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,
- str is equal to any one of the N words
- str is equal to reverse of any of the words.
- 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 wordsbool 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 Codeint 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 stringsimport java.util.*;class GFG{// Function to check if str can be made using// given wordsstatic 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 Codepublic 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 stringsusing System; using System.Collections.Generic; class GFG{// Function to check if str can be made using// given wordsstatic 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 Codepublic 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
Time Complexity: O(n), where n represents the size of the given vector.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
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