How do you solve Ambiguous Coordinates on LeetCode?

Ambiguous Coordinates (LeetCode #816) can be solved using Brute Force. The optimal approach is Brute Force with undefined time complexity and undefined space complexity. Key insight: The brute-force approach generates all possible splits of the string into two parts for the x and y coordinates. It checks each split to see if the resulting numbers are valid according to the problem constraints.

What is the time complexity of Ambiguous Coordinates?

The optimal solution for Ambiguous Coordinates runs in undefined time and uses undefined space. undefined

What algorithm or data structure is used to solve Ambiguous Coordinates?

Ambiguous Coordinates uses the following concepts: String, Backtracking, Enumeration. The recommended patterns to study are: .

#816

Ambiguous Coordinates

Medium

StringBacktrackingEnumeration

LeetCode ↗

Approaches

💡

Intuition

Time Space

The brute-force approach generates all possible splits of the string into two parts for the x and y coordinates. It checks each split to see if the resulting numbers are valid according to the problem constraints.

⚙️

Algorithm

5 steps

1Step 1: Remove the parentheses from the input string.
2Step 2: Iterate through all possible split points in the string to divide it into two parts for x and y coordinates.
3Step 3: For each split, generate all valid combinations of x and y by adding a decimal point where applicable.
4Step 4: Validate each combination to ensure it adheres to the rules (e.g., no leading zeros).
5Step 5: Collect all valid combinations into a result list.

solution.py30 lines

1# Full working Python code
2from typing import List
3
4def ambiguousCoordinates(s: str) -> List[str]:
5    s = s[1:-1]  # Remove parentheses
6    result = []
7    for i in range(1, len(s)):
8        x_part = s[:i]
9        y_part = s[i:]
10        for x in generate_valid_numbers(x_part):
11            for y in generate_valid_numbers(y_part):
12                result.append(f'({x}, {y}')
13    return result
14
15def generate_valid_numbers(part: str) -> List[str]:
16    valid_numbers = []
17    n = len(part)
18    if n == 0:
19        return valid_numbers
20    if part == '0':
21        valid_numbers.append('0')
22        return valid_numbers
23    if part[0] != '0':
24        valid_numbers.append(part)
25    for i in range(1, n):
26        left = part[:i]
27        right = part[i:]
28        if right and right[-1] != '0':
29            valid_numbers.append(left + '.' + right)
30    return valid_numbers

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