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Check if a two character string can be made using given words

  • Difficulty Level : Easy
  • Last Updated : 31 May, 2021

Given a string of two characters and n distinct words of two characters. The task is to find if it is possible to arrange given words in such a way that the concatenated string has the given two character string as a substring. We can append a word multiple times.
Examples: 
 

Input : str = "ya"
        words[] = {"ah", "oy", "to", "ha"} 
Output : YES
We can join "oy" and then "ah", and
then "ha" to form the string "oyahha" 
which contains the string "ya".
 So, the answer is "YES"

Input : str[] = "ha"
        words[] = "ah"
Output :YES
The string "ahah" contains "ha" 
as a substring.

Input : str = "hp"
       words[] = {"ht", "tp"|
Output :NO
We can't produce a string containing
"hp" as a sub-string. Note that we
can join "ht" and then "tp" producing
"http", but it doesn't contain the 
"hp" as a sub-string.

 

If we look at the given examples carefully, we can see that our answer will be “YES” if any of the following conditions is true, 
 

  1. str is equal to any one of the N words
  2. str is equal to reverse of any of the words.
  3. It first letter of str is equal to last letter of any of the given N strings and last letter is equal to the first letter of any of the given N strings.

Otherwise our output will always be NO.
Below is the implementation of the above approach.
 

C++




// CPP code to check if a two character string can
// be made using given strings
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if str can be made using
// given words
bool makeAndCheckString(vector<string> words, string str)
{
    int n = words.size();
    bool first = false, second = false;
 
    for (int i = 0; i < n; i++) {
 
        // If str itself is present
        if (words[i] == str)
            return true;
     
        // Match first character of str
        // with second of word and vice versa
        if (str[0] == words[i][1])
            first = true;           
        if (str[1] == words[i][0])
            second = true;
 
        // If both characters found.
        if (first && second)
            return true;
    }
     
    return false;
}
 
// Driver Code
int main()
{
    string str = "ya";        
    vector<string> words = { "ah", "oy", "to", "ha"};    
    if (makeAndCheckString(words, str))
       cout << "Yes";
    else
       cout << "No";
    return 0;
}

Java




// Java code to check if a two character string can
// be made using given strings
import java.util.*;
 
class GFG
{
 
// Function to check if str can be made using
// given words
static boolean makeAndCheckString(Vector<String> words,
                                            String str)
{
    int n = words.size();
    boolean first = false, second = false;
 
    for (int i = 0; i < n; i++)
    {
 
        // If str itself is present
        if (words.get(i) == str)
            return true;
     
        // Match first character of str
        // with second of word and vice versa
        if (str.charAt(0) == words.get(i).charAt(1))
            first = true;        
        if (str.charAt(1) == words.get(i).charAt(0))
            second = true;
 
        // If both characters found.
        if (first && second)
            return true;
    }
     
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
    String str = "ya";
    String[] array = { "ah", "oy", "to", "ha"};
    Vector<String> words = new Vector<String>(Arrays.asList(array));    
    if (makeAndCheckString(words, str))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 code to check if a two character string can 
# be made using given strings
 
# Function to check if str can be made using
# given words
def makeAndCheckString(words, str):
    n = len(words)
    first = second = False
 
    for i in range(n):
        # If str itself is present
        if words[i]==str:
            return True
 
        # Match first character of str
        # with second of word and vice versa
        if str[0] == words[i][1]:
            first = True
        if str[1] == words[i][0]:
            second = True
 
        # If both characters found.
        if first and second:
            return True
     
    return False
     
# Driver Code
str = 'ya'
words = ['ah', 'oy', 'to', 'ha']
if makeAndCheckString(words, str):
    print('YES')
else:
    print('NO')
 
# This code is contributed 
# by SamyuktaSHegde

C#




// C# code to check if a two character string can
// be made using given strings
using System;
using System.Collections.Generic;
 
class GFG
{
 
// Function to check if str can be made using
// given words
static bool makeAndCheckString(List<String> words,
                                            String str)
{
    int n = words.Count;
    bool first = false, second = false;
 
    for (int i = 0; i < n; i++)
    {
 
        // If str itself is present
        if (words[i] == str)
            return true;
     
        // Match first character of str
        // with second of word and vice versa
        if (str[0] == words[i][1])
            first = true;        
        if (str[1] == words[i][0])
            second = true;
 
        // If both characters found.
        if (first && second)
            return true;
    }
     
    return false;
}
 
// Driver Code
public static void Main(String[] args)
{
    String str = "ya";
    String[] array = { "ah", "oy", "to", "ha"};
    List<String> words = new List<String>(array);    
    if (makeAndCheckString(words, str))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by Princi Singh

PHP




<?php
// PHP code to check if a two character string can
// be made using given strings
 
// Function to check if str can be made using
// given words
function makeAndCheckString($words, $str)
{
    $n = sizeof($words) ;
    $first = false ;
    $second = false;
 
    for ($i = 0; $i < $n; $i++) {
 
        // If str itself is present
        if ($words[$i] == $str)
            return true;
     
        // Match first character of str
        // with second of word and vice versa
        if ($str[0] == $words[$i][1])
            $first = true;            
        if ($str[1] == $words[$i][0])
            $second = true;
 
        // If both characters found.
        if ($first && $second)
            return true;
    }
     
    return false;
}
 
    // Driver Code 
    $str = "ya";        
    $words = array( "ah", "oy", "to", "ha") ;
    if (makeAndCheckString($words, $str))
        echo "Yes";
    else
        echo "No";
 
    // This code is contributed by Ryuga
?>

Javascript




<script>
    // Javascript code to check if a two character string can
    // be made using given strings
     
    // Function to check if str can be made using
    // given words
    function makeAndCheckString(words, str)
    {
        let n = words.length;
        let first = false, second = false;
 
        for (let i = 0; i < n; i++)
        {
 
            // If str itself is present
            if (words[i] == str)
                return true;
 
            // Match first character of str
            // with second of word and vice versa
            if (str[0] == words[i][1])
                first = true;        
            if (str[1] == words[i][0])
                second = true;
 
            // If both characters found.
            if (first && second)
                return true;
        }
 
        return false;
    }
     
    let str = "ya";
    let words = [ "ah", "oy", "to", "ha"];   
    if (makeAndCheckString(words, str))
        document.write("YES");
    else
        document.write("NO");
     
    // This code is contributed by suresh07.
</script>

Output: 
 



YES

This article is contributed by Sarthak Kohli. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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