Build confidence with an intuition-first walkthrough focused on two pointers fundamentals.
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 <= 1000s consists of only lowercase English letters.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Two Pointers · Greedy
"egcfe"
"abcd"
"seven"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2697: Lexicographically Smallest Palindrome
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);
}
}
// Accepted solution for LeetCode #2697: Lexicographically Smallest Palindrome
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)
}
# Accepted solution for LeetCode #2697: Lexicographically Smallest Palindrome
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)
// Accepted solution for LeetCode #2697: Lexicographically Smallest Palindrome
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()
}
}
// Accepted solution for LeetCode #2697: Lexicographically Smallest Palindrome
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('');
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.
Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.
Review these before coding to avoid predictable interview regressions.
Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.
Usually fails on: A valid pair can be skipped when only one side should move.
Fix: Move exactly one pointer per decision branch based on invariant.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.