Decoded String at Index

class Solution:
    def decodeAtIndex(self, encoded_str: str, k: int) -> str:
        tape_length = 0

        # First, find the length of the decoded tape
        for ch in encoded_str:
            if ch.isdigit():
                tape_length *= int(ch)
                tape_length += 1

        # Iterate through the encoded string in reverse
        for ch in reversed(encoded_str):
            k %= tape_length
            if k == 0 and ch.isalpha():
                return ch

            # Update the tape_length based on the character
            if ch.isdigit():
                tape_length //= int(ch)
                tape_length -= 1

Identifying Problem Isomorphism

“Decoded String at Index” requires you to find the character that would be at the k-th position after expanding a string according to a certain set of rules, which involves both string processing and understanding of sequences.

An approximate isomorphic problem is “Decode String”. In this problem, you’re also given an encoded string with certain rules, and you have to return the decoded string. The concept of decoding according to certain rules is common to both problems. In the “Decode String” problem, you return the entire decoded string, not just a character at a specific position. This makes “Decode String” simpler than “Decoded String at Index”.

A more complex problem is “Count Encodings”. In this problem, you’re given a digit sequence and you have to count the number of ways the sequence can be decoded into a valid string based on a set of rules. It’s more complex because, unlike the previous two problems, it asks for the count of all possible valid strings, not just producing a single decoded string or character. This requires more complex logic and often involves dynamic programming.

From simple to more complex:

  • Decode String
  • Decoded String at Index
  • Count Encodings

All these problems involve decoding a sequence or string according to certain rules.

10 Prerequisite LeetCode Problems

The following are a good preparation:

  1. “28. Implement strStr()” - This problem provides practice on basic string manipulation, which is a skill required in the main problem.

  2. “125. Valid Palindrome” - In this problem, you need to inspect a string from both ends, a technique that can also be used in the main problem.

  3. “344. Reverse String” - This problem provides practice on basic string manipulation and gives you a sense of how strings can be traversed backwards.

  4. “541. Reverse String II” - This problem is a step up from “344. Reverse String” and gives further practice on string manipulation.

  5. “657. Robot Return to Origin” - This problem requires you to keep track of a position while iterating through a string, much like the main problem.

  6. “709. To Lower Case” - This problem deals with transformation of characters in a string, a basic skill required in the main problem.

  7. “771. Jewels and Stones” - This problem requires counting occurrences of characters in a string, a technique also required in the main problem.

  8. “917. Reverse Only Letters” - This problem provides practice on string manipulation, particularly focusing on specific character types, similar to the main problem.

  9. “925. Long Pressed Name” - The string manipulation and character matching techniques in this problem will be useful for solving the main problem.

  10. “937. Reorder Data in Log Files” - This problem involves complex string manipulation and can help to understand how strings can be processed in different ways.

These cover string manipulation and character matching, which are needed to solve the “Decoded String at Index” problem.

Clarification Questions

  1. Is the input string s guaranteed to contain at least one digit?
  2. Is the input string s always in lowercase, or can it contain uppercase letters as well?
  3. Is the integer k always positive and within the bounds of the length of the decoded string?
  4. What should the behavior be if the input string s contains characters outside of the specified lowercase letters and digits 2 through 9? Should we assume this won’t happen?
  5. Is there any pattern in which letters and digits appear in the string s, or can they be randomly mixed?
  6. Are there any memory constraints for solving this problem?
  7. What’s the expected time complexity for solving this problem?
  8. Can the string s contain leading zeros before a digit? For example, is “02” a valid representation for the digit “2”?
  9. Will the digit in the string s ever be ‘1’, or will it always be between ‘2’ and ‘9’?
  10. Is it safe to assume that the length of the decoded string will always be less than 2^63, as mentioned in the problem statement?

These clarification questions can help in understanding the edge cases and constraints better, which is crucial for developing an optimized solution.

Problem Analysis and Key Insights

  1. Avoid Full Decoding: Due to constraints, especially the maximum value of k, fully decoding the string would not be practical. We need an efficient way to find the kth letter.

  2. Rule-Based: The decoding follows a set of specific rules which can be leveraged algorithmically. This hints at the possibility of using a stack or recursion to keep track of the current state.

  3. Index-Based Retrieval: We are not interested in the entire decoded string but only in the kth character. We should focus on algorithms that can directly give us this character.

  4. String and Digit Interplay: The challenge involves dealing with both letters and digits in a string, so our solution has to handle both effectively.

  5. Memory Constraints: The problem specifies that the decoded string can be very long (up to 2^63), so it’s crucial to find a solution that doesn’t rely on storing the entire decoded string in memory.

  6. Length Repetition: The digit tells us to repeat the entire tape, which gives a pattern. This can be used to predict the length of the decoded string at various stages, without actually decoding it.

  7. Starts with a Letter: The string always starts with a letter, which means the initial tape will never be empty when a digit is encountered.

  8. String Constraints: The string contains only lowercase English letters and digits 2 through 9. We don’t have to account for uppercase letters, other symbols, or the digit ‘1’.

  9. 1-Indexed: The problem uses 1-indexing for the kth position, which is different from typical 0-indexing in programming. This needs to be accounted for in the algorithm.

Key Takeaways:

  • We need an optimized, likely non-linear solution that avoids full string decoding.
  • The constraints and rules are specific enough to suggest using a stack or recursive approach for efficient problem-solving.
  • Due to the repetitive nature of the problem, there are patterns that can be exploited to find the kth character without going through the entire string.

Problem Boundary

The scope of this problem is defined by the following key aspects:

  1. Algorithmic Complexity: The problem requires an algorithmic solution that can handle string decoding and index-based retrieval efficiently, adhering to specific constraints.

  2. Data Structures: Utilizing appropriate data structures like a stack or employing recursion is within the scope to achieve an efficient solution.

