How to Break a Palindrome String by Replacing a Character?

  • 时间：2020-09-11 08:23:45
  • 分类：网络文摘
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Given a palindromic string palindrome, replace exactly one character by any lowercase English letter so that the string becomes the lexicographically smallest possible string that isn’t a palindrome. After doing so, return the final string. If there is no way to do so, return the empty string.

Example 1:
Input: palindrome = “abccba”
Output: “aaccba”

Example 2:
Input: palindrome = “a”
Output: “”

Constraints:
1 <= palindrome.length <= 1000
palindrome consists of only lowercase English letters.

Hints:
How to detect if there is impossible to perform the replacement? Only when the length = 1.
Change the first non ‘a’ character to ‘a’.
What if the string has only ‘a’?
Change the last character to ‘b’.

Algorithms to Break the Palindrome String

If the string is a one-character Palindrome, then it isn’t possible to do so since every one-character string is a Palindrome. Then, we can scan the first half of the Palindrome to see if it is all ‘a’. If it is, then we change the last character to ‘b’. Otherwise, we change the first non ‘a’ character to ‘a’. This will make a non Palindrome with the smallest lexicographically order.

C++:

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class Solution {
public:
    string breakPalindrome(string palindrome) {
        if (palindrome.size() == 1) {
            return "";
        }
        for (int i = 0; i < palindrome.size() / 2; ++ i) {
            if (palindrome[i] != 'a') {
                return palindrome.substr(0, i) + 'a' + 
                    palindrome.substr(i + 1);
                // or the following should work as well
                // palindrome[i] = 'a';
                // return palindrome;
            }
        }
        return palindrome.substr(0, palindrome.size() - 1) + 'b';
        // or the following should work
        // palindrome[palindrome.size() - 1] = 'b';
        // return palindrome;
    }
};
class Solution {
public:
    string breakPalindrome(string palindrome) {
        if (palindrome.size() == 1) {
            return "";
        }
        for (int i = 0; i < palindrome.size() / 2; ++ i) {
            if (palindrome[i] != 'a') {
                return palindrome.substr(0, i) + 'a' + 
                    palindrome.substr(i + 1);
                // or the following should work as well
                // palindrome[i] = 'a';
                // return palindrome;
            }
        }
        return palindrome.substr(0, palindrome.size() - 1) + 'b';
        // or the following should work
        // palindrome[palindrome.size() - 1] = 'b';
        // return palindrome;
    }
};

Java:

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class Solution {
    public String breakPalindrome(String palindrome) {
        if (palindrome.length() == 1) return "";
        for (int i = 0; i < palindrome.length() / 2; ++ i) {
            if (palindrome.charAt(i) != 'a') {
                return palindrome.substring(0, i) + 'a' +
                    palindrome.substring(i + 1);
            }
        }
        return palindrome.substring(0, palindrome.length() - 1) + 'b';
    }
}
class Solution {
    public String breakPalindrome(String palindrome) {
        if (palindrome.length() == 1) return "";
        for (int i = 0; i < palindrome.length() / 2; ++ i) {
            if (palindrome.charAt(i) != 'a') {
                return palindrome.substring(0, i) + 'a' +
                    palindrome.substring(i + 1);
            }
        }
        return palindrome.substring(0, palindrome.length() - 1) + 'b';
    }
}

Python (s[:-1] to trim the last character):

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class Solution:
    def breakPalindrome(self, palindrome: str) -> str:
        if len(palindrome) == 1:
            return ""
        for i in range(len(palindrome) // 2):
            if palindrome[i] != 'a':
                return palindrome[:i] + 'a' + palindrome[i + 1:]
        return palindrome[:-1] + 'b';
class Solution:
    def breakPalindrome(self, palindrome: str) -> str:
        if len(palindrome) == 1:
            return ""
        for i in range(len(palindrome) // 2):
            if palindrome[i] != 'a':
                return palindrome[:i] + 'a' + palindrome[i + 1:]
        return palindrome[:-1] + 'b';

Javascript:

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/**
 * @param {string} palindrome
 * @return {string}
 */
var breakPalindrome = function(palindrome) {
    if (palindrome.length === 1) {
        return "";
    }
    for (let i = 0; i < Math.floor(palindrome.length / 2); ++ i) {
        if (palindrome[i] != 'a') {
            return palindrome.substr(0, i) + 'a' + 
                palindrome.substr(i + 1);
        }
    }
    return palindrome.substr(0, palindrome.length - 1) + 'b';
};
/**
 * @param {string} palindrome
 * @return {string}
 */
var breakPalindrome = function(palindrome) {
    if (palindrome.length === 1) {
        return "";
    }
    for (let i = 0; i < Math.floor(palindrome.length / 2); ++ i) {
        if (palindrome[i] != 'a') {
            return palindrome.substr(0, i) + 'a' + 
                palindrome.substr(i + 1);
        }
    }
    return palindrome.substr(0, palindrome.length - 1) + 'b';
};

All implementations run O(N) in the worst case. O(1) constant is required.

–EOF (The Ultimate Computing & Technology Blog) —

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