extract words from the sentence
    //for(String e : a)
    for(var i=0;i<a.length;i++)
    {
        v.push(a[i]);
    }
    // Traverse each word
    for(var i = 0; i < v.length; i++)
    {
         
        // Traverse the characters of word
        for(var j = 0; j < v[i].length; j++)
        {
             
            // If prefix does not match
            if (v[i].charAt(j) != Word[j])
                break;
                 
            // Otherwise
            else if (j == Word.length- 1)
             
                // Return  the word
                return v[i];
        }
    }
     
    // Return -1 if not present
    return "-1";
}
   
var s = "Welcome to Geeksforgeeks";
var word = "Gee";
document.write(isPrefixOfWord(s, word));
 
//This code in contributed by SoumikMondal
</script>


Output: 

Geeksforgeeks

 

Time Complexity: O(L), where L denotes the length of the string S 
Auxiliary Space: O(L)

Approach: Follow the steps below to find which word has the given prefix:

1. Using the split() function, to extract word from the sentences

2. We use in operator to check presence of substring 

C++




#include <iostream>
#include <string>
#include <vector>
#include <sstream>
 
using namespace std;
 
string prefix(string sentence, string word)
{
   
    // Converting the string into an array of words
  // using the stringstream and getline() method
    vector<string> l;
    stringstream ss(sentence);
    string temp;
    while (ss >> temp) {
        l.push_back(temp);
    }
 
    // Initializing the variable to store
  // the last word containing the prefix
    string c = "-1";
 
    // Looping through each word in the array
    for (string i : l)
    {
       
        // Checking if the prefix is a substring of
      // the current word using the find() method
        if (i.find(word) != string::npos) {
            c = i;
        }
    }
 
    // Returning the last word containing the prefix or -1 if not found
    return c;
}
 
// Driver code
int main() {
    string s = "Welcome to Geeksforgeeks";
    string word = "Gee";
 
    cout << prefix(s, word) << endl; // Outputs "Geeksforgeeks"
 
    return 0;
}


Java




import java.util.*;
 
public class Main {
    static String prefix(String sentence, String word) {
        // Converting the string into an array of words using the split() method
        String[] l = sentence.split(" ");
 
        // Initializing the variable to store the last word containing the prefix
        String c = "-1";
 
        // Looping through each word in the array
        for (String i : l) {
            // Checking if the prefix is a substring of the current word using the contains() method
            if (i.contains(word)) {
                c = i;
            }
        }
 
        // Returning the last word containing the prefix or -1 if not found
        return c;
    }
 
    // Driver code
    public static void main(String[] args) {
        String s = "Welcome to Geeksforgeeks";
        String word = "Gee";
 
        System.out.println(prefix(s, word)); // Outputs "Geeksforgeeks"
    }
}


Python3




def prefix(sentence, word):
  # converting string into list
  l=sentence.split()
  c=-1
  for i in l:
    # in operator to check substring
    if word in i:
      c=i
  # Returning the substring or -1
  return c
# input string
s = "Welcome to Geeksforgeeks"
#prefix
word = "Gee"
print(prefix(s, word))
# This code is contributed by Asif shaik


C#




//C# code
using System;
 
public class Program
{
    // Function to return the last word that contains the given prefix
    static string Prefix(string sentence, string word)
    {
        // Converting the string into an array of words using the Split() method
        string[] l = sentence.Split(" ");
 
        // Initializing the variable to store the last word containing the prefix
        string c = "-1";
 
        // Looping through each word in the array
        foreach (string i in l)
        {
            // Checking if the prefix is a substring of the current word using the Contains() method
            if (i.Contains(word))
            {
                c = i;
            }
        }
 
        // Returning the last word containing the prefix or -1 if not found
        return c;
    }
 
    // Driver code
    public static void Main()
    {
        string s = "Welcome to Geeksforgeeks";
        string word = "Gee";
 
        Console.WriteLine(Prefix(s, word)); // Outputs "Geeksforgeeks"
    }
}


Javascript




function prefix(sentence, word) {
  // Converting the string into an array of words using the split() method
  let l = sentence.split(" ");
 
  // Initializing the variable to store the last word containing the prefix
  let c = -1;
 
  // Looping through each word in the array
  for (let i of l) {
    // Checking if the prefix is a substring of the current word using the includes() method
    if (i.includes(word)) {
      c = i;
    }
  }
 
  // Returning the last word containing the prefix or -1 if not found
  return c;
}
 
// Input string
let s = "Welcome to Geeksforgeeks";
// Prefix
let word = "Gee";
 
console.log(prefix(s, word)); // Outputs "Geeksforgeeks"


Output

Geeksforgeeks

Time Complexity: O(L), where L denotes the length of the string S 
Auxiliary Space: O(L)



Last Updated : 15 Mar, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads