How do I approach the Letter Case Permutation problem?

Use backtracking to explore all case combinations, but optimize by pruning the search space and focusing only on alphabetic characters.

What is the time complexity of solving the Letter Case Permutation problem?

The time complexity is O(2^k), where `k` is the number of alphabetic characters in the string, since each character has two case options.

Can we optimize the space complexity of Letter Case Permutation?

Yes, by constructing the string incrementally during the backtracking process, you can avoid storing intermediate results and optimize space usage.

Why are pruning techniques important in the Letter Case Permutation problem?

Pruning prevents unnecessary recursion for digits and non-alphabetic characters, reducing computation and improving performance, especially for longer strings.

What variations can be applied to the Letter Case Permutation problem?

You could introduce additional constraints, explore iterative methods, or handle strings with a mix of non-alphabetic characters for more complex challenges.

#784

Medium

auto_awesome回溯·pruning

LeetCode 题解工作台

字母大小写全排列

给定一个字符串 s ，通过将字符串 s 中的每个字母转变大小写，我们可以获得一个新的字符串。返回所有可能得到的字符串集合。以任意顺序返回输出。示例 1：输入： s = "a1b2" 输出： ["a1b2", "a1B2", "A1b2", "A1B2"] 示例 2: 输入: s = "…

字符串回溯位运算

题目描述

给定一个字符串 s ，通过将字符串 s 中的每个字母转变大小写，我们可以获得一个新的字符串。

返回 所有可能得到的字符串集合 。以 任意顺序 返回输出。

示例 1：

输入：s = "a1b2"
输出：["a1b2", "a1B2", "A1b2", "A1B2"]

示例 2:

输入: s = "3z4"
输出: ["3z4","3Z4"]

提示:

1 <= s.length <= 12
s 由小写英文字母、大写英文字母和数字组成

lightbulb

解题思路

方法一：DFS

由于 $s$ 中的每个字母都可以转换为大写或小写，因此可以使用 DFS 深度优先搜索的方法，枚举所有可能的情况。

具体地，从左到右遍历字符串 $s$ ，对于遍历到的每个字母，可以选择将其转变为大写或小写，然后继续遍历后面的字母。当遍历到字符串的末尾时，得到一个转换方案，将该方案加入答案即可。

转变大小写的方法可以使用位运算实现。对于一个字母，小写形式与大写形式的 ASCII 码之差为 $32$ ，因此，我们可以通过将该字母的 ASCII 码与 $32$ 进行异或运算来实现大小写转换。

时间复杂度 $O(n \times 2^n)$ ，其中 $n$ 是字符串 $s$ 的长度。对于每个字母，我们可以选择将其转换为大写或小写，因此一共有 $2^n$ 种转换方案。对于每种转换方案，我们需要 $O(n)$ 的时间生成一个新的字符串。

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class Solution:
    def letterCasePermutation(self, s: str) -> List[str]:
        def dfs(i: int) -> None:
            if i >= len(t):
                ans.append("".join(t))
                return
            dfs(i + 1)
            if t[i].isalpha():
                t[i] = chr(ord(t[i]) ^ 32)
                dfs(i + 1)

        t = list(s)
        ans = []
        dfs(0)
        return ans

speed

复杂度分析

指标	值
时间	and space complexity depend on the number of alphabetic characters in the string. The number of possible permutations grows exponentially with each letter that has two case options. Specifically, for a string of length `n`, with `k` alphabetic characters, the time and space complexity are both O(2^k), as each letter has two choices (uppercase or lowercase).
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Candidate demonstrates knowledge of backtracking and its application to case permutations.
question_mark
Ability to prune the search space and optimize the recursion depth for efficiency.
question_mark
Candidate can explain the trade-offs between recursive and iterative solutions.

warning

常见陷阱

外企场景

error
Not correctly handling digits or non-alphabetic characters, which should remain unchanged.
error
Failure to optimize the backtracking approach by pruning unnecessary recursion calls.
error
Not understanding the exponential time complexity caused by processing alphabetic characters.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Allowing more than one non-alphabetic character in the string.
arrow_right_alt
Exploring iterative methods rather than backtracking for generating permutations.
arrow_right_alt
Adding constraints on the number of operations or runtime limits to challenge the approach.

help

常见问题

外企场景

继续练习

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