How do you solve Decoded String at Index on LeetCode?

Decoded String at Index (LeetCode #880) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The decoded string can grow exponentially, making full construction impractical.

What is the brute force solution for Decoded String at Index?

The brute force approach for Decoded String at Index is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves fully decoding the string and then accessing the k-th character. This is straightforward but inefficient for large inputs due to potentially massive string sizes. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Decoded String at Index?

Decoded String at Index uses the following concepts: String, Stack. The recommended patterns to study are: String Manipulation, Dynamic Programming.

What are the interview tips for Decoded String at Index?

Always clarify constraints and edge cases with the interviewer. Think about space and time efficiency, especially with potentially large outputs. Practice explaining your thought process clearly while coding.

What are common mistakes when solving Decoded String at Index?

Forgetting to handle the modulo operation when k exceeds the current length. Assuming that the decoded string can be constructed in memory.

#880

Decoded String at Index

Medium

StringStackString ManipulationDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

Instead of fully decoding the string, we can calculate the length of the decoded string dynamically and find the k-th character without constructing the entire string. This is efficient and handles large inputs gracefully.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable to keep track of the total length of the decoded string.
2Step 2: Iterate through the encoded string in reverse to find the effective length of the decoded string.
3Step 3: If the current character is a digit, divide the length by that digit to find the length of the previous segment.
4Step 4: If the current character is a letter and the current length is greater than or equal to k, check if this letter is the k-th character.
5Step 5: If found, return the letter; otherwise, continue until the beginning of the string.

solution.py15 lines

1def decodeAtIndex(s, k):
2    length = 0
3    for char in s:
4        if char.isdigit():
5            length *= int(char)
6        else:
7            length += 1
8    for char in reversed(s):
9        k %= length
10        if k == 0 and char.isalpha():
11            return char
12        if char.isdigit():
13            length //= int(char)
14        else:
15            length -= 1

ℹ

Complexity note: The complexity is O(n) because we only traverse the string a couple of times, and we do not store the entire decoded string, making it efficient.

1The decoded string can grow exponentially, making full construction impractical.
2Using reverse traversal allows us to find the k-th character without full expansion.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.