You are given an encoded string s. To decode the string to a tape, the encoded string is read one character at a time and the following steps are taken:
d, the entire current tape is repeatedly written d - 1 more times in total.Given an integer k, return the kth letter (1-indexed) in the decoded string.
Example 1:
Input: s = "leet2code3", k = 10 Output: "o" Explanation: The decoded string is "leetleetcodeleetleetcodeleetleetcode". The 10th letter in the string is "o".
Example 2:
Input: s = "ha22", k = 5 Output: "h" Explanation: The decoded string is "hahahaha". The 5th letter is "h".
Example 3:
Input: s = "a2345678999999999999999", k = 1 Output: "a" Explanation: The decoded string is "a" repeated 8301530446056247680 times. The 1st letter is "a".
Constraints:
2 <= s.length <= 100s consists of lowercase English letters and digits 2 through 9.s starts with a letter.1 <= k <= 109k is less than or equal to the length of the decoded string.263 letters.The key challenge in #880 Decoded String at Index is that the decoded string can grow extremely large, making it impossible to construct it explicitly. Instead of building the full string, the idea is to track the length of the decoded string as we iterate through the input.
When encountering a character, the decoded length increases by one. When encountering a digit d, it means the current decoded sequence is repeated d times, so the length multiplies accordingly. Once the total length reaches or exceeds the target index k, we can work backwards through the string to determine which character contributes to that position.
During this reverse traversal, we reduce k using modulo with the current decoded length and adjust the length when encountering digits. When a letter is found that matches the reduced index, it is the answer. This strategy avoids building the full decoded string.
Time Complexity: O(n) with a forward and backward pass. Space Complexity: O(1) since only counters are maintained.
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Length Tracking with Reverse Traversal | O(n) | O(1) |
NeetCode
This approach involves simulating the decoding process in reverse. Instead of constructing the entire decoded string, we keep track of its potential length. The core idea is to process the string in reverse while maintaining the effective length given the repeat operations.
When a digit is encountered, it denotes how many times the current segment should be repeated to match the original decoding logic. Conversely, letters are treated based on their position in this accumulated length.
By working backwards, when the kth character location matches a letter’s position, it indicates we navigated back through the expansions accurately to find the original letter corresponding to the kth position.
Time Complexity: O(n), where n is the length of the string s, as it requires scanning through the string and possibly rescanning for finding the exact letter.
Space Complexity: O(1), no extra space is used apart from constants.
1def decodeAtIndex(s: str, k: int) -> str:
2 size = 0
3
4 # Calculate the length of the decoded string
5 for char in s:
6 if char.isdigit():
7 size *= int(char)
8 else:
9 size += 1
10
11 for char in reversed(s):
12 k %= size # where's the kth character in the current pattern
13 if k == 0 and char.isalpha():
14 return char
15 if char.isdigit():
16 size //= int(char)
17 else:
18 size -= 1The Python implementation computes the length of the decoded string indirectly by iterating over the encoded characters. When digits are encountered, they scale the accumulated size. By reverting the transformations in the reverse iteration, the character’s position (k) is resolved modulo size to back-calculate its source during the construction. If k mod size becomes zero next to a letter, it confirms the correct backward mapping of the kth position corresponds to that letter.
This alternative approach iteratively simulates the expansion of the tape until the kth position is intelligibly reached. We avoid producing the content itself but calculate how long the tape would be if fully decoded. Upon encountering the kth target, we can infer shortly by aligning tape lengths with character identities. Essentially, this approach is an iterative construct of the maximal state, allowing us to peel back layers in a controlled manner.
Time Complexity: O(n), iterates over the encoded string twice.
Space Complexity: O(1), uses constant extra space.
1function decodeAtIndex(s, k) {
2 let size = 0
This approach involves calculating the length of the decoded string dynamically and using that information to find the k-th character by backtracking. Instead of generating the complete decoded string, we track the total effective length as we simulate the decoding process.
Time Complexity: O(n), where n is the length of the string, as we iterate through it twice.
Space Complexity: O(1), as only a fixed amount of space is used.
1
This approach is quite similar to the backtracking approach, but emphasizes deeper understanding by dissecting the string iteratively and understanding the effective final character through targeted calculation at each stage of string development without evaluating unneeded sections.
Time Complexity: O(n). Every character in the string is utilized only once in theory analysis.
Space Complexity: O(1).
1
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Yes, variations of string decoding and index tracking problems are common in technical interviews at companies like Amazon, Google, and Meta. This problem tests understanding of string processing, mathematical reasoning, and space optimization.
The optimal approach avoids building the full decoded string. Instead, compute the length of the decoded string while scanning the input and then traverse the string in reverse to determine which character corresponds to the target index.
The problem mainly relies on string traversal and arithmetic operations rather than heavy data structures. Some explanations mention a stack conceptually, but an optimized solution only uses variables to track decoded length and index adjustments.
The decoded string can grow exponentially because digits repeat the existing sequence multiple times. This can result in extremely large strings that exceed memory limits, so length tracking is used instead.
In this JavaScript solution, we iteratively scan to expand the tape size as dictated by the input string. Each character’s impact on the tape accumulating length is considered until the target k value can be mapped definitively backward. The deviation to discover a letter when it resolves directly to size multiples (mod k condition) signals those fit criteria.
The function decodeAtIndex calculates the size of the theoretical decoded string without actually creating it. It iterates through the string, calculating this size. As you move backwards, update k using k %= size. If you find a letter where k becomes zero, you’ve found the k-th character.
Strategically following size imperative within pragmatic encoded char-safety construct non-unravelled extends culmination effect measurable here as specifics become available technique-on-demand.