检查是否可以使用给定的单词字符串两个字符

给定一个由两个字符组成的字符串和由两个字符组成的 n 个不同单词。任务是找出是否有可能以这样一种方式排列给定的单词，即连接的字符串将给定的两个字符作为子字符串。我们可以多次附加一个单词。
例子：

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.

如果我们仔细查看给定的示例，我们可以看到，如果以下任何条件为真，我们的答案将是“是”，

str 等于 N 个单词中的任何一个
str 等于任何单词的反转。
str 的第一个字母等于任何给定 N 个字符串的最后一个字母，最后一个字母等于任何给定 N 个字符串的第一个字母。

否则我们的输出将永远是 NO。
下面是上述方法的实现。

C++

// CPP code to check if a two character string can
// be made using given strings
#include 
using namespace std;
 
// Function to check if str can be made using
// given words
bool makeAndCheckString(vector 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 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 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 words = new Vector(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 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 words = new List(array);    
    if (makeAndCheckString(words, str))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by Princi Singh

PHP

Javascript

输出：

YES