1415. The k-th Lexicographical String of All Happy Strings of Length n - Detailed Explanation
Problem Statement
Given two integers n and k, a happy string is defined as a string that:
- Consists only of the characters
'a','b', and'c'. - Does not have any two consecutive characters that are the same (i.e., it does not contain
"aa","bb", or"cc").
Your task is to return the k-th lexicographically smallest happy string of length n. If there are fewer than k happy strings of length n, return an empty string ("").
Example 1
- Input: n = 1, k = 3
- Happy Strings of Length 1:
["a", "b", "c"] - Output:
"c" - Explanation: The 3rd string in lexicographical order is
"c".
Example 2
- Input: n = 1, k = 4
- Happy Strings of Length 1:
["a", "b", "c"] - Output:
"" - Explanation: There are only 3 happy strings of length 1. Since k = 4 is greater than 3, the output is an empty string.
Example 3
- Input: n = 3, k = 9
- Happy Strings of Length 3:
- Strings starting with
'a':"aba","abc","aca","acb" - Strings starting with
'b':"bab","bac","bca","bcb" - Strings starting with
'c':"cab","cac","cba","cbc"
- Strings starting with
- Sorted Order:
"aba""abc""aca""acb""bab""bac""bca""bcb""cab""cac""cba""cbc"
- Output:
"cab" - Explanation: The 9th string in lexicographical order is
"cab".
Constraints
- 1 ≤ n ≤ 10
- 1 ≤ k ≤ 100
Hints
-
Brute Force Approach:
- Think about generating all possible strings of length n using the characters
'a','b', and'c'. - While generating, skip any string that has two identical consecutive characters.
- Once all valid happy strings are generated, sort them lexicographically and pick the k-th string.
- Think about generating all possible strings of length n using the characters
-
Optimized DFS Approach:
- Use recursion (or DFS/backtracking) to generate strings in lexicographical order.
- Instead of generating all strings and then sorting, you can keep a counter and stop as soon as you reach the k-th valid string.
- This avoids unnecessary generation once the desired string is found.
Approach 1: Brute Force Generation
Explanation
-
Idea:
Use backtracking to generate every possible happy string of length n. Check the condition (no two adjacent characters are the same) while building each string. -
Steps:
- Start with an empty string.
- At each step, try adding
'a','b', or'c'only if it does not match the previous character. - Once a string of length n is generated, add it to a list.
- After generating all strings, sort the list lexicographically.
- Return the k-th string (if it exists), otherwise return
"".
-
Visual Example (n = 3):
Imagine a tree where the root is an empty string. From the root:
- First level (choose first character):
"a","b","c"
- Second level (append a new character avoiding repetition):
- From
"a": possible are"ab"and"ac". - From
"b": possible are"ba"and"bc". - From
"c": possible are"ca"and"cb".
- From
- Third level:
- For
"ab": can extend to"aba"and"abc". - Continue similarly for other branches.
- For
- First level (choose first character):
-
Drawback:
Even though this approach works within the given constraints, it may be less efficient if n were larger.
Approach 2: Optimized Lexicographical DFS (Backtracking)
Explanation
-
Idea:
Generate the happy strings in lexicographical order using recursion and backtracking without storing all strings. -
Steps:
- Start with an empty string and explore choices in order:
'a','b', then'c'. - Append a character only if it does not match the last character of the current string.
- Maintain a counter. Each time you complete a string of length n, decrement k.
- When k reaches zero, record the current string as the answer and stop further recursion.
- Start with an empty string and explore choices in order:
-
Advantage:
This method avoids generating all possible happy strings and stops early once the k-th string is found, making it more efficient. -
Visual Example (n = 3, k = 9):
- The recursive process will first generate strings starting with
'a'(e.g.,"aba","abc","aca","acb"), then strings starting with'b', and finally those starting with'c'. - While generating, it counts each valid string until the 9th string (
"cab") is reached, and then terminates early.
- The recursive process will first generate strings starting with
Code Solutions
Python Code
Java Code
Complexity Analysis
Brute Force Approach
- Time Complexity:
- The total number of happy strings for length n is at most 3 * 2^(n-1) (since the first character has 3 options and each subsequent character has 2 options).
- Generating all strings and then sorting them leads to a complexity of approximately O(3 * 2^(n-1) * n * log(3 * 2^(n-1))).
- Space Complexity:
- O(n) for recursion stack and temporary string storage.
- O(3 * 2^(n-1)) for storing all strings (if all are stored).
Optimized DFS Approach
- Time Complexity:
- In the best case, the algorithm stops once the k-th string is found. In the worst-case scenario, it might traverse a large portion of the valid strings.
- Worst-case complexity is O(k * n).
- Space Complexity:
- O(n) for recursion stack and the current string.
Step-by-Step Walkthrough and Visual Examples
-
Start:
Begin with an empty string. The DFS will try'a', then'b', then'c'in that order. -
First Level:
- If the first character is
'a', the string becomes"a". - Next, for
"a", available choices for the second character are'b'and'c'(cannot add'a'because it would create"aa").
- If the first character is
-
Second Level:
- For
"a" → "ab", the third character can be'a'or'c'(forming"aba"and"abc"). - Similarly, for
"a" → "ac", generate"aca"and"acb".
- For
-
Counting:
- Each time a string of length n is completed, decrement k.
- When k becomes zero, the current string is the answer and the recursion stops.
-
Visual Tree (n = 3):
"" / | \ "a" "b" "c" / \ / \ / \ "ab" "ac" ... ... / \ / \ "aba" "abc" "aca" "acb"
- The DFS visits strings in lexicographical order. For instance, after visiting
"aba", it goes to"abc", and so on, until the 9th string is found.
Common Mistakes and Edge Cases
Common Mistakes
-
Incorrect Lexicographical Order:
Not generating or sorting the strings in lexicographical order can lead to an incorrect k-th string. -
Overlooking the Happy String Condition:
Failing to check for consecutive identical characters during string generation. -
Not Handling k Out-of-Range:
Forgetting to return an empty string if k exceeds the total number of valid happy strings.
Edge Cases
-
n = 1:
Only 3 possible strings exist:"a","b", and"c". If k > 3, return"". -
k Too Large:
For a given n, if k is greater than the total number of happy strings (which is at most 3 * 2^(n-1)), return an empty string.
Alternative Variations and Related Problems
Alternative Variations
-
Different Character Set:
What if the allowed characters change (for example, to include more letters)? -
Modified Conditions:
Instead of avoiding consecutive duplicates, consider other conditions such as no repeated substring patterns. -
Counting Problems:
Determine the total number of happy strings of length n without generating them.
Related Problems
-
Kth Permutation Sequence:
Find the kth permutation of a set of numbers. -
Letter Combinations of a Phone Number:
Generate all possible letter combinations from a digit string. -
Generate Parentheses:
Generate all valid combinations of parentheses. -
Combination Sum:
Find all possible combinations that sum to a target value.
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