How do I avoid generating the entire decoded string in Decoded String at Index?

Use a stack to keep track of characters and their repeat counts, allowing you to traverse the encoded string efficiently and determine the k-th letter without expanding the entire string.

What is the primary pattern in Decoded String at Index?

The primary pattern involves stack-based state management to handle the repetition of substrings defined by digits, allowing efficient decoding and indexing.

How does stack-based management help in Decoded String at Index?

Stack-based management helps by storing intermediate characters and repetition counts, which allows you to compute the k-th letter directly without creating the full string.

What are the time and space complexities of solving Decoded String at Index?

The time complexity is O(n), where n is the length of the encoded string, and the space complexity is O(n) due to stack usage for state management.

Can Decoded String at Index be solved with other methods?

While there are other ways to approach the problem, stack-based state management is the most efficient for handling large strings and keeping memory use low.

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LeetCode 题解工作台

索引处的解码字符串

给定一个编码字符串 s 。请你找出解码字符串并将其写入磁带。解码时，从编码字符串中每次读取一个字符，并采取以下步骤：如果所读的字符是字母，则将该字母写在磁带上。如果所读的字符是数字（例如 d ），则整个当前磁带总共会被重复写 d-1 次。现在，对于给定的编码字符串 s 和索引 k ，查…

字符串栈

题目描述

给定一个编码字符串 s 。请你找出 解码字符串 并将其写入磁带。解码时，从编码字符串中 每次读取一个字符 ，并采取以下步骤：

如果所读的字符是字母，则将该字母写在磁带上。
如果所读的字符是数字（例如 d），则整个当前磁带总共会被重复写 d-1 次。

现在，对于给定的编码字符串 s 和索引 k，查找并返回解码字符串中的第 k 个字母。

示例 1：

输入：s = "leet2code3", k = 10
输出："o"
解释：
解码后的字符串为 "leetleetcodeleetleetcodeleetleetcode"。
字符串中的第 10 个字母是 "o"。

示例 2：

输入：s = "ha22", k = 5
输出："h"
解释：
解码后的字符串为 "hahahaha"。第 5 个字母是 "h"。

示例 3：

输入：s = "a2345678999999999999999", k = 1
输出："a"
解释：
解码后的字符串为 "a" 重复 8301530446056247680 次。第 1 个字母是 "a"。

提示：

2 <= s.length <= 100
s 只包含小写字母与数字 2 到 9 。
s 以字母开头。
1 <= k <= 10⁹
题目保证 k 小于或等于解码字符串的长度。
解码后的字符串保证少于 2⁶³ 个字母。

lightbulb

解题思路

方法一：逆向思维

我们可以先计算出解码字符串的总长度 $m$ ，然后从后向前遍历字符串，每次更新 $k$ 为 $k \bmod m$ ，直到 $k$ 为 $0$ 且当前字符为字母，返回当前字符。否则，如果当前字符为数字，则将 $m$ 除以该数字。如果当前字符为字母，则将 $m$ 减 $1$ 。

时间复杂度 $O(n)$ ，其中 $n$ 为字符串的长度。空间复杂度 $O(1)$ 。

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class Solution:
    def decodeAtIndex(self, s: str, k: int) -> str:
        m = 0
        for c in s:
            if c.isdigit():
                m *= int(c)
            else:
                m += 1
        for c in s[::-1]:
            k %= m
            if k == 0 and c.isalpha():
                return c
            if c.isdigit():
                m //= int(c)
            else:
                m -= 1

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
Look for understanding of stack-based state management.
question_mark
Evaluate how the candidate optimizes space and avoids full string expansion.
question_mark
Assess the candidate's ability to work with large numbers and memory constraints.

warning

常见陷阱

外企场景

error
Mistaking the problem as requiring full string expansion, which leads to inefficient memory use.
error
Not properly managing stack state and traversal, resulting in incorrect indexing or performance issues.
error
Overcomplicating the problem and missing the straightforward stack-based solution.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Handling very large k values efficiently without running out of memory.
arrow_right_alt
Modifying the approach to handle different patterns of repetition beyond digits 2-9.
arrow_right_alt
Adapting the method to return substrings instead of single characters.

help

常见问题

外企场景

继续练习

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