String Without AAA or BBB

String Without AAA or BBB - Problem

String Greedy Medium

Given two integers a and b, return any string s such that:

s has length a + b and contains exactly a 'a' letters, and exactly b 'b' letters
The substring 'aaa' does not occur in s
The substring 'bbb' does not occur in s

You need to construct a valid string that uses all the given characters while avoiding three consecutive identical characters.

Input & Output

Example 1 — Basic Case

$ Input: a = 1, b = 2

› Output: "bab"

💡 Note: We have 1 'a' and 2 'b's. Since b > a, we can start with 'b', then 'a', then 'b' to get "bab". No 'aaa' or 'bbb' patterns exist.

Example 2 — Equal Counts

$ Input: a = 2, b = 1

› Output: "aab"

💡 Note: We have 2 'a's and 1 'b'. Since a > b, we can place "aa" then "b" to get "aab". This uses all characters without forbidden patterns.

Example 3 — Large Difference

$ Input: a = 1, b = 4

› Output: "bbabb"

💡 Note: With 4 'b's and 1 'a', we need to carefully place 2 'b's, then 'a', then remaining 'b's to avoid 'bbb' pattern.

Constraints

0 ≤ a ≤ 100
0 ≤ b ≤ 100
1 ≤ a + b ≤ 100

Visualization

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Asked in

G Google 35 f Facebook 28 a Amazon 22

The key insight is to always prioritize the more abundant character while ensuring no three consecutive identical characters. The greedy approach works optimally: choose the character with higher remaining count, but switch when we've placed 2 consecutive. Time: O(a+b), Space: O(1).

Common Approaches

✓ Brute Force - Generate All Permutations

⏱️ Time: O((a+b)!) Space: O(a+b)

Generate every possible permutation of 'a' and 'b' characters, then check each one to see if it contains 'aaa' or 'bbb' substrings. Return the first valid string found.

Greedy Construction

⏱️ Time: O(a+b) Space: O(1)

Construct the string character by character by always choosing the character we have more of, but never placing more than 2 consecutive identical characters. When we have significantly more of one character, place 2 of it followed by 1 of the other.

Smart Greedy with Lookahead

⏱️ Time: O(a+b) Space: O(1)

Enhanced greedy strategy that places 2 of the abundant character when we have significantly more of it, followed by 1 of the other character. This creates better balance and uses characters more efficiently.

Brute Force - Generate All Permutations — Algorithm Steps

Generate all permutations of a 'a's and b 'b's
For each permutation, check if it contains 'aaa' or 'bbb'
Return the first valid permutation

Visualization

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Step-by-Step Walkthrough

Generate

Create all permutations of characters

Check

Validate each permutation for 'aaa' or 'bbb'

Return

First valid permutation found

Code -

solution.c — C

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

static char result[201];

char* solution(int a, int b) {
    int pos = 0;
    
    while (a > 0 || b > 0) {
        // Check if last two chars are the same
        if (pos >= 2 && result[pos-1] == result[pos-2]) {
            // Must place the opposite character
            if (result[pos-1] == 'a' && b > 0) {
                result[pos++] = 'b';
                b--;
            } else if (result[pos-1] == 'b' && a > 0) {
                result[pos++] = 'a';
                a--;
            } else {
                break;  // Can't continue
            }
        } else {
            // Choose which character to place based on counts
            if (a > b && a > 0) {
                // Place 'a', possibly double if we have many more
                if (a > b + 1 && (pos == 0 || result[pos-1] != 'a')) {
                    result[pos++] = 'a';
                    result[pos++] = 'a';
                    a -= 2;
                } else {
                    result[pos++] = 'a';
                    a--;
                }
            } else if (b > a && b > 0) {
                // Place 'b', possibly double if we have many more
                if (b > a + 1 && (pos == 0 || result[pos-1] != 'b')) {
                    result[pos++] = 'b';
                    result[pos++] = 'b';
                    b -= 2;
                } else {
                    result[pos++] = 'b';
                    b--;
                }
            } else {
                // Equal counts or one is zero, alternate
                if (a > 0) {
                    result[pos++] = 'a';
                    a--;
                } else if (b > 0) {
                    result[pos++] = 'b';
                    b--;
                }
            }
        }
    }
    
    result[pos] = '\0';
    return result;
}

int main() {
    int a, b;
    scanf("%d", &a);
    scanf("%d", &b);
    char* res = solution(a, b);
    printf("%s\n", res);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O((a+b)!)

Generating all permutations of a+b characters takes factorial time

✓ Linear Growth

Space Complexity

O(a+b)

Space for storing one permutation and recursion stack

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Generate All Permutations — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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