检查给定的字符串是否可以通过仅删除单一类型的字符来制作回文 |第 2 组

给定一个字符串S ，任务是是否可以制作一个字符串 删除相同字符的出现后的回文，任意次数

例子：

Input: S = “abczdzacb“
Output: Yes
Explanation: Remove first and second occurrence of character ‘a’. 
String S becomes “bczdzcb”, which is a palindrome.

Input: S = “madem”
Output: No
Explanation: Since only we can remove 1 character of any frequency only once. 
There is no such character removing which a palindrome can be made.

方法：这个问题可以利用回文的性质来解决。很明显，从整个中删除相同字符的出现不会影响回文性质。因此，也可以删除该字符的全部出现。因此，如果它们不相同，请从两端检查字符串，然后在删除两端的整个字符出现后检查字符串。如果任何删除满足回文条件，那么答案是Yes else No 。

请按照以下步骤解决问题：

• 使用(0, N/2)范围内的循环进行迭代
• 现在检查两端的字符是否相等
  • 如果字符不相同
    • 从两端删除整个出现的字符后检查它是否是回文。
    • 如果从任何一端删除所有出现的字符是回文，则打印“YES”并停止迭代。
• 如果在整个遍历之后，没有找到这样的回文，则打印“NO” 。

下面是上述方法的实现：

C++

// C++ code for the above approach
#include 
using namespace std;
 
// Function to check whether the string
// is palindrome after removal or
// neglecting character c
bool check_palindrome(string str, char c)
{
    int n = str.length(), i = 0, j = n - 1;
 
    while (i < j) {
 
        // If it is same as c neglect
        // this and move front
        if (str[i] == c)
            i++;
 
        // If it is same as c neglect
        // this and move back
        else if (str[j] == c)
            j--;
 
        // If they are not same it is
        // not a palindrome so return 0
        else if (str[i] != str[j])
            return 0;
 
        // It can be a palindrome so move
        // and check for remaining string
        else
            i++, j--;
    }
 
    return 1;
}
 
// Function to check if it is possible to
// form a palindrome after removal of
// any number of same characters once
string make_palindrome(string str)
{
    bool is_palindrome = 1;
    int n = str.length();
 
    // If n==1 || n==2 it is always possible
    if (n == 1 || n == 2) {
        return "YES";
    }
 
    // Check the character from both the ends
    // of the string
    for (int i = 0; i < n / 2; ++i) {
 
        // If the characters are not equal
        if (str[i] != str[n - 1 - i]) {
 
            // Remove str[i] and check if
            // it is a palindrome or
            // Remove str[n-i-1] and check
            // if it is a palindrome
            is_palindrome
                = check_palindrome(str, str[i])
                  || check_palindrome(
                         str, str[n - 1 - i]);
            break;
        }
    }
    if (is_palindrome)
        return "Yes";
 
    else
        return "No";
}
 
// Driver Code
int main()
{
    string S = "madem";
 
    string res = make_palindrome(S);
    cout << (res);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
   
// Function to check whether the string
// is palindrome after removal or
// neglecting character c
static boolean check_palindrome(String str, char c)
{
    int n = str.length(), i = 0, j = n - 1;
 
    while (i < j)
    {
 
        // If it is same as c neglect
        // this and move front
        if (str.charAt(i) == c)
            i++;
 
        // If it is same as c neglect
        // this and move back
        else if (str.charAt(j) == c)
            j--;
 
        // If they are not same it is
        // not a palindrome so return 0
        else if (str.charAt(i) != str.charAt(j))
            return false;
 
        // It can be a palindrome so move
        // and check for remaining string
        else {
            i++;
            j--;
        }
    }
 
    return true;
}
 
// Function to check if it is possible to
// form a palindrome after removal of
// any number of same characters once
static String make_palindrome(String str)
{
    boolean is_palindrome = true;
    int n = str.length();
 
    // If n==1 || n==2 it is always possible
    if (n == 1 || n == 2) {
        return "YES";
    }
 
    // Check the character from both the ends
    // of the string
    for (int i = 0; i < n / 2; ++i) {
 
        // If the characters are not equal
        if (str.charAt(i) != str.charAt(n - 1 - i)) {
 
            // Remove str[i] and check if
            // it is a palindrome or
            // Remove str[n-i-1] and check
            // if it is a palindrome
            is_palindrome
                = check_palindrome(str, str.charAt(i))
                  || check_palindrome(
                         str, str.charAt(n - 1 - i));
            break;
        }
    }
    if (is_palindrome)
        return "Yes";
 
    else
        return "No";
}
 
// Driver Code
public static void main(String args[])
{
    String S = "madem";
 
    String res = make_palindrome(S);
    System.out.println(res);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Python3

# Python code for the above approach
 
# Function to check whether the string
# is palindrome after removal or
# neglecting character c
def check_palindrome (str, c):
    n = len(str)
    i = 0
    j = n - 1
 
    while (i < j):
 