  3. String Manipulation: The problem primarily revolves around manipulating strings based on given rules. Handling both characters and digits effectively is essential.

  4. Constraints: The problem scope is bounded by specific constraints on the length of the string, the value of k, and the characters that the string can contain.

  5. Optimization: Given the constraints, the focus is on optimizing the solution for both time and memory.

  6. Edge Cases: Addressing edge cases like maximum/minimum values for k and how to handle repeated segments efficiently is within the scope.

  7. Output: The final output is a single character from the decoded string at the kth position, which dictates the nature of the solution to be targeted.

  8. Mathematical Reasoning: Some level of mathematical understanding may be required to predict repetitive patterns and lengths without actually building the full string.

  9. Indexing: The 1-based index for k is a specific detail that must be accounted for, altering typical 0-based indexing logic.

  10. Error Handling: While not explicitly mentioned, ensuring that the input adheres to the specified constraints may also fall within the problem’s scope, depending on context.

In summary, the scope involves devising an optimized algorithmic approach that leverages the specific rules and constraints for string decoding and character retrieval.

Establishing the boundary of the problem involves clearly identifying what the problem asks for and what constraints and rules must be adhered to. Here’s how to delineate these boundaries:

What’s Included

  1. String Input: The problem operates on an input string that contains lowercase letters and digits from 2 to 9.

  2. Decoding Rules: The decoding operation has strict rules about repeating the tape (decoded string so far) or adding a character to it based on whether the current character is a letter or a digit.

  3. Index Retrieval: The goal is to find the kth character in the decoded string, where k is 1-indexed.

  4. Constraints: The string length, value of k, and other constraints set the bounds for the problem’s input and output.

  5. Optimization: The solution must be optimized to handle the constraints, particularly those related to the length of the decoded string and the value of k.

  6. Output: The output is a single character at the kth position in the decoded string.

What’s Excluded

  1. Full Decoding: Generating the entire decoded string is out of scope due to impracticality and efficiency concerns.

  2. Multiple Queries: The problem doesn’t ask for a solution to multiple queries of k, only a single k value.

  3. Error Handling: The problem assumes that the input will adhere to the given constraints, so robust error handling is likely outside its boundary.

  4. Additional Features: Any extensions like supporting more characters, case sensitivity, or additional decoding rules are not part of the problem.

  5. Storage: There’s no mention of storing the decoded string or any of its parts, so it’s not within the boundary.

  6. External Libraries: Given it’s an algorithmic problem, the use of specialized external libraries for string manipulation or otherwise is likely outside the scope.

By understanding what is included and what is excluded, you establish a clear boundary around the problem. This helps to focus your problem-solving efforts and ensures you’re developing a solution that fully meets the problem’s requirements while ignoring irrelevant details.

Problem Classification

The problem falls under the domain of String Manipulation and Algorithms.


  1. Input: An encoded string s composed of letters and digits 2 through 9.
  2. Input: An integer k indicating the 1-indexed position of the letter to return.
  3. Operation: Decode the string s based on specific rules.
  • If the character is a letter, write it to the tape.
  • If the character is a digit d, repeat the entire tape d - 1 more times.
  1. Output: The kth letter in the decoded string.
  2. Constraints:
  • The string s will have a length between 2 and 100.
  • The string s will consist of lowercase English letters and digits 2 through 9.
  • The string s will start with a letter.
  • k is between 1 and 10^9.
  • k is guaranteed to be less than or equal to the length of the decoded string.

This is a problem of string decoding following specific rules. It also involves index-based retrieval. Given that it includes constraints on the string length, the problem leans toward requiring an optimized algorithmic solution.

The constraints, particularly the maximum value for k, strongly suggest that generating the fully decoded string is impractical. Therefore, the problem likely needs a clever algorithmic approach to find the kth letter without fully decoding the string.

It’s a problem that combines elements of string manipulation and algorithm optimization. Therefore, it could be classified as an “Optimized String Decoding” problem.

Distilling the Problem to Its Core Elements

Fundamental Concept

The fundamental concept this problem is based on is “String Decoding with Repetition”. It combines principles of string manipulation and index-based retrieval within constraints of time and memory complexity.

Simplest Description

Imagine you have a string that contains letters and numbers. Each letter gets written down as is. When you see a number, you repeat everything you’ve written down that many times. Now, given a position, what letter would be at that spot after all this repeating?

Core Problem

The core problem is to find the letter at the kth position in a specially decoded string without actually decoding the entire string, due to efficiency reasons.

Key Components

  1. Input String: Contains both letters and numbers.
  2. Decoding Rules: Letters get written down, numbers cause repetition.
  3. Position k: The location in the decoded string whose character we want.
  4. Constraints: String length, value of k, and character types.

Minimal Set of Operations

  1. Initialize a variable to keep track of the “virtual length” of the decoded string.
  2. Traverse the input string from end to start, updating the “virtual length” based on characters and digits encountered.
  3. Use the “virtual length” and rules to backtrack from the end to the start of the string to find the kth character without full decoding.

By understanding these aspects, you get a clear idea of what the problem entails, what it is fundamentally based on, and how you can go about solving it in an optimized way.

Visual Model of the Problem

Visualizing this problem can help in understanding its intricacies. Here are some ways to visualize the problem statement:

Tape Analogy

Imagine a blank tape or an empty list where you will “write down” the characters. As you read each character from the input string, the tape will get filled based on the decoding rules. This visualization helps you understand how the string expands with each digit encountered.

Before decoding:

  • Tape: [ ]
  • Input: "leet2code3"


  1. Read ’l’: Tape becomes ['l']
  2. Read ’e’: Tape becomes ['l', 'e']
  3. Read ‘2’: Tape becomes ['l', 'e', 'e', 't', 'l', 'e', 'e', 't'] (repeated once more)
  4. Read ‘c’: Tape becomes ['l', 'e', 'e', 't', 'l', 'e', 'e', 't', 'c']
  5. Read ‘3’: Tape expands by repeating itself 3 times in total.

