grid47 Exploring patterns and algorithms
Aug 17, 2024
7 min read
Solution to LeetCode 816: Ambiguous Coordinates Problem
You are given a string s representing coordinates that have been encoded by removing commas, decimal points, and spaces. Your task is to restore the coordinates to their original form by considering all possible valid ways the digits could be split into two parts: a left part representing the x-coordinate and a right part representing the y-coordinate. Both parts may optionally contain a decimal point. The final coordinates must follow specific rules: no leading zeros are allowed, and the decimal point must be placed correctly. Return a list of all possible valid coordinates, formatted as ‘(x, y)’, where x and y represent the respective coordinates.
Problem
Approach
Steps
Complexity
Input: You are given a string s that encodes two coordinates in the format '(x, y)'. The string consists of digits and is enclosed by parentheses. The commas, decimal points, and spaces have been removed.
Example: Input: s = '(78910)'
Constraints:
• The length of s is between 4 and 12 characters.
• The first character of s is '(' and the last character is ')'.
• All other characters are digits.
Output: Return a list of strings, where each string represents a possible valid coordinate in the format '(x, y)', with exactly one space after the comma between the x-coordinate and the y-coordinate.
• The output list should contain all possible valid coordinate representations.
Goal: To generate all possible valid coordinates from the given encoded string by splitting the string into two parts and checking the validity of each part.
Steps:
• Split the string s at every possible position to create two substrings, each representing a potential x and y coordinate.
• For each substring, check if it represents a valid integer or float coordinate (no leading zeros, correct placement of decimal points).
• Return all valid coordinates in the specified format.
Goal: The solution should handle strings of length between 4 and 12 and ensure that no invalid coordinates (with leading zeros or incorrect decimal placement) are included in the result.
Steps:
• Do not include coordinates that have invalid decimal points or leading zeros.
• Ensure that each output coordinate follows the format '(x, y)'.
Assumptions:
• The input string will always start with '(' and end with ')'.
• The input will only contain digits after removing the commas and spaces.
• Input: Input: s = '(12345)'
• Explanation: The string '(12345)' can be split into several valid coordinates: '(1, 23.45)', '(1, 2345)', '(12, 3.45)', '(12, 345)', '(123, 4.5)', and '(123, 45)'.
Approach: We approach the problem by iterating over all possible positions to split the string s into two parts: one for the x-coordinate and the other for the y-coordinate. For each split, we generate all valid representations of the x and y coordinates and combine them into the desired format.
Observations:
• This is a string manipulation problem that requires careful handling of decimal points and leading zeros.
• We need to check each possible split and ensure that the resulting coordinates are valid based on the problem constraints.
Steps:
• Loop through all positions to split the string s into two parts.
• For each split, extract the left and right parts and generate all valid representations of both coordinates (x and y).
• Check for valid formats (no leading zeros unless the number is '0', and handle decimal points correctly).
• Format each valid pair as '(x, y)' and add it to the result list.
Empty Inputs:
• Input is not allowed to be empty, as the length of s is guaranteed to be between 4 and 12.
Large Inputs:
• Ensure the solution can handle strings of maximum length (12).
Special Values:
• Test cases where coordinates contain leading zeros (e.g., '(00123)') or decimal points placed incorrectly (e.g., '(001.23)') should be handled correctly.
Constraints:
• No leading zeros are allowed unless the coordinate is '0'.
vector<string> ambiguousCoordinates(string S) {
int n = S.size();
vector<string> res;
for (int i =1; i < n -2; ++i) {
vector<string> A = f(S.substr(1, i)), B = f(S.substr(i +1, n -2- i));
for (auto&a : A) for (auto&b : B) res.push_back("("+ a +", "+ b +")");
}
return res;
}
vector<string> f(string S) {
int n = S.size();
if (n ==0|| (n >1&& S[0] =='0'&& S[n -1] =='0')) return {};
if (n >1&& S[0] =='0') return {"0."+ S.substr(1)};
if (n ==1|| S[n -1] =='0') return {S};
vector<string> res = {S};
for (int i =1; i < n; ++i) res.push_back(S.substr(0, i) +'.'+ S.substr(i));
return res;
}
1 : Function Definition
vector<string> ambiguousCoordinates(string S) {
This line defines the function ambiguousCoordinates, which takes a string S as input and returns a vector of strings representing all valid coordinates formed from the input string.
2 : Variable Declaration
int n = S.size();
This line calculates the size of the input string and stores it in the variable n.
3 : Variable Declaration
vector<string> res;
A vector res is declared to store the resulting coordinate pairs.
4 : Function Call
vector<string> A = f(S.substr(1, i)), B = f(S.substr(i +1, n -2- i));
This line splits the string into two substrings and passes them to the helper function f, which generates all valid numbers from those substrings.
5 : Nested Loop
for (auto&a : A) for (auto&b : B) res.push_back("("+ a +", "+ b +")");
These nested loops iterate over the possible combinations of the two substrings A and B, adding the resulting pairs of coordinates to the result vector res.
6 : Return
return res;
This returns the vector res, containing all valid coordinate pairs formed from the input string.
7 : Function Definition
vector<string> f(string S) {
This is a helper function f, which takes a string S and returns a vector of valid numbers that can be formed from it, considering decimal placements.
8 : Variable Declaration
int n = S.size();
This line calculates the size of the string S and stores it in variable n.
This checks if the string is empty or contains invalid decimal numbers (like '00'). If true, it returns an empty vector.
10 : Condition
if (n >1&& S[0] =='0') return {"0."+ S.substr(1)};
If the string starts with '0' and has more than one character, it returns a valid decimal number like '0.x' by placing a decimal point after '0'.
11 : Condition
if (n ==1|| S[n -1] =='0') return {S};
If the string is a single character or ends with '0', it returns the string as it is (no decimal needed).
12 : Variable Declaration
vector<string> res = {S};
A vector res is initialized with the original string S.
13 : Loop
for (int i =1; i < n; ++i) res.push_back(S.substr(0, i) +'.'+ S.substr(i));
This loop adds decimal points at different positions within the string and stores the results in the vector res.
14 : Return
return res;
This returns the vector res containing all valid numbers formed by adding decimal points at various positions.
Best Case: O(n^2), where n is the length of the string s. This is because we loop through all possible splits and check the validity of the resulting substrings.
Average Case: O(n^2)
Worst Case: O(n^2), as we check all splits and generate all valid coordinate combinations.
Description: The time complexity is quadratic due to the nested iteration for splitting and checking substrings.
Best Case: O(n), in case the input string is very simple and only a few valid combinations exist.
Worst Case: O(n^2), as the result list can contain up to n^2 valid coordinate combinations.
Description: The space complexity is quadratic because we store all possible valid coordinate combinations.
