LeetCode #1328 — MEDIUM

Break a Palindrome

Move from brute-force thinking to an efficient approach using greedy strategy.

The Problem

Problem Statement

Given a palindromic string of lowercase English letters palindrome, replace exactly one character with any lowercase English letter so that the resulting string is not a palindrome and that it is the lexicographically smallest one possible.

Return the resulting string. If there is no way to replace a character to make it not a palindrome, return an empty string.

A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, a has a character strictly smaller than the corresponding character in b. For example, "abcc" is lexicographically smaller than "abcd" because the first position they differ is at the fourth character, and 'c' is smaller than 'd'.

Example 1:

Input: palindrome = "abccba"
Output: "aaccba"
Explanation: There are many ways to make "abccba" not a palindrome, such as "zbccba", "aaccba", and "abacba".
Of all the ways, "aaccba" is the lexicographically smallest.

Example 2:

Input: palindrome = "a"
Output: ""
Explanation: There is no way to replace a single character to make "a" not a palindrome, so return an empty string.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: Given a palindromic string of lowercase English letters palindrome, replace exactly one character with any lowercase English letter so that the resulting string is not a palindrome and that it is the lexicographically smallest one possible. Return the resulting string. If there is no way to replace a character to make it not a palindrome, return an empty string. A string a is lexicographically smaller than a string b (of the same length) if in the first position where a and b differ, a has a character strictly smaller than the corresponding character in b. For example, "abcc" is lexicographically smaller than "abcd" because the first position they differ is at the fourth character, and 'c' is smaller than 'd'.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Greedy

Example 1

"abccba"

Example 2

"a"

Step 02

Core Insight

What unlocks the optimal approach

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'.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1328: Break a Palindrome
class Solution {
    public String breakPalindrome(String palindrome) {
        int n = palindrome.length();
        if (n == 1) {
            return "";
        }
        char[] s = palindrome.toCharArray();
        int i = 0;
        while (i < n / 2 && s[i] == 'a') {
            ++i;
        }
        if (i == n / 2) {
            s[n - 1] = 'b';
        } else {
            s[i] = 'a';
        }
        return String.valueOf(s);
    }
}

// Accepted solution for LeetCode #1328: Break a Palindrome
func breakPalindrome(palindrome string) string {
	n := len(palindrome)
	if n == 1 {
		return ""
	}
	i := 0
	s := []byte(palindrome)
	for i < n/2 && s[i] == 'a' {
		i++
	}
	if i == n/2 {
		s[n-1] = 'b'
	} else {
		s[i] = 'a'
	}
	return string(s)
}

# Accepted solution for LeetCode #1328: Break a Palindrome
class Solution:
    def breakPalindrome(self, palindrome: str) -> str:
        n = len(palindrome)
        if n == 1:
            return ""
        s = list(palindrome)
        i = 0
        while i < n // 2 and s[i] == "a":
            i += 1
        if i == n // 2:
            s[-1] = "b"
        else:
            s[i] = "a"
        return "".join(s)

// Accepted solution for LeetCode #1328: Break a Palindrome
impl Solution {
    pub fn break_palindrome(palindrome: String) -> String {
        let n = palindrome.len();
        if n == 1 {
            return "".to_string();
        }
        let mut s: Vec<char> = palindrome.chars().collect();
        let mut i = 0;

        while i < n / 2 && s[i] == 'a' {
            i += 1;
        }

        if i == n / 2 {
            s[n - 1] = 'b';
        } else {
            s[i] = 'a';
        }

        s.into_iter().collect()
    }
}

// Accepted solution for LeetCode #1328: Break a Palindrome
function breakPalindrome(palindrome: string): string {
    const n = palindrome.length;
    if (n === 1) {
        return '';
    }
    const s = palindrome.split('');
    let i = 0;
    while (i < n >> 1 && s[i] === 'a') {
        i++;
    }
    if (i == n >> 1) {
        s[n - 1] = 'b';
    } else {
        s[i] = 'a';
    }
    return s.join('');
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

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.