After decoding:

  • Tape: ['l', 'e', 'e', 't', 'l', 'e', 'e', 't', 'c', 'o', 'd', 'e', 'l', 'e', 'e', 't', 'l', 'e', 'e', 't', 'c', 'o', 'd', 'e', ...]
  • k = 10, Character at 10th position is ‘o’

Number Line

Draw a number line to represent the “virtual length” of the tape. As you go through the string, mark how the length would change after each digit is encountered.


  • Before Start: ------0
  • After ’leet’: ------4
  • After ‘2’: ------8 (4 * 2)
  • After ‘code’: ------12 (8 + 4)
  • After ‘3’: ------36 (12 * 3)

Now you know the ‘virtual length’ of the decoded string is 36, and you can use this information to find the character at the kth position without decoding the entire string.

Nested Boxes (For Recursion)

For a more complex string, you could visualize the decoding process as nested boxes, where each box expands into smaller boxes based on the digits encountered. This can help if you are using a recursive approach.

Using these visual aids can clarify how the decoded string evolves, how its length changes with each character or digit, and where you can expect to find the kth character.

Problem Restatement

You’re given a string that consists of letters and digits. The letters get directly written to an imaginary “tape,” while the digits instruct you to repeat the existing content on the tape a certain number of times. Your task is to identify the character that will appear at the kth position on this tape after fully decoding the string.


  1. The string will only have lowercase letters and digits ranging from 2 to 9.
  2. The string will always start with a letter.
  3. You need to find the character at the kth position, which is 1-indexed.


  1. The length of the input string will be between 2 and 100 characters.
  2. The value of k will be between 1 and 10^9.
  3. It’s guaranteed that k will be smaller than the total length of the decoded tape.
  4. The total length of the decoded tape will be less than 2^63.

The essence of the problem is to decode a string based on specific rules and retrieve the character at a given position, all while operating within defined constraints.

Abstract Representation of the Problem

This problem is about mapping sequences and transformation rules to efficiently find an element at a specific index. The mapping operation is dictated by a sequence of alphabetic characters and digits. Each alphabetic character extends the sequence linearly, while each digit triggers a repetitive transformation on the existing sequence. The goal is to find the character at a given index k in the transformed sequence.

Abstract Representation

  1. Initial Sequence S: A finite sequence S of length n containing elements from two distinct sets: alphabets A and digits D.

  2. Transform Function T: A function T(S, x) that takes the current sequence S and an element x. If x belongs to A, it appends x to S. If x belongs to D, it duplicates S x-1 more times.

  3. Index k: An integer k, 1 <= k <= m, where m is the length of the sequence after all transformations.

  4. Objective Function O: A function O(S, k) that returns the kth element of the sequence S after all transformations have been applied according to T.


  • Length of initial sequence S: 2 <= n <= 100
  • Character set for S: A U D where A = lowercase English alphabets, D = {2, 3, …, 9}
  • k: 1 <= k <= 10^9
  • The length of the transformed sequence m will be < 2^63

The problem can thus be abstracted as the optimized computation of O(S, k) based on a given initial sequence S and transformation rules encapsulated in T.


There are some specialized terms that are crucial for understanding this problem.

  1. Decoding: Transformation of an encoded string into its original or usable form. In this problem, decoding involves expanding the string based on the rules given for alphabets and digits.

  2. 1-Indexed: The position k starts from 1 instead of 0, which is typical in programming but can cause off-by-one errors if not handled correctly.

  3. Constraints: Conditions or limitations imposed on the input and the workings of the problem. In this case, constraints like string length, range of k, and character types set the boundary.

  4. String Manipulation: Operations like appending, slicing, and length calculation on strings. Here, the manipulation doesn’t happen on the actual string but on its abstract or “virtual” representation.

  5. Time Complexity: The computational complexity that describes the time taken to solve a problem as a function of the size of the input to the program. The constraint that k can go up to 10^9 makes it essential to solve the problem in an optimized manner.

  6. Virtual Length: A conceptual term used to denote the length of the string if it were fully decoded. This is useful for avoiding actual decoding but still determining the character at the kth position.

  7. Backtracking: A general algorithm for finding solutions by exploring each possibility until the solution is found. Here, backtracking may be used to efficiently find the kth character by “rewinding” the virtual tape.

  8. Repetition Factor: The number by which the existing sequence is repeated. It’s derived from the digits in the string.

  9. Base Case: The simplest, smallest instance of the problem, used in recursion to stop the recursive chain. Here, the base case would be when the ‘virtual length’ is reduced to a value smaller or equal to `k’.

Understanding these terms helps in not only grasping the problem statement but also in formulating an effective solution.

Problem Simplification and Explanation

Key Concepts:

  1. Encoded String: A mix of letters and numbers. Letters go straight to an imaginary line (or “tape”), while numbers tell you to repeat what’s already on the line.

  2. Decoding: Transforming the encoded string into its expanded form.

  3. Position k: You need to find which letter ends up at this spot on the line.

How Concepts Interact:

  • You start reading the encoded string one character at a time.
  • If it’s a letter, you write it onto the line.
  • If it’s a number, you make copies of what’s already on the line.

Finally, you find out which letter sits at position k on the line.


Imagine you are a construction worker building a wall using two types of bricks:

  1. Alphabet bricks (each has a letter on it)
  2. Number bricks (each has a number on it)

You start with an empty wall and begin placing bricks from left to right:

  • When you get an alphabet brick, you simply add it to the wall.
  • When you get a number brick, you take the existing wall and duplicate it, stacking it right next to itself as many times as the number on the brick minus one.

You continue this process until you’ve used all the bricks. Your job is to tell which alphabet brick is at the kth position from the left in this final wall structure.

