检查是否可以通过连接两个给定字符串的子字符串来形成回文字符串

给定两个字符串str1和str2 ，任务是检查是否可以通过串联str1和str2的两个子字符串来形成回文字符串。

例子：

Input: str1 = “abcd”, str2 = “acba”
Output: Yes
Explanation:
There are five possible cases where concatenation of two substrings from str1 and str2 gives palindromic string:
“ab” + “a” = “aba”
“ab” + “ba” = “abba”
“bc” + “cb” = “bccb”
“bc” + “b” = “bcb”
“cd” + “c” = “cdc”

Input: str1 = “pqrs”, str2 = “abcd”
Output: No
Explanation:
There is no possible concatenation of sub-strings from given strings which gives palindromic string.

天真的方法：
解决问题的最简单方法是生成str1和str2 的每个可能的子字符串，并将它们组合起来生成所有可能的连接。对于每个串联，检查它是否是回文的。如果发现是真的，打印“是” 。否则，打印“否” 。
时间复杂度： O(N ² * M ² * (N+M))，其中 N 和 M 分别是 str1 和 str2 的长度。
辅助空间： O(1)

有效的方法：
为了优化上述方法，需要进行以下观察：

If the given strings possess at least one common character, then they will always form a palindromic string on concatenation of the common character from both the strings. 
Illustration:
str1 = “abc”, str2 = “fad” 
Since ‘a’ is common in both strings, a palindromic string “aa” can be obtained. 

请按照以下步骤解决问题：

• 初始化一个布尔数组以标记两个字符串中每个字母的存在。
• 遍历str1并将索引(str1[i] – ‘a’)标记为真。
• 现在，遍历str2并检查是否有任何索引(str2[i] – ‘a’)已标记为true ，打印“Yes” 。
• 完全遍历str2 后，如果没有找到公共字符，则打印“No”。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
#include 
using namespace std;
 
// Function to check if a palindromic
// string can be formed from the
// substring of given strings
bool check(string str1, string str2)
{
    // Boolean array to mark
    // presence of characters
    vector mark(26, false);
 
    int n = str1.size(),
        m = str2.size();
 
    for (int i = 0; i < n; i++) {
 
        mark[str1[i] - 'a'] = true;
    }
 
    // Check if any of the character
    // of str2 is already marked
    for (int i = 0; i < m; i++) {
 
        // If a common character
        // is found
        if (mark[str2[i] - 'a'])
            return true;
    }
 
    // If no common character
    // is found
    return false;
}
 
// Driver Code
int main()
{
 
    string str1 = "abca",
        str2 = "efad";
 
    if (check(str1, str2))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.Arrays;
 
class GFG{
     
// Function to check if a palindromic
// string can be formed from the
// substring of given strings
public static boolean check(String str1,
                            String str2)
{
     
    // Boolean array to mark
    // presence of characters
    boolean[] mark = new boolean[26];
    Arrays.fill(mark, false);
     
    int n = str1.length(),
        m = str2.length();
 
    for(int i = 0; i < n; i++)
    {
        mark[str1.charAt(i) - 'a'] = true;
    }
 
    // Check if any of the character
    // of str2 is already marked
    for(int i = 0; i < m; i++)
    {
 
        // If a common character
        // is found
        if (mark[str2.charAt(i) - 'a'])
            return true;
    }
 
    // If no common character
    // is found
    return false;
}
 
// Driver code
public static void main(String[] args)
{
    String str1 = "abca",
    str2 = "efad";
 
    if (check(str1, str2))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program to implement
# the above approach
     
# Function to check if a palindromic
# string can be formed from the
# substring of given strings
def check(str1, str2):
     
    # Boolean array to mark
    # presence of characters
    mark = [False for i in range(26)]
     
    n = len(str1)
    m = len(str2)
     
    for i in range(n):
        mark[ord(str1[i]) - ord('a')] = True
     
    # Check if any of the character
    # of str2 is already marked
    for i in range(m):
         
        # If a common character
        # is found
        if (mark[ord(str2[i]) - ord('a')]):
            return True;
 
    # If no common character
    # is found
    return False
     
# Driver code
if __name__=="__main__":
     
    str1 = "abca"
    str2 = "efad"
 
    if (check(str1, str2)):
        print("Yes");
    else:
        print("No");
 
# This code is contributed by rutvik_56

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
     
// Function to check if a palindromic
// string can be formed from the
// substring of given strings
public static bool check(String str1,
                        String str2)
{
     
    // Boolean array to mark
    // presence of characters
    bool[] mark = new bool[26];
     
    int n = str1.Length,
        m = str2.Length;
 
    for(int i = 0; i < n; i++)
    {
        mark[str1[i] - 'a'] = true;
    }
 
    // Check if any of the character
    // of str2 is already marked
    for(int i = 0; i < m; i++)
    {
 
        // If a common character
        // is found
        if (mark[str2[i] - 'a'])
            return true;
    }
 
    // If no common character
    // is found
    return false;
}
 
// Driver code
public static void Main(String[] args)
{
    String str1 = "abca",
    str2 = "efad";
 
    if (check(str1, str2))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by amal kumar choubey

Javascript

 
// Javascript Program to implement
// the above approach
 
// Function to check if a palindromic
// string can be formed from the
// substring of given strings
function check(str1, str2)
{
    // Boolean array to mark
    // presence of characters
    var mark = Array(26).fill(false);
 
    var n = str1.length,
        m = str2.length;
 
    for (var i = 0; i < n; i++) {
 
        mark[str1[i] - 'a'] = true;
    }
 
    // Check if any of the character
    // of str2 is already marked
    for (var i = 0; i < m; i++) {
 
        // If a common character
        // is found
        if (mark[str2[i] - 'a'])
            return true;
    }
 
    // If no common character
    // is found
    return false;
}
 
// Driver Code
var str1 = "abca",
    str2 = "efad";
if (check(str1, str2))
    document.write( "Yes");
else
    document.write( "No");
 
// This code is contributed by noob2000.

输出：

Yes

时间复杂度： O(max(N, M)) 其中 N 和 M 分别是 str1 和 str2 的长度。
辅助空间： O(1)

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