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.

Python3JavaC++GoTypeScriptRust

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

class Solution {
    public String makeSmallestPalindrome(String s) {
        char[] cs = s.toCharArray();
        for (int i = 0, j = cs.length - 1; i < j; ++i, --j) {
            cs[i] = cs[j] = (char) Math.min(cs[i], cs[j]);
        }
        return new String(cs);
    }
}

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

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

function makeSmallestPalindrome(s: string): string {
    const cs = s.split('');
    for (let i = 0, j = s.length - 1; i < j; ++i, --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 cs: Vec<char> = s.chars().collect();
        let n = cs.len();
        for i in 0..n / 2 {
            let j = n - 1 - i;
            cs[i] = std::cmp::min(cs[i], cs[j]);
            cs[j] = cs[i];
        }
        cs.into_iter().collect()
    }
}

2697. Lexicographically Smallest Palindrome

Description

Solutions

Solution 1: Greedy + Two Pointers

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