This analogy simplifies the problem by translating it into a physical task involving bricks (letters and numbers) and a wall (the decoded string). You’re essentially building this wall based on certain rules, and you need to identify which brick ends up at a particular position.


There are several aspects in the problem statement and constraints that can be exploited for an efficient solution:

  1. Limited Character Set: The string consists of only lowercase English letters and digits 2-9. This limited range allows for efficient mapping and processing.

  2. Starting Character: The string starts with a letter, not a digit. This simplifies the decoding logic, as it means we always start writing a letter onto the “tape” first.

  3. Size Constraints: The length of the initial string is capped at 100 characters. This is small enough to allow for pre-processing without a substantial computational burden.

  4. 1-Indexed k: k is 1-indexed and is at most 10^9. This numerical range helps us to know that we don’t have to decode the whole string to find the kth character. We can stop as soon as we reach or pass k.

  5. Digit Range: Digits can only be between 2 and 9. This places a clear upper bound on the multiplication factor for the tape length. Knowing this can be useful for optimizing the decoding steps.

  6. Virtual Length: Given that the “tape” can be astronomically large but needs to be smaller than 2^63, calculating its virtual length could be more efficient than actually decoding it. This length can be calculated dynamically as we go through the string.

  7. Repetition Patterns: Any sequence before a digit is a unit that will be repeated. These units often form nested structures due to the sequential digits in the string. Observing this pattern might help us skip redundant calculations.

  8. Sequential Decoding: We can decode the string in sequential portions instead of all at once. This allows us to only process as much of the string as needed to find the kth character, making it possible to solve the problem in less time.

  9. Problem Decomposition: The digits allow the problem to be broken down into smaller sub-problems. Each digit creates a repeat sequence that we can analyze independently before merging it back into the larger sequence.

By exploiting these characteristics, we can aim for a solution that is both time-efficient and memory-efficient.

Analyzing the constraints yields some important insights for solving the problem efficiently:

  1. Memory Efficiency: The constraints on the length of the initial string (up to 100 characters) and the nature of the string (only lower-case letters and digits 2-9) suggest that memory usage is not likely to be an issue. However, decoding the entire string would not be feasible, given that k can go up to 10^9 and the decoded string can have up to 2^63 characters.

  2. Focus on k: The constraint that k is at most 10^9 indicates that you don’t have to decode the entire string. You just need to find out what the kth character would be. This enables early termination of any decoding algorithm as soon as k is reached.

  3. Digit Limitation: The digits range from 2 to 9, meaning the expansion of the tape is somewhat limited at each step. This insight can be useful for calculating the “virtual length” of the tape at each step without decoding it.

  4. String Start: The string starts with a letter, providing a simplified starting point for any decoding algorithm. You won’t have to worry about edge cases where the string starts with a number, making the logic straightforward from the first character.

  5. Bound on Decoded Length: The decoded string is guaranteed to have fewer than 2^63 letters. This is useful to avoid overflow errors and to understand the maximum extent to which the string can grow.

  6. Fixed Character Set: The string consists of only lower-case letters and single-digit numbers. This makes it easier to parse the string and apply the decoding rules without having to deal with various data types or large numbers.

  7. Optimization Requirement: The constraint on k being up to 10^9 suggests that an efficient algorithm is required. Any solution with a time complexity based on expanding the entire string would be impractical.

By understanding these constraints, you can tailor your solution to be both effective and efficient, taking advantage of these specific conditions to simplify the problem-solving process.

Case Analysis

Case 1: Minimal String Size (Edge Case)

  • Input: s = “a2”, k = 1
  • Output: “a”
  • Analysis: The smallest string size allowed by the constraints is 2. Here, “a2” decodes to “aa”. The kth character is “a”.

Case 2: Large k Value (Edge Case)

  • Input: s = “ab2c2”, k = 8
  • Output: “c”
  • Analysis: The string decodes to “ababcababc”, and k is at the last position. Be mindful of 1-indexing.

Case 3: Single Repetition

  • Input: s = “abc2”, k = 4
  • Output: “a”
  • Analysis: The string decodes to “abcabc”. Notice how the repeated part (“abc”) starts again at the 4th position.

Case 4: Nested Repetitions

  • Input: s = “a2b2”, k = 6
  • Output: “b”
  • Analysis: The string decodes to “aabbaabb”. Nested repetitions are common, and understanding their structure can optimize the solution.

Case 5: Consecutive Numbers

  • Input: s = “abc2de3”, k = 15
  • Output: “d”
  • Analysis: The string decodes to “abcabcdeedee”. Consecutive numbers may not interact but stack sequentially.

Case 6: Single Character with Large Repetition (Boundary Condition)

  • Input: s = “a9”, k = 1
  • Output: “a”
  • Analysis: The string decodes to “a” repeated 9 times. Despite the large repetition, the kth character is still the first one, showcasing the importance of not decoding the whole string.

Case 7: Number at the End (Potential Pitfall)

  • Input: s = “abc2d3”, k = 10
  • Output: “c”
  • Analysis: The string decodes to “abcabcdabc”. Notice that the last number (‘3’) does not have any effect on the kth character.

Case 8: Entire Sequence Repeated

  • Input: s = “abc3”, k = 9
  • Output: “c”
  • Analysis: The string decodes to “abcabcabc”. A single trailing digit expands the entire string, not just a part of it.

By analyzing these test cases, we can identify several key points:

  1. 1-Indexed k: Always remember that the index starts at 1.
  2. Early Termination: You don’t need to decode the entire string, only up to k.
  3. Nested and Sequential Numbers: Pay attention to how numbers repeat what’s before them, either nested or in sequence.
  4. Trailing Numbers: Numbers at the end might not affect the solution if k is small.

These insights will help in crafting a solution that is robust and takes into account all possible scenarios and edge cases.

