Lexicographically Smallest Beautiful String

Lexicographically Smallest Beautiful String - Problem

String Greedy Hard

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.

Input & Output

Example 1 — Basic Increment

$ Input: s = "abcz", k = 4

› Output: "acba"

💡 Note: Cannot increment 'z', so go to 'c' and increment to 'd', but "abd" contains palindrome. Try 'b'→'c', then build valid suffix: "acba"

Example 2 — Simple Case

$ Input: s = "abc", k = 4

› Output: "abd"

💡 Note: Can increment last character 'c' to 'd'. Result "abd" has no palindromic substrings and is beautiful.

Example 3 — No Solution

$ Input: s = "zzab", k = 2

› Output: ""

💡 Note: All characters are at maximum for k=2, and no valid increment possible. Return empty string.

Constraints

1 ≤ n ≤ 10⁵
1 ≤ k ≤ 4
s consists of the first k letters of the English lowercase alphabet
s is beautiful

Visualization

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Understanding the Visualization

Input Analysis

Beautiful string s with k allowed characters

Find Increment

Locate rightmost position we can increment

Build Result

Create lexicographically smallest valid suffix

Key Takeaway

🎯 Key Insight: Work backwards to find the rightmost incrementable position, then build the smallest valid suffix greedily

Asked in

G Google 25 f Meta 18 Apple 12

The key insight is to work backwards from the rightmost position to find where we can increment a character, then greedily build the lexicographically smallest valid suffix. Best approach is greedy increment and fix. Time: O(n²), Space: O(n)

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Try All Next Strings	O(k^n * n²)	O(n)	Generate all lexicographically larger strings until finding a beautiful one
Greedy - Increment and Fix	O(n²)	O(n)	Find rightmost position to increment, then build valid suffix greedily

Brute Force - Try All Next Strings — Algorithm Steps

Generate next lexicographically larger string
Check if it's beautiful (no palindromic substrings)
If not beautiful, try next string
Repeat until found or exhausted

Visualization

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Step-by-Step Walkthrough

Start

Begin with input string

Generate

Create next lexicographically larger string

Test

Check if string is beautiful (no palindromic substrings)

Repeat

Continue until beautiful string found

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

int isPalindrome(char* str, int start, int end) {
    while (start < end) {
        if (str[start] != str[end]) return 0;
        start++;
        end--;
    }
    return 1;
}

int isBeautiful(char* s) {
    int n = strlen(s);
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            if (isPalindrome(s, i, j)) {
                return 0;
            }
        }
    }
    return 1;
}

int nextString(char* s, int k) {
    int n = strlen(s);
    int i = n - 1;
    while (i >= 0) {
        if (s[i] < 'a' + k - 1) {
            s[i]++;
            return 1;
        } else {
            s[i] = 'a';
            i--;
        }
    }
    return 0;
}

char* solution(char* s, int k) {
    static char result[1001];
    strcpy(result, s);
    
    while (nextString(result, k)) {
        if (isBeautiful(result)) {
            return result;
        }
    }
    
    strcpy(result, "");
    return result;
}

int main() {
    char s[1001];
    int k;
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    scanf("%d", &k);
    
    char* result = solution(s, k);
    printf("%s\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(k^n * n²)

k^n possible strings to check, n² time to validate each for palindromes

⚠ Quadratic Growth

Space Complexity

O(n)

String storage and recursion stack

⚡ Linearithmic Space

24.5K Views

Medium Frequency

~35 min Avg. Time

892 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Try All Next Strings — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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