LeetCode #2663 — HARD

Lexicographically Smallest Beautiful String

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

A string is beautiful if:

It consists of the first k letters of the English lowercase alphabet.
It does not contain any substring of length 2 or more which is a palindrome.

You are given a beautiful string s of length n and a positive integer k.

Return the lexicographically smallest string of length n, which is larger than s and is beautiful. If there is no such string, return an empty string.

A string a is lexicographically larger than a string b (of the same length) if in the first position where a and b differ, a has a character strictly larger than the corresponding character in b.

For example, "abcd" is lexicographically larger than "abcc" because the first position they differ is at the fourth character, and d is greater than c.

Example 1:

Input: s = "abcz", k = 26
Output: "abda"
Explanation: The string "abda" is beautiful and lexicographically larger than the string "abcz".
It can be proven that there is no string that is lexicographically larger than the string "abcz", beautiful, and lexicographically smaller than the string "abda".

Example 2:

Input: s = "dc", k = 4
Output: ""
Explanation: It can be proven that there is no string that is lexicographically larger than the string "dc" and is beautiful.

Constraints:

1 <= n == s.length <= 10⁵
4 <= k <= 26
s is a beautiful string.

Step 01

Brute Force Baseline

Problem summary: A string is beautiful if: It consists of the first k letters of the English lowercase alphabet. It does not contain any substring of length 2 or more which is a palindrome. You are given a beautiful string s of length n and a positive integer k. Return the lexicographically smallest string of length n, which is larger than s and is beautiful. If there is no such string, return an empty string. A string a is lexicographically larger than a string b (of the same length) if in the first position where a and b differ, a has a character strictly larger than the corresponding character in b. For example, "abcd" is lexicographically larger than "abcc" because the first position they differ is at the fourth character, and d is greater than c.

Baseline thinking

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

Pattern signal: Greedy

Example 1

"abcz"
26

Example 2

"dc"
4

Core Insight

What unlocks the optimal approach

If the string does not contain any palindromic substrings of lengths 2 and 3, then the string does not contain any palindromic substrings at all.
Iterate from right to left and if it is possible to increase character at index i without creating any palindromic substrings of lengths 2 and 3, then increase it.
After increasing the character at index i, set every character after index i equal to character a. With this, we will ensure that we have created a lexicographically larger string than s, which does not contain any palindromes before index i and is lexicographically the smallest.
Finally, we are just left with a case to fix palindromic substrings, which come after index i. This can be done with a similar method mentioned in the second hint.

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

Largest constraint values

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 #2663: Lexicographically Smallest Beautiful String
class Solution {
    public String smallestBeautifulString(String s, int k) {
        int n = s.length();
        char[] cs = s.toCharArray();
        for (int i = n - 1; i >= 0; --i) {
            int p = cs[i] - 'a' + 1;
            for (int j = p; j < k; ++j) {
                char c = (char) ('a' + j);
                if ((i > 0 && cs[i - 1] == c) || (i > 1 && cs[i - 2] == c)) {
                    continue;
                }
                cs[i] = c;
                for (int l = i + 1; l < n; ++l) {
                    for (int m = 0; m < k; ++m) {
                        c = (char) ('a' + m);
                        if ((l > 0 && cs[l - 1] == c) || (l > 1 && cs[l - 2] == c)) {
                            continue;
                        }
                        cs[l] = c;
                        break;
                    }
                }
                return String.valueOf(cs);
            }
        }
        return "";
    }
}

// Accepted solution for LeetCode #2663: Lexicographically Smallest Beautiful String
func smallestBeautifulString(s string, k int) string {
	cs := []byte(s)
	n := len(cs)
	for i := n - 1; i >= 0; i-- {
		p := int(cs[i] - 'a' + 1)
		for j := p; j < k; j++ {
			c := byte('a' + j)
			if (i > 0 && cs[i-1] == c) || (i > 1 && cs[i-2] == c) {
				continue
			}
			cs[i] = c
			for l := i + 1; l < n; l++ {
				for m := 0; m < k; m++ {
					c = byte('a' + m)
					if (l > 0 && cs[l-1] == c) || (l > 1 && cs[l-2] == c) {
						continue
					}
					cs[l] = c
					break
				}
			}
			return string(cs)
		}
	}
	return ""
}

# Accepted solution for LeetCode #2663: Lexicographically Smallest Beautiful String
class Solution:
    def smallestBeautifulString(self, s: str, k: int) -> str:
        n = len(s)
        cs = list(s)
        for i in range(n - 1, -1, -1):
            p = ord(cs[i]) - ord('a') + 1
            for j in range(p, k):
                c = chr(ord('a') + j)
                if (i > 0 and cs[i - 1] == c) or (i > 1 and cs[i - 2] == c):
                    continue
                cs[i] = c
                for l in range(i + 1, n):
                    for m in range(k):
                        c = chr(ord('a') + m)
                        if (l > 0 and cs[l - 1] == c) or (l > 1 and cs[l - 2] == c):
                            continue
                        cs[l] = c
                        break
                return ''.join(cs)
        return ''

// Accepted solution for LeetCode #2663: Lexicographically Smallest Beautiful String
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2663: Lexicographically Smallest Beautiful String
// class Solution {
//     public String smallestBeautifulString(String s, int k) {
//         int n = s.length();
//         char[] cs = s.toCharArray();
//         for (int i = n - 1; i >= 0; --i) {
//             int p = cs[i] - 'a' + 1;
//             for (int j = p; j < k; ++j) {
//                 char c = (char) ('a' + j);
//                 if ((i > 0 && cs[i - 1] == c) || (i > 1 && cs[i - 2] == c)) {
//                     continue;
//                 }
//                 cs[i] = c;
//                 for (int l = i + 1; l < n; ++l) {
//                     for (int m = 0; m < k; ++m) {
//                         c = (char) ('a' + m);
//                         if ((l > 0 && cs[l - 1] == c) || (l > 1 && cs[l - 2] == c)) {
//                             continue;
//                         }
//                         cs[l] = c;
//                         break;
//                     }
//                 }
//                 return String.valueOf(cs);
//             }
//         }
//         return "";
//     }
// }

// Accepted solution for LeetCode #2663: Lexicographically Smallest Beautiful String
function smallestBeautifulString(s: string, k: number): string {
    const cs: string[] = s.split('');
    const n = cs.length;
    for (let i = n - 1; i >= 0; --i) {
        const p = cs[i].charCodeAt(0) - 97 + 1;
        for (let j = p; j < k; ++j) {
            let c = String.fromCharCode(j + 97);
            if ((i > 0 && cs[i - 1] === c) || (i > 1 && cs[i - 2] === c)) {
                continue;
            }
            cs[i] = c;
            for (let l = i + 1; l < n; ++l) {
                for (let m = 0; m < k; ++m) {
                    c = String.fromCharCode(m + 97);
                    if ((l > 0 && cs[l - 1] === c) || (l > 1 && cs[l - 2] === c)) {
                        continue;
                    }
                    cs[l] = c;
                    break;
                }
            }
            return cs.join('');
        }
    }
    return '';
}

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 log n)

Space

O(1)

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.