Visualizing the examples can provide a clearer understanding of the problem’s mechanics. You can think of the decoded string as a tape being constructed step-by-step based on the input string. Here’s how to visualize some of the cases:

Case 1: Minimal String Size

  • Input: s = “a2”, k = 1
  • Tape: “a -> aa”
  • Visualization: a | a

Case 2: Large k Value

  • Input: s = “ab2c2”, k = 8
  • Tape: “a -> ab -> abab -> ababc -> ababcababc”
  • Visualization: a | b | a | b | c | a | b | c

Case 3: Single Repetition

  • Input: s = “abc2”, k = 4
  • Tape: “a -> ab -> abc -> abcabc”
  • Visualization: a | b | c | a

Case 4: Nested Repetitions

  • Input: s = “a2b2”, k = 6
  • Tape: “a -> aa -> aab -> aabaab”
  • Visualization: a | a | b | a | a | b

Case 5: Consecutive Numbers

  • Input: s = “abc2de3”, k = 15
  • Tape: “a -> ab -> abc -> abcabc -> abcabcde -> abcabcdeedee”
  • Visualization: a | b | c | a | b | c | d | e | e | d | e | e

Case 6: Single Character with Large Repetition

  • Input: s = “a9”, k = 1
  • Tape: “a -> aaaaaaaaa”
  • Visualization: a | a | a | a | a | a | a | a | a

Case 7: Number at the End

  • Input: s = “abc2d3”, k = 10
  • Tape: “a -> ab -> abc -> abcabc -> abcabcd -> abcabcddab”
  • Visualization: a | b | c | a | b | c | d | d | a | b

Case 8: Entire Sequence Repeated

  • Input: s = “abc3”, k = 9
  • Tape: “a -> ab -> abc -> abcabcabc”
  • Visualization: a | b | c | a | b | c | a | b | c

In each visualization, the vertical bars (|) separate the individual characters on the tape. The sequence expands from left to right based on the rules given in the problem statement. This allows you to visually track how the tape changes as you read each character of the input string, and more importantly, where the kth character would land on this tape.

Analyzing different cases gives us several key insights:

  1. Avoid Full Decoding: You don’t have to decode the entire string to find the kth element. This can save computation time.

  2. Sequential and Nested Numbers: Numbers can act on previously decoded segments of the string. Understanding this helps in optimizing the solution. Nested numbers have a multiplicative effect on the repeated segments.

  3. 1-Indexed: The kth element is 1-indexed, not 0-indexed. This is a small but crucial detail to remember during implementation.

  4. Early Termination: Depending on the value of k, the decoding process can often be terminated early, even before the entire input string is read.

  5. Trailing Digits: A trailing digit repeats everything that comes before it. This can have a multiplicative effect on the length of the decoded string.

  6. Consecutive Digits: Consecutive digits in the input string do not interact with each other but affect only their preceding characters.

These insights are instrumental in devising an efficient algorithm for the problem. They can help you rule out naive approaches and guide you toward more optimized solutions.

Identification of Applicable Theoretical Concepts

This problem lends itself to several mathematical and algorithmic concepts that can simplify or optimize its solution:

  1. Recursion and Divide-and-Conquer: This problem has a recursive nature where each segment of the string might be a repetition of a smaller segment. This naturally fits the divide-and-conquer paradigm, where each segment can be solved independently and combined to solve the larger problem.

  2. Dynamic Programming/Memoization: If we choose to traverse the string multiple times to find different kth elements, a memoization approach can be beneficial to store previously computed segments.

  3. String Manipulation: Knowledge of efficient string manipulation techniques can be essential here, particularly when repeated substrings are involved.

  4. Stack Data Structure: A stack can be useful to manage the substrings that are being repeated, especially when there are nested repetitions. This allows us to revert back to previous states easily.

  5. Big-O Analysis: Given the constraints, understanding the time and space complexity is essential. For example, decoding the entire string would be infeasible, so an understanding of computational complexity can guide us toward more efficient solutions.

  6. Arithmetic Sequences: When a segment is repeated, the length of the string grows in an arithmetic progression. This mathematical concept can help us quickly calculate positions without explicitly repeating the string.

  7. Modular Arithmetic: Since we are interested in the kth position, modular arithmetic can sometimes simplify the problem, allowing us to find the equivalent position in a shorter, repeated segment.

  8. Counting and Enumeration: Keeping a running count of the length of the decoded string at various stages can facilitate quick access to the kth element.

These mathematical and algorithmic techniques can significantly simplify the problem and lead to more efficient solutions. The key is to identify which of these can be most effectively applied given the specific constraints and requirements of the problem.

Simple Explanation

Imagine you have a toy typewriter that prints out letters and numbers on a long piece of tape. This toy typewriter is special because when it prints a number, it repeats all the letters that came before it that many times.

So, if you type “cat2”, the tape will show “catcat”. If you type “dog3”, the tape will show “dogdogdog”.

Now, let’s say we type a string like “cat2dog3”, and the typewriter will print “catcatdogdogdog”. The challenge is to figure out what letter will be at a certain position on this tape. So, if I ask you what’s the 10th letter on this tape, you’d say it’s ‘o’ because if you count from the start, ‘o’ is at the 10th spot.

In simpler terms, it’s like playing a game where you repeat some words a certain number of times, and then you have to guess what word comes in a particular position in the final long sentence you’ve created.

Problem Breakdown and Solution Methodology

Approach to Solving the Problem

  1. Initialize Variables: Start with an empty tape (an empty string) and set a variable to keep track of the length of the decoded string so far.

  2. Iterate Through the String: Loop through each character in the given string.

    • Character is a Letter: If the character is a letter, append it to the tape and update the length.

    • Character is a Digit: Multiply the tape by the digit, and update the length. This mimics the tape being rewritten.

  3. Check for Early Termination: After every operation, check if the length of the tape surpasses k. If it does, stop the loop to save computation time.

