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2697. Lexicographically Smallest Palindrome

Description

You are given a string s consisting of lowercase English letters, and you are allowed to perform operations on it. In one operation, you can replace a character in s with another lowercase English letter.

Your task is to make s a palindrome with the minimum number of operations possible. If there are multiple palindromes that can be made using the minimum number of operations, make the lexicographically smallest one.

A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, string a has a letter that appears earlier in the alphabet than the corresponding letter in b.

Return the resulting palindrome string.

 

Example 1:

Input: s = "egcfe"
Output: "efcfe"
Explanation: The minimum number of operations to make "egcfe" a palindrome is 1, and the lexicographically smallest palindrome string we can get by modifying one character is "efcfe", by changing 'g'.

Example 2:

Input: s = "abcd"
Output: "abba"
Explanation: The minimum number of operations to make "abcd" a palindrome is 2, and the lexicographically smallest palindrome string we can get by modifying two characters is "abba".

Example 3:

Input: s = "seven"
Output: "neven"
Explanation: The minimum number of operations to make "seven" a palindrome is 1, and the lexicographically smallest palindrome string we can get by modifying one character is "neven".

 

Constraints:

• 1 <= s.length <= 1000
• s consists of only lowercase English letters.

Solutions

Solution 1: Greedy + Two Pointers

We use two pointers $i$ and $j$ to point to the beginning and end of the string, initially $i = 0$, $j = n - 1$.

Next, each time we greedily modify $s[i]$ and $s[j]$ to their smaller value to make them equal. Then we move $i$ one step forward and $j$ one step backward, and continue this process until $i \ge j$. At this point, we have obtained the smallest palindrome string.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public String makeSmallestPalindrome(String s) {
          char[] cs = s.toCharArray();
          for (int i = 0, j = cs.length - 1; i < j; ++i, --j) {
              if (cs[i] != cs[j]) {
                  cs[i] = cs[j] = cs[i] < cs[j] ? cs[i] : cs[j];
              }
          }
          return String.valueOf(cs);
      }
  }
  

• class Solution {
  public:
      string makeSmallestPalindrome(string s) {
          for (int i = 0, j = s.size() - 1; i < j; ++i, --j) {
              if (s[i] != s[j]) {
                  s[i] = s[j] = s[i] < s[j] ? s[i] : s[j];
              }
          }
          return s;
      }
  };
  

• class Solution:
      def makeSmallestPalindrome(self, s: str) -> str:
          i, j = 0, len(s) - 1
          cs = list(s)
          while i < j:
              if s[i] != s[j]:
                  cs[i] = cs[j] = min(s[i], s[j])
              i, j = i + 1, j - 1
          return "".join(cs)
  
  

• func makeSmallestPalindrome(s string) string {
  	cs := []byte(s)
  	for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
  		if cs[i] != cs[j] {
  			if cs[i] < cs[j] {
  				cs[j] = cs[i]
  			} else {
  				cs[i] = cs[j]
  			}
  		}
  	}
  	return string(cs)
  }
  

• function makeSmallestPalindrome(s: string): string {
      const cs = s.split('');
      for (let i = 0, j = s.length - 1; i < j; ++i, --j) {
          if (s[i] !== s[j]) {
              cs[i] = cs[j] = s[i] < s[j] ? s[i] : s[j];
          }
      }
      return cs.join('');
  }
  
  

• impl Solution {
      pub fn make_smallest_palindrome(s: String) -> String {
          let mut b: Vec<u8> = s.bytes().collect();
          let mut i = 0;
          let mut j = b.len() - 1;
  
          while i < j {
              if b[i] != b[j] {
                  if b[i] < b[j] {
                      b[j] = b[i];
                  } else {
                      b[i] = b[j];
                  }
              }
  
              i += 1;
              j -= 1;
          }
  
          String::from_utf8(b).unwrap()
      }
  }
  

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