LeetCode 2697. Lexicographically Smallest Palindrome Solution in Java, C++, Python & More | Explanation + Code

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.

PythonJavaC++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)(code-box)

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); } }(code-box)

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; } };(code-box)

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) }(code-box)

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(''); }(code-box)

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() } }(code-box)