通过插入给定字符使字符串非回文

给定一个字符串S和一个字符X ，任务是通过在字符串S 中插入字符X来生成一个非回文字符串。如果无法获得非回文字符串，则打印“-1” 。

例子：

Input: S = “ababab”, X = ‘a’
Output: “aababab”
Explanation: Inserting the character ‘a’ at the beginning of the string S modifies the string to “aababab”, which is non-palindromic.

Input: S = “rrr”, X = ‘r’
Output: -1

方法：根据以下观察可以解决给定的问题，如果字符串仅包含字符X ，则不可能使字符串非回文。否则，字符串可以是非回文的。请按照以下步骤解决问题：

1. 如果在字符串S字符X的计数是相同的字符串S的长度，然后打印“-1”。
2. 否则，检查字符串S和字符X 的连接是否为回文。如果发现为真，则打印字符串S+X 。否则打印(X + S) 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to check if a
// string is palindromic or not
bool Palindrome(string str)
{
    // Traverse the string str
    for (int i = 0, j = str.length() - 1;
         i < j; i++, j--) {
 
        // Check if i-th character from
        // both ends are the same or not
        if (str[i] != str[j])
            return false;
    }
 
    // Return true, as str is palindrome
    return true;
}
 
// Function to make the non-palindromic
// string by inserting the character X
void NonPalindrome(string str, char X)
{
 
    // If all the characters
    // in the string are X
    if (count(str.begin(), str.end(), X)
        == str.length()) {
 
        cout << "-1";
        return;
    }
 
    // Check if X + str is
    // palindromic or not
    if (Palindrome(X + str))
        cout << str + X << endl;
    else
        cout << X + str << endl;
}
 
// Driver Code
int main()
{
    string S = "geek";
    char X = 's';
    NonPalindrome(S, X);
 
    return 0;
}

Java

// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
public class GFG {
 
    // Function to check if a
    // string is palindromic or not
    static boolean Palindrome(String str)
    {
        // Traverse the string str
        for (int i = 0, j = str.length() - 1; i < j;
             i++, j--) {
 
            // Check if i-th character from
            // both ends are the same or not
            if (str.charAt(i) != str.charAt(j))
                return false;
        }
 
        // Return true, as str is palindrome
        return true;
    }
 
    // Function to make the non-palindromic
    // string by inserting the character X
    static void NonPalindrome(String str, char X)
    {
 
        // stores the count of char X in str
        int count = 0;
        for (int i = 0; i < str.length(); i++)
            if (str.charAt(i) == X)
                count++;
 
        // If all the characters
        // in the string are X
        if (count == str.length()) {
 
            System.out.println("-1");
            return;
        }
 
        // Check if X + str is
        // palindromic or not
        if (Palindrome(X + str))
            System.out.println(str + X);
        else
            System.out.println(X + str);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
 
        String S = "geek";
        char X = 's';
        NonPalindrome(S, X);
    }
}
 
// This code is contributed by Kingash.

Python3

# Python3 program for the above approach
 
# Function to check if a
# string is palindromic or not
def Palindrome(str):
     
    if str == str[::-1]:
        return True
 
    # Return true, as str is palindrome
    return False
 
# Function to make the non-palindromic
# string by inserting the character X
def NonPalindrome(str, X):
 
    # If all the characters
    # in the string are X
    if (str.count(X) == len(str)):
        print("-1")
        return
 
    # Check if X + str is
    # palindromic or not
    if (Palindrome(X + str)):
        print(str + X)
    else:
        print(X + str)
 
# Driver Code
if __name__ == '__main__':
     
    S = "geek"
    X = 's'
     
    NonPalindrome(S, X)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
 
using System;
using System.Linq;
class GFG {
    // Function to check if a
    // string is palindromic or not
    static bool Palindrome(string str)
    {
        // Traverse the string str
        for (int i = 0, j = str.Length - 1; i < j;
             i++, j--) {
 
            // Check if i-th character from
            // both ends are the same or not
            if (str[i] != str[j])
                return false;
        }
 
        // Return true, as str is palindrome
        return true;
    }
 
    // Function to make the non-palindromic
    // string by inserting the character X
    static void NonPalindrome(string str, char X)
    {
 
        // If all the characters
        // in the string are X
        if (str.Count(p => p == X) == str.Length) {
 
            Console.Write("-1");
            return;
        }
 
        // Check if X + str is
        // palindromic or not
        if (Palindrome(X + str))
            Console.WriteLine(str + X);
        else
            Console.WriteLine(X + str);
    }
 
    // Driver Code
    public static void Main()
    {
        string S = "geek";
        char X = 's';
        NonPalindrome(S, X);
    }
}
 
// This code is contributed by ukasp.

Javascript

 
// javascript program for the above approach
 
// Function to check if a
// string is palindromic or not
function Palindrome( str)
{
    // Traverse the string str
    for (let i = 0, j = str.length - 1; i < j;
        i++, j--) {
 
        // Check if i-th character from
        // both ends are the same or not
        if (str.charAt(i) != str.charAt(j))
            return false;
    }
 
    // Return true, as str is palindrome
    return true;
}
 
// Function to make the non-palindromic
// string by inserting the character X
function NonPalindrome( str, X)
{
 
    // stores the count of char X in str
    var count = 0;
    for (let i = 0; i < str.length; i++)
        if (str.charAt(i) == X)
            count++;
 
    // If all the characters
    // in the string are X
    if (count == str.length) {
 
        document.write("-1");
        return;
    }
 
    // Check if X + str is
    // palindromic or not
    if (Palindrome(X + str))
        document.write(str + X);
    else
        document.write(X + str);
}
 
 
// Driver Code
 
var S = "geek";
var X = 's';
NonPalindrome(S, X);
 

输出：

sgeek

时间复杂度： O(N)
辅助空间： O(1)

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