What is the time complexity of Check if an Original String Exists Given Two Encoded Strings?

The optimal solution for Check if an Original String Exists Given Two Encoded Strings runs in O(n) time and uses O(n) space. The optimal solution runs in linear time as it processes each segment only once, making it efficient for large inputs.

What is the brute force solution for Check if an Original String Exists Given Two Encoded Strings?

The brute force approach for Check if an Original String Exists Given Two Encoded Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible original strings from the encoded strings and checking if they can match. This is straightforward but inefficient since it explores all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check if an Original String Exists Given Two Encoded Strings?

Check if an Original String Exists Given Two Encoded Strings uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Two Pointers, String Parsing.

What are the interview tips for Check if an Original String Exists Given Two Encoded Strings?

Always clarify the requirements and constraints of the problem before diving into coding. Think about edge cases and how they might affect your algorithm. Explain your thought process clearly as you code, as this helps the interviewer understand your reasoning.

What are common mistakes when solving Check if an Original String Exists Given Two Encoded Strings?

Failing to handle edge cases where segments are entirely numeric or entirely alphabetic. Not considering the implications of numeric segments on subsequent segments.

#2060

Check if an Original String Exists Given Two Encoded Strings

Hard

StringDynamic ProgrammingTwo PointersString Parsing

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a two-pointer technique to compare the segments of both encoded strings efficiently. It checks if the segments can match by considering the possible lengths of numeric segments.

⚙️

Algorithm

3 steps

1Step 1: Parse both strings into segments of letters and numbers.
2Step 2: Use two pointers to traverse both segments simultaneously.
3Step 3: For each numeric segment, check if it can be matched with the corresponding letter segments by considering all possible splits.

solution.py23 lines

1def canEncode(s1, s2):
2    def parse(s):
3        import re
4        return re.findall(r'\D+|\d+', s)
5    segments1 = parse(s1)
6    segments2 = parse(s2)
7    i, j = 0, 0
8    while i < len(segments1) and j < len(segments2):
9        if segments1[i].isdigit() and segments2[j].isdigit():
10            return False
11        if segments1[i].isdigit():
12            length = int(segments1[i])
13            if length < len(segments2[j]):
14                return False
15            segments2[j] = segments2[j][length:]
16            if not segments2[j]:
17                j += 1
18        else:
19            if segments1[i] != segments2[j]:
20                return False
21            i += 1
22            j += 1
23    return i == len(segments1) and j == len(segments2)

ℹ

Complexity note: The optimal solution runs in linear time as it processes each segment only once, making it efficient for large inputs.

1Understanding how to parse and interpret segments is crucial for solving this problem.
2Numeric segments can represent multiple possible lengths, leading to various interpretations.

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