Ambiguous Coordinates

Let’s write a function that will find all the possible combinations of coordinates by applying the rules described in the problem statement. The challenge is to find all the valid ways to insert commas and decimal points back into the original string.

The approach can be broken down into the following steps:

  1. Remove Parentheses: Strip the parentheses from the string.
  2. Split Combinations: Create all the possible splits for the comma.
  3. Generate Numbers: For each split, generate all valid ways to insert the decimal points into the first and second numbers.
  4. Filter Invalid Numbers: Exclude numbers with extraneous zeroes, such as “00”, “0.0”, “0.00”, “1.0”, “001”, “00.01”, etc.
  5. Combine and Return Results: Combine all the valid combinations into the final format “(x, y)” and return the result.

Here is the code:

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class Solution:
    def ambiguousCoordinates(self, s: str) -> List[str]:
        def make_numbers(num_str):
            if not num_str:
                return []
            if len(num_str) == 1:
                return [num_str]
            if num_str[0] == '0':
                if num_str[-1] == '0':
                    return []
                return ['0.' + num_str[1:]]
            if num_str[-1] == '0':
                return [num_str]
            return [num_str[:i] + '.' + num_str[i:] for i in range(1, len(num_str))] + [num_str]

        s = s[1:-1]
        n = len(s)
        results = []
        for i in range(1, n):
            left_nums = make_numbers(s[:i])
            right_nums = make_numbers(s[i:])
            for left in left_nums:
                for right in right_nums:
                    results.append(f'({left}, {right})')
        return results

This code first defines a helper function make_numbers to handle the logic of generating valid numbers with decimal points. Then, it iterates through all possible splits of the input string, using the helper function to form all valid coordinate combinations, and returns the final list of results.

The time complexity of this code is (O(n^3)), as we have a nested loop within a loop that iterates over the length of the input string, and within the inner loop, we make string slices and concatenations. Here, (n) is the length of the input string (s).