#1415

Medium

auto_awesomeBacktracking search with pruning

LeetCode Problem Workspace

The k-th Lexicographical String of All Happy Strings of Length n

Given n and k, find the k-th lexicographical happy string of length n using backtracking search with pruning.

Problem Statement

A happy string is defined as a string where no two adjacent characters are the same. For example, strings like 'abc', 'ac', and 'bca' are happy strings, but strings like 'aa', 'ababbc', and 'baa' are not. Given an integer n, you are tasked with generating all happy strings of length n and sorting them lexicographically.

The problem asks you to return the k-th string from the lexicographically sorted list of all happy strings of length n. If k exceeds the number of happy strings possible for the given n, return an empty string. Implementing this solution efficiently requires using a backtracking approach with pruning to avoid unnecessary computations.

Examples

Example 1

Input: n = 1, k = 3

Output: "c"

The list ["a", "b", "c"] contains all happy strings of length 1. The third string is "c".

Example 2

Input: n = 1, k = 4

Output: ""

There are only 3 happy strings of length 1.

Example 3

Input: n = 3, k = 9

Output: "cab"

There are 12 different happy string of length 3 ["aba", "abc", "aca", "acb", "bab", "bac", "bca", "bcb", "cab", "cac", "cba", "cbc"]. You will find the 9th string = "cab"

Constraints

1 <= n <= 10
1 <= k <= 100

Solution Approach

Backtracking Search

Use backtracking to recursively generate all possible happy strings. While constructing strings, ensure no two adjacent characters are the same, pruning branches of the recursion tree where the adjacent characters would violate this condition.

Pruning to Optimize Search

Instead of generating all happy strings of length n upfront, use pruning to avoid generating strings that are guaranteed to be invalid. This allows for more efficient computation, especially when k is much smaller than the total number of valid strings.

Return the k-th String

Once the happy strings are generated, sort them lexicographically and return the k-th string. If k exceeds the number of valid happy strings, return an empty string.

Complexity Analysis

Metric	Value
Time	O(n)
Space	O(1)

Time complexity is O(n) due to the backtracking search generating each string step by step. Space complexity is O(1) since the algorithm does not store the entire list of happy strings, only the current string being constructed.

What Interviewers Usually Probe

Candidate demonstrates the ability to think recursively and apply pruning effectively.
Candidate's understanding of backtracking search is sound, especially regarding early termination of invalid branches.
Candidate efficiently manages time and space complexity considerations when implementing the solution.

Common Pitfalls or Variants

Common pitfalls

Failing to prune branches during the backtracking search, leading to excessive computation.
Incorrectly handling the case where k exceeds the total number of valid strings.
Not ensuring adjacent characters are different during string construction, leading to invalid happy strings.

Follow-up variants

Extend the problem to handle strings with larger lengths, testing the scalability of the solution.
Allow for custom sets of characters to be used for generating happy strings.
Implement the solution iteratively rather than recursively to explore different approaches.

FAQ

What is the primary approach to solving the 'k-th Lexicographical String of All Happy Strings of Length n' problem?

The primary approach is backtracking search with pruning, where you recursively build the strings and stop generating invalid ones early.

How do you ensure the strings are happy strings?

During the backtracking search, you ensure that no two adjacent characters in the string are the same, which guarantees the string is happy.

What should you do if k exceeds the number of valid happy strings?

If k exceeds the total number of valid strings, the solution should return an empty string as there are not enough valid strings to satisfy the request.

What is the time complexity of the solution?

The time complexity is O(n) due to the recursive generation of strings, with pruning improving the efficiency of the search.

What role does pruning play in this problem?

Pruning prevents the algorithm from generating strings that would be invalid, thus improving efficiency by avoiding unnecessary computations.

terminal

Solution

Solution 1: DFS

We use a string $\textit{s}$ to record the current string, initially an empty string. Then, we design a function $\text{dfs}$ to generate all happy strings of length $n$.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

class Solution:
    def getHappyString(self, n: int, k: int) -> str:
        def dfs():
            if len(s) == n:
                ans.append("".join(s))
                return
            if len(ans) >= k:
                return
            for c in "abc":
                if not s or s[-1] != c:
                    s.append(c)
                    dfs()
                    s.pop()

        ans = []
        s = []
        dfs()
        return "" if len(ans) < k else ans[k - 1]

Solution 2: Mathematics

We can directly calculate what the $k$-th happy string is, without generating all happy strings.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

class Solution:
    def getHappyString(self, n: int, k: int) -> str:
        def dfs():
            if len(s) == n:
                ans.append("".join(s))
                return
            if len(ans) >= k:
                return
            for c in "abc":
                if not s or s[-1] != c:
                    s.append(c)
                    dfs()
                    s.pop()

        ans = []
        s = []
        dfs()
        return "" if len(ans) < k else ans[k - 1]

Continue Practicing

#1307 verbal arithmetic puzzle #1286 iterator for combination #1255 maximum score words formed by letters