How do I approach the Decrypt String from Alphabet to Integer Mapping problem?

Start by scanning from right to left. Check if the character is part of a two-digit number with '#', or a single digit, and map accordingly.

What should I do when encountering a '#' in the Decrypt String problem?

Treat it as a delimiter to identify a two-digit number that needs to be mapped to a letter. Skip the next character after handling the mapping.

What are the common pitfalls in solving the Decrypt String from Alphabet to Integer Mapping problem?

The main issues include misinterpreting the '#' symbol, failing to skip characters after decoding two-digit numbers, and inefficiently reversing the result string.

How can I optimize the solution for larger input sizes in the Decrypt String problem?

Optimize by scanning the string once from right to left and constructing the result string directly, avoiding unnecessary reversals or extra passes.

Is there any specific pattern I should focus on when solving the Decrypt String problem?

The pattern to focus on is scanning the string from right to left while handling two-digit numbers with '#'. This ensures that you map correctly and avoid skipping characters.

#1309

Easy

auto_awesomeString-driven solution strategy

LeetCode 题解工作台

解码字母到整数映射

给你一个字符串 s ，它由数字（ '0' - '9' ）和 '#' 组成。我们希望按下述规则将 s 映射为一些小写英文字符：字符（ 'a' - 'i' ）分别用（ '1' - '9' ）表示。字符（ 'j' - 'z' ）分别用（ '10#' - '26#' ）表示。返回映射之后形成的新字符串…

字符串

题目描述

给你一个字符串 s，它由数字（'0' - '9'）和 '#' 组成。我们希望按下述规则将 s 映射为一些小写英文字符：

字符（'a' - 'i'）分别用（'1' - '9'）表示。
字符（'j' - 'z'）分别用（'10#' - '26#'）表示。

返回映射之后形成的新字符串。

题目数据保证映射始终唯一。

示例 1：

输入：s = "10#11#12"
输出："jkab"
解释："j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".

示例 2：

输入：s = "1326#"
输出："acz"

提示：

1 <= s.length <= 1000
s[i] 只包含数字（'0'-'9'）和 '#' 字符。
s 是映射始终存在的有效字符串。

lightbulb

解题思路

方法一：模拟

我们直接模拟即可。

遍历字符串 $s$ ，对于当前遍历到的下标 $i$ ，如果 $i + 2 < n$ 且 $s[i + 2]$ 为 #，则将 $s[i]$ 和 $s[i + 1]$ 组成的字符串转换为整数，加上 a 的 ASCII 码值减去 1，然后转换为字符，添加到结果数组中，并将 $i$ 增加 3；否则，将 $s[i]$ 转换为整数，加上 a 的 ASCII 码值减去 1，然后转换为字符，添加到结果数组中，并将 $i$ 增加 1。

最后将结果数组转换为字符串返回即可。

时间复杂度 $O(n)$ ，空间复杂度 $O(n)$ 。其中 $n$ 为字符串 $s$ 的长度。

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class Solution:
    def freqAlphabets(self, s: str) -> str:
        ans = []
        i, n = 0, len(s)
        while i < n:
            if i + 2 < n and s[i + 2] == "#":
                ans.append(chr(int(s[i : i + 2]) + ord("a") - 1))
                i += 3
            else:
                ans.append(chr(int(s[i]) + ord("a") - 1))
                i += 1
        return "".join(ans)

speed

复杂度分析

指标	值
时间	Depends on the final approach
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Candidate scans the string from right to left.
question_mark
Candidate correctly handles '#' for two-digit numbers.
question_mark
Candidate avoids unnecessary string reversals, constructing the result efficiently.

warning

常见陷阱

外企场景

error
Forgetting to skip characters after decoding two-digit numbers with '#'.
error
Misinterpreting the '#' symbol, either as part of a digit or skipping it.
error
Inefficiently reversing the string after decoding, instead of building it from right to left.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Handling strings with only single-digit characters.
arrow_right_alt
Adapting the solution to work with uppercase letters instead of lowercase.
arrow_right_alt
Dealing with larger strings and optimizing for performance.

help

常见问题

外企场景

继续练习

#1307 口算难题 #1312 让字符串成为回文串的最少插入次数 #1316 不同的循环子字符串