  4. Find the kth Character: Use the updated tape length to backtrack and find the kth character.


Think of this like making a bead necklace where some beads have a multiplier effect. If you place a “3x” bead, the pattern before it triples. You’re trying to find out which bead will be at the kth position in the final necklace.

Changes in Parameters

  • String Length: Longer strings require more computation, but the early termination check can mitigate this.

  • Value of k: A smaller k allows for earlier termination, making the solution more efficient. A larger k might require processing the entire string.

  • Digits in String: The larger the digit, the faster the tape length grows, potentially reaching k faster.

Working Example

Let’s use the example: s = "leet2code3", k = 10.

  1. Start with an empty tape: tape = "", tape_length = 0.

  2. Iterate through the string:

    • l: append to tape -> tape = "l", tape_length = 1.

    • e: append to tape -> tape = "le", tape_length = 2.

    • 2: multiply tape -> tape = "leetleet", tape_length = 8.

    • c: append to tape -> tape = "leetleetc", tape_length = 9.

    • o: append to tape -> tape = "leetleetco", tape_length = 10.

    • We’ve reached a tape length of 10, which is equal to k. Stop.

  3. The 10th character is ‘o’, which is the answer.

By breaking the problem down this way, we simplify each step and improve efficiency by avoiding unnecessary operations.

Inference of Problem-Solving Approach from the Problem Statement

  1. Encoded String: This is the input string containing letters and digits. Knowing it’s encoded informs us that straightforward decoding will be needed, which can be done via iteration.

  2. Decode: This operation specifies how to interpret the encoded string. The decoding rules directly lead to a strategy involving string manipulation.

  3. Tape: This is an abstract representation of the decoded string. Keeping track of the ’tape’ allows us to avoid decoding the entire string and optimize the search for the kth character.

  4. Letter: When the character is a letter, it’s appended to the ’tape’. This straightforward rule forms the basic operation of our algorithm.

  5. Digit: When encountered, this causes the entire current ’tape’ to be repeated. Understanding this leads us to consider the multiplication operation for strings and keeping track of the ’tape’ length.

  6. kth Letter: This is the target output. Knowing that we’re looking for a specific index position informs us to maintain a count and allows for early termination of the algorithm.

  7. 1-Indexed: The term informs us that counting starts from 1, not 0. This is a minor but crucial detail when trying to find the kth character.

  8. Constraints: These guide the efficiency and limitations of our solution. For instance, knowing that k is less than or equal to 10^9 informs us that we can’t afford to decode the entire string due to time complexity concerns.

Each of these terms and concepts guides specific aspects of the approach to solving the problem, from the choice of algorithm to minor but crucial details like indexing.

Drawing tables or diagrams can greatly aid in visualizing the problem’s properties and behavior. Here’s how you can visualize key elements:

Table for Tracking Steps

A table can help track what happens to the tape and its length at each step:

StepCharacterOperationTapeTape Length

Visual Diagram

You can also draw a tree-like diagram that shows how the tape evolves at each step:

Step 1: Append 'l'          l
Step 2: Append 'e'          le
Step 3: Append 'e'          lee
Step 4: Append 't'          leet
Step 5: Multiply by 2       leetleet
Step 6: Append 'c'          leetleetc
Step 7: Append 'o'          leetleetco

Diagram for Decoding

To represent the decoding operation when a digit is encountered, you could visualize it like a droplet impacting a body of water, creating ripples (repetitions):

Initial Tape: leet
                  * (Digit 2 encountered)
                /   \
         (1st copy)  (2nd copy)
               leet   leet

Through tables and diagrams, you can visualize the step-by-step changes, the impact of each operation, and how close you are to reaching the kth character, which aids in understanding the problem and in debugging your solution.

How did you infer from the problem statement that this problem can be solved using ?

Simple Explanation of the Proof

I’m having trouble understanding the proof of this algorithm. Could you explain it in a way that’s easy to understand?

Stepwise Refinement

1. Highlevel Steps

  1. Initialize Variables: Start with an empty ’tape’ and set the tape length to zero.
  2. Iterate through Encoded String: Loop through each character in the encoded string s.
    • Append Letter: If it’s a letter, append it to the ’tape’ and update the tape length.
    • Multiply for Digit: If it’s a digit, multiply the tape by that digit and update the tape length.
  3. Check for kth Letter: At each step, check if the tape length has reached or surpassed k. If it has, return the kth letter.

2. Granular, Actionable Steps

  1. Declare a variable tape as an empty string and another variable tape_length as zero.
  2. Use a for loop to iterate through each character c in the string s.
    • Inside the loop:
      1. Check if c is a letter using c.isalpha().
      • If true, append c to tape and increment tape_length by 1.
      1. Else, convert c to an integer d.
      • Update tape to tape * d and tape_length to tape_length * d.
  3. At each update to tape_length, check if tape_length >= k.
    • If true, calculate the position in the last copy of the original ’tape’ where the kth letter is and return it.

3. Independent Parts

  1. The operation to append a letter to the ’tape’ is independent of the multiplication operation and can be isolated.
  2. The check for the kth character can also be isolated as it only depends on the ’tape’ and the variable k.

4. Repeatable Patterns

  1. Letter Appending: Appending a letter to the ’tape’ and updating the length is a frequent operation.
  2. String Multiplication: The operation to multiply the ’tape’ when a digit is encountered is another pattern.
  3. Length Check: Continuously checking if tape_length >= k after each update to tape is a pattern.

By breaking down the problem this way, you set up a structured approach to solve the problem in manageable pieces.

Solution Approach and Analysis

Detailed Approach to Solving the Problem

Let’s compare the encoded string to a construction blueprint and the tape to the actual construction. We start with an empty construction site (empty tape) and follow the blueprint (encoded string) step-by-step to build the structure (decoded string).


