#880

Medium

auto_awesomeStack-based state management

LeetCode Problem Workspace

Decoded String at Index

Decode the string and find the k-th letter efficiently using stack-based state management in this problem.

Problem Statement

Given an encoded string s consisting of letters and digits, decode the string using the specified pattern where each digit represents a repetition of the previous substring. You are tasked with determining the k-th letter in the decoded string without fully expanding it.

The decoded string is constructed by repeating the substrings as defined by the digits following each group of characters. Given an integer k, return the k-th letter (1-indexed) of the decoded string. The problem can be solved efficiently without generating the entire decoded string by leveraging stack-based state management.

Examples

Example 1

Input: s = "leet2code3", k = 10

Output: "o"

The decoded string is "leetleetcodeleetleetcodeleetleetcode". The 10th letter in the string is "o".

Example 2

Input: s = "ha22", k = 5

Output: "h"

The decoded string is "hahahaha". The 5th letter is "h".

Example 3

Input: s = "a2345678999999999999999", k = 1

Output: "a"

The decoded string is "a" repeated 8301530446056247680 times. The 1st letter is "a".

Constraints

2 <= s.length <= 100
s consists of lowercase English letters and digits 2 through 9.
s starts with a letter.
1 <= k <= 109
It is guaranteed that k is less than or equal to the length of the decoded string.
The decoded string is guaranteed to have less than 263 letters.

Solution Approach

Stack-based Decompression

We use a stack to keep track of the characters and their repeat counts. For each character and digit pair, we calculate how far the desired k-th index lies within the expanded version of the string. This allows for efficient traversal of the encoded string.

Efficient Index Traversal

Instead of decoding the entire string, we maintain a running length of the decoded string and adjust our search space based on the current repetition. This ensures that we can determine the k-th letter without ever needing to expand the full string.

Space Optimization

The solution relies on keeping track of only the necessary parts of the encoded string, thus avoiding unnecessary memory consumption. By calculating the lengths of each substring dynamically, we ensure that our solution remains space-efficient.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on how we manage stack operations and the size of the numbers involved in repetition. With an efficient traversal strategy, we avoid generating the full string, achieving a time complexity of O(n), where n is the length of the encoded string. The space complexity remains O(n) due to the stack storing intermediate state information.

What Interviewers Usually Probe

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

Common Pitfalls or Variants

Common pitfalls

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

Follow-up variants

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

FAQ

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.

terminal

Solution

Solution 1: Reverse Thinking

We can first calculate the total length $m$ of the decoded string, then traverse the string from back to front. Each time, we update $k$ to be $k \bmod m$, until $k$ is $0$ and the current character is a letter, then we return the current character. Otherwise, if the current character is a number, we divide $m$ by this number. If the current character is a letter, we subtract $1$ from $m$.

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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

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