Longest Happy String

Longest Happy String - Problem

A string s is called happy if it satisfies the following conditions:

s only contains the letters 'a', 'b', and 'c'.
s does not contain any of "aaa", "bbb", or "ccc" as a substring.
s contains at most a occurrences of the letter 'a'.
s contains at most b occurrences of the letter 'b'.
s contains at most c occurrences of the letter 'c'.

Given three integers a, b, and c, return the longest possible happy string. If there are multiple longest happy strings, return any of them. If there is no such string, return the empty string "".

A substring is a contiguous sequence of characters within a string.

Input & Output

Example 1 — Basic Case

$ Input: a = 1, b = 1, c = 7

› Output: "ccaccbcc"

💡 Note: We have 7 c's, 1 a, and 1 b. Using a greedy approach: add cc, then a (to avoid 3 c's), then cc, then b, then cc. Result: ccaccbcc uses 6 c's, 1 a, 1 b = 8 characters total.

Example 2 — Equal Counts

$ Input: a = 2, b = 2, c = 1

› Output: "aabbc"

💡 Note: With equal counts of a and b, we can use both twice. Pattern: a-a-b-b-c uses all characters optimally.

Example 3 — No Solution

$ Input: a = 0, b = 2, c = 1

› Output: "bbc"

💡 Note: Only b and c available. We can create b-b-c without violating the three consecutive rule.

Constraints

0 ≤ a, b, c ≤ 100
a + b + c > 0

Visualization

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Asked in

f Facebook 25 G Google 18 a Amazon 15

The key insight is to always use the character with the highest remaining count while ensuring no three consecutive identical characters. The optimal approach uses a priority queue (max heap) to efficiently track and select characters. Time: O((a+b+c) log 3), Space: O(1).

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Brute Force Recursion

⏱️ Time: O(3^(a+b+c)) Space: O(a+b+c)

Generate all possible strings by recursively trying to add each character ('a', 'b', 'c') at each position. Check if the resulting string is valid (no three consecutive characters) and keep track of the longest valid string found.

Greedy Approach

⏱️ Time: O((a+b+c) * log 3) Space: O(1)

At each step, choose the character with the highest remaining count. If the last two characters are the same, choose the character with the second highest count to avoid three consecutive characters.

Priority Queue (Heap)

⏱️ Time: O((a+b+c) * log 3) Space: O(1)

Maintain a max heap of character counts. At each step, extract the character with highest count. If last two characters are same, extract second highest. This approach efficiently handles the selection without repeatedly sorting.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

⚡ Linearithmic

Space Complexity

✓ Linear Space

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Smart Actions

💡 Explanation

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