  1. Initialize Variables:

    • Start with an empty construction site (tape="", tape_length=0).
  2. Follow the Blueprint:

    • Loop through each character (c) in the encoded string s.
  3. Check the Character:

    • If the character is a letter, it’s like adding a single brick to our construction.

      • Append the letter to tape and increment tape_length by 1.
    • If the character is a digit, it’s like duplicating our existing structure.

      • Multiply tape by the digit and update tape_length.
  4. Stopping Condition:

    • As we build, check if our construction (tape_length) has reached or exceeded the target (k).
      • If so, find the kth brick (character) in the construction and stop.

Changes in Problem Parameters:

  1. Longer Encoded Strings:

    • As the blueprint gets longer, the construction will take more time. We must be careful to optimize our approach, especially when dealing with digits that can exponentially increase the tape_length.
  2. Larger k:

    • If k is large, you could reach a point where the loop might take a long time to find the kth character, especially if the encoded string has many letters before reaching the first digit.


Let’s consider s = "leet2code3" and k = 10. Our aim is to find the 10th letter in the decoded string.

  1. tape = "", tape_length = 0
  2. Loop starts:
    • ’l’: tape = "l", tape_length = 1
    • ’e’: tape = "le", tape_length = 2
    • ’e’: tape = "lee", tape_length = 3
    • ’t’: tape = "leet", tape_length = 4
    • ‘2’: tape = "leetleet", tape_length = 8
    • ‘c’: tape = "leetleetc", tape_length = 9
    • ‘o’: tape = "leetleetco", tape_length = 10

Here, tape_length has reached k, which is 10. So, the kth character is ‘o’.

By following this methodical approach, we can decode the string while keeping an eye on the kth character, stopping when we have enough information to determine it.

Identify Invariant

An invariant in the context of this problem would be the relationship between the tape_length and the value of k. Specifically, our loop continues as long as tape_length is less than k. Once tape_length reaches or exceeds k, we can stop and find the kth character in the tape.

The invariant acts as a guardrail or a condition that remains true throughout the operation of the algorithm. It guides the execution and provides a criterion for termination, ensuring that we don’t perform unnecessary computations or overshoot our target. This helps in keeping the algorithm efficient.

Identify Loop Invariant

What is the loop invariant in this problem?

Is invariant and loop invariant the same for this problem?

Thought Process

Basic Thought Process

  1. Understand the Problem: The first step is to fully grasp what the problem is asking. Here, we’re asked to decode an encoded string according to certain rules and find the kth letter in the decoded string.

  2. Identify Cues: The cues here are the characters in the string s and the integer k. The characters could either be alphabets or digits, and the k gives us the position of the character we need to find in the decoded string.

  3. Plan the Approach: The next step is to plan how to tackle the problem. You could loop through the string and construct the decoded string, but that could be inefficient for large values. A better approach is to keep a count of the length of the decoded string, and stop when it reaches k.

  4. Work with Invariants: Our invariant is that the loop continues until the length of the tape (tape_length) reaches or exceeds k. This serves as a guardrail to stop unnecessary computation.

  5. Implement the Code: The final step is to convert the approach into Python code.

Python Code

Code is shitgpt. Generates buggy code.

Key Insights

  • We don’t need to construct the entire decoded string; we just need its length.
  • The invariant guides us when to stop decoding, making the algorithm efficient.
  • We can find the kth character by iterating over the string in reverse after the first loop, using the modulo operation to quickly jump back in the string.

This way, we keep our operations to a minimum, making it a more efficient solution.

Establishing Preconditions and Postconditions

  1. Parameters:

    • What are the inputs to the method?
    • What types are these parameters?
    • What do these parameters represent in the context of the problem?
  2. Preconditions:

    • Before this method is called, what must be true about the state of the program or the values of the parameters?
    • Are there any constraints on the input parameters?
    • Is there a specific state that the program or some part of it must be in?
  3. Method Functionality:

    • What is this method expected to do?
    • How does it interact with the inputs and the current state of the program?
  4. Postconditions:

    • After the method has been called and has returned, what is now true about the state of the program or the values of the parameters?
    • What does the return value represent or indicate?
    • What side effects, if any, does the method have?
  5. Error Handling:

    • How does the method respond if the preconditions are not met?
    • Does it throw an exception, return a special value, or do something else?

Problem Decomposition

  1. Problem Understanding:

    • Can you explain the problem in your own words? What are the key components and requirements?
  2. Initial Breakdown:

    • Start by identifying the major parts or stages of the problem. How can you break the problem into several broad subproblems?
  3. Subproblem Refinement:

    • For each subproblem identified, ask yourself if it can be further broken down. What are the smaller tasks that need to be done to solve each subproblem?
  4. Task Identification:

    • Within these smaller tasks, are there any that are repeated or very similar? Could these be generalized into a single, reusable task?
  5. Task Abstraction:

    • For each task you’ve identified, is it abstracted enough to be clear and reusable, but still makes sense in the context of the problem?
  6. Method Naming:

    • Can you give each task a simple, descriptive name that makes its purpose clear?
  7. Subproblem Interactions:

    • How do these subproblems or tasks interact with each other? In what order do they need to be performed? Are there any dependencies?

From Brute Force to Optimal Solution

Buggy code ahead:

Brute Force Solution


A naive brute-force solution is to expand the string fully and then return the kth letter.


def decodeAtIndex(s: str, k: int) -> str:
    tape = ""
    for char in s:
        if char.isalpha():
            tape += char
            tape = tape * int(char)
    return tape[k-1]


  1. Time Complexity: The time complexity of this approach is very high. In the worst case, it could be (O(2^{63})) since the decoded string can have (< 2^{63}) letters.
  2. Space Complexity: The space complexity is also the same as the time complexity because we’re storing the entire string.

Optimized Solution

The brute force solution is impractical due to its high time and space complexity. We can use two passes to solve this problem efficiently.