        # If it is same as c neglect
        # this and move front
        if (str[i] == c):
            i += 1
 
        # If it is same as c neglect
        # this and move back
        elif (str[j] == c):
            j -= 1
 
        # If they are not same it is
        # not a palindrome so return 0
        elif (str[i] != str[j]):
            return 0
 
        # It can be a palindrome so move
        # and check for remaining string
        else:
            i += 1
            j -= 1
 
 
    return 1
 
# Function to check if it is possible to
# form a palindrome after removal of
# any number of same characters once
def make_palindrome (str):
    is_palindrome = 1
    n = len(str)
 
    # If n==1 || n==2 it is always possible
    if (n == 1 or n == 2):
        return "YES"
     
    # Check the character from both the ends
    # of the string
    for i in range(n // 2):
 
        # If the characters are not equal
        if (str[i] != str[n - 1 - i]):
 
            # Remove str[i] and check if
            # it is a palindrome or
            # Remove str[n-i-1] and check
            # if it is a palindrome
            is_palindrome = check_palindrome(str, str[i]) or check_palindrome(str, str[n - 1 - i])
            break
         
    if (is_palindrome):
        return "Yes"
 
    else:
        return "No"
 
# Driver Code
S = "madem"
res = make_palindrome(S)
print(res)
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
class GFG
{
   
// Function to check whether the string
// is palindrome after removal or
// neglecting character c
static bool check_palindrome(string str, char c)
{
    int n = str.Length, i = 0, j = n - 1;
 
    while (i < j)
    {
 
        // If it is same as c neglect
        // this and move front
        if (str[i] == c)
            i++;
 
        // If it is same as c neglect
        // this and move back
        else if (str[j] == c)
            j--;
 
        // If they are not same it is
        // not a palindrome so return 0
        else if (str[i] != str[j])
            return false;
 
        // It can be a palindrome so move
        // and check for remaining string
        else {
            i++;
            j--;
        }
    }
 
    return true;
}
 
// Function to check if it is possible to
// form a palindrome after removal of
// any number of same characters once
static string make_palindrome(string str)
{
    bool is_palindrome = true;
    int n = str.Length;
 
    // If n==1 || n==2 it is always possible
    if (n == 1 || n == 2) {
        return "YES";
    }
 
    // Check the character from both the ends
    // of the string
    for (int i = 0; i < n / 2; ++i) {
 
        // If the characters are not equal
        if (str[i] != str[n - 1 - i]) {
 
            // Remove str[i] and check if
            // it is a palindrome or
            // Remove str[n-i-1] and check
            // if it is a palindrome
            is_palindrome
                = check_palindrome(str, str[i])
                  || check_palindrome(
                         str, str[n - 1 - i]);
            break;
        }
    }
    if (is_palindrome)
        return "Yes";
 
    else
        return "No";
}
 
// Driver Code
public static void Main()
{
    string S = "madem";
 
    string res = make_palindrome(S);
    Console.Write(res);
}
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

    // JavaScript code for the above approach
 
    // Function to check whether the string
    // is palindrome after removal or
    // neglecting character c
    const check_palindrome = (str, c) => {
        let n = str.length, i = 0, j = n - 1;
 
        while (i < j) {
 
            // If it is same as c neglect
            // this and move front
            if (str[i] == c)
                i++;
 
            // If it is same as c neglect
            // this and move back
            else if (str[j] == c)
                j--;
 
            // If they are not same it is
            // not a palindrome so return 0
            else if (str[i] != str[j])
                return 0;
 
            // It can be a palindrome so move
            // and check for remaining string
            else
                i++, j--;
        }
 
        return 1;
    }
 
    // Function to check if it is possible to
    // form a palindrome after removal of
    // any number of same characters once
    const make_palindrome = (str) => {
        let is_palindrome = 1;
        let n = str.length;
 
        // If n==1 || n==2 it is always possible
        if (n == 1 || n == 2) {
            return "YES";
        }
 
        // Check the character from both the ends
        // of the string
        for (let i = 0; i < parseInt(n / 2); ++i) {
 
            // If the characters are not equal
            if (str[i] != str[n - 1 - i]) {
 
                // Remove str[i] and check if
                // it is a palindrome or
                // Remove str[n-i-1] and check
                // if it is a palindrome
                is_palindrome
                    = check_palindrome(str, str[i])
                    || check_palindrome(
                        str, str[n - 1 - i]);
                break;
            }
        }
        if (is_palindrome)
            return "Yes";
 
        else
            return "No";
    }
 
    // Driver Code
 
    let S = "madem";
 
    let res = make_palindrome(S);
    document.write(res);
 
// This code is contributed by rakeshsahni
 

输出

No

时间复杂度： O(N)，其中 N 是字符串的长度。
空间复杂度： O(1)