  1. First Pass: Compute the length of the decoded string without actually building it.
  2. Second Pass: Go backward and reduce k until it points to a character in the string.


class Solution:
    def decodeAtIndex(self, encoded_str: str, k: int) -> str:
        tape_length = 0

        # Forward loop to find tape_length
        for c in encoded_str:
            if c.isalpha():
                tape_length += 1
                tape_length *= int(c)

        # Reverse loop to find the kth character
        for c in reversed(encoded_str):
            k %= tape_length

            if k == 0 and c.isalpha():
                return c

            if c.isalpha():
                tape_length -= 1
                tape_length //= int(c)


  1. We don’t actually build the decoded string, saving space.
  2. We perform two passes, one forward and one backward, through the encoded string.

Time Complexity:

Each pass runs in (O(n)), where (n) is the length of the encoded string. So, the total time complexity is (O(n)).

Space Complexity:

We only use variables to store the current length and the current value of (k), so the space complexity is (O(1)).

The optimized solution dramatically reduces both the time and space complexity compared to the brute-force solution.

Code Explanation and Design Decisions

  1. Identify the initial parameters and explain their significance in the context of the problem statement or the solution domain.

  2. Discuss the primary loop or iteration over the input data. What does each iteration represent in terms of the problem you’re trying to solve? How does the iteration advance or contribute to the solution?

  3. If there are conditions or branches within the loop, what do these conditions signify? Explain the logical reasoning behind the branching in the context of the problem’s constraints or requirements.

  4. If there are updates or modifications to parameters within the loop, clarify why these changes are necessary. How do these modifications reflect changes in the state of the solution or the constraints of the problem?

  5. Describe any invariant that’s maintained throughout the code, and explain how it helps meet the problem’s constraints or objectives.

  6. Discuss the significance of the final output in relation to the problem statement or solution domain. What does it represent and how does it satisfy the problem’s requirements?

Remember, the focus here is not to explain what the code does on a syntactic level, but to communicate the intent and rationale behind the code in the context of the problem being solved.

Coding Constructs

Consider the code for the solution of this problem.

  1. What are the high-level problem-solving strategies or techniques being used by this code?

  2. If you had to explain the purpose of this code to a non-programmer, what would you say?

  3. Can you identify the logical elements or constructs used in this code, independent of any programming language?

  4. Could you describe the algorithmic approach used by this code in plain English?

  5. What are the key steps or operations this code is performing on the input data, and why?

  6. Can you identify the algorithmic patterns or strategies used by this code, irrespective of the specific programming language syntax?

Language Agnostic Coding Drills

Your mission is to deconstruct this code into the smallest possible learning units, each corresponding to a separate coding concept. Consider these concepts as unique coding drills that can be individually implemented and later assembled into the final solution.

  1. Dissect the code and identify each distinct concept it contains. Remember, this process should be language-agnostic and generally applicable to most modern programming languages.

  2. Once you’ve identified these coding concepts or drills, list them out in order of increasing difficulty. Provide a brief description of each concept and why it is classified at its particular difficulty level.

  3. Next, describe the problem-solving approach that would lead from the problem statement to the final solution. Think about how each of these coding drills contributes to the overall solution. Elucidate the step-by-step process involved in using these drills to solve the problem. Please refrain from writing any actual code; we’re focusing on understanding the process and strategy.

Targeted Drills in Python

Now that you’ve identified and ordered the coding concepts from a complex software code in the previous exercise, let’s focus on creating Python-based coding drills for each of those concepts.

  1. Begin by writing a separate piece of Python code that encapsulates each identified concept. These individual drills should illustrate how to implement each concept in Python. Please ensure that these are suitable even for those with a basic understanding of Python.

  2. In addition to the general concepts, identify and write coding drills for any problem-specific concepts that might be needed to create a solution. Describe why these drills are essential for our problem.

  3. Once all drills have been coded, describe how these pieces can be integrated together in the right order to solve the initial problem. Each drill should contribute to building up to the final solution.

Remember, the goal is to not only to write these drills but also to ensure that they can be cohesively assembled into one comprehensive solution.


Similar Problems

Here are 10 problems that use similar underlying concepts:

  1. Decode String (394)
    Why: It also deals with decoding a string based on some rules, and you have to calculate the length of the final string or find a specific character in the decoded string.

  2. Basic Calculator (224)
    Why: This problem requires you to evaluate a string expression, similar to how you would decode a string in the original problem.

  3. Evaluate Reverse Polish Notation (150)
    Why: Involves parsing and evaluating a string, but this time it’s in postfix notation. It tests your ability to maintain state and evaluate.

  4. Implement Queue using Stacks (232)
    Why: Like decoding a string, you are transforming one data structure into another but must adhere to specific rules.

  5. Maximum Subarray (53)
    Why: You are again dealing with arrays and searching for a specific condition that satisfies a rule, similar to finding the kth character.

  6. Valid Parentheses (20)
    Why: You need to iterate through the string and keep track of its state (open and closed brackets) in a stack, somewhat similar to keeping track of tape length and k value.

  7. First Missing Positive (41)
    Why: Involves an array and searching for a specific condition (the first missing positive number) under given rules, much like searching for the kth character in the decoded string.

  8. Remove Duplicates from Sorted Array (26)
    Why: You are transforming an array based on a specific rule, similar to decoding a string.

  9. Container With Most Water (11)
    Why: It involves iterating through an array to find two lines that together with the x-axis forms a container that can hold the most water. It’s about searching for a condition that satisfies a rule.

  10. Longest Substring Without Repeating Characters (3)
    Why: Involves traversing a string and maintaining a state to find a substring that satisfies a particular condition. This is similar to finding the kth character in the decoded string based on certain rules.

Each of these problems requires you to maintain a state and update it as you traverse through a string or array, and you often have to find a particular element or condition that satisfies a set of rules.