Ambiguous Coordinates (Leetcode 816)

Difficulty: Medium Link: Day 13: May Leetcode Challenge

Problem Description

We had some 2-dimensional coordinates, like "(1, 3)" or "(2, 0.5)".  Then, we removed all commas, decimal points, and spaces, and ended up with the string s.  Return a list of strings representing all possibilities for what our original coordinates could have been.

Our original representation never had extraneous zeroes, so we never started with numbers like “00”, “0.0”, “0.00”, “1.0”, “001”, “00.01”, or any other number that can be represented with less digits.  Also, a decimal point within a number never occurs without at least one digit occuring before it, so we never started with numbers like “.1”.

The final answer list can be returned in any order.  Also note that all coordinates in the final answer have exactly one space between them (occurring after the comma.)

Examples

Example 1:
Input: s = "(123)"
Output: ["(1, 23)", "(12, 3)", "(1.2, 3)", "(1, 2.3)"]
Example 2:
Input: s = "(00011)"
Output:  ["(0.001, 1)", "(0, 0.011)"]
Explanation: 
0.0, 00, 0001 or 00.01 are not allowed.
Example 3:
Input: s = "(0123)"
Output: ["(0, 123)", "(0, 12.3)", "(0, 1.23)", "(0.1, 23)", "(0.1, 2.3)", "(0.12, 3)"]
Example 4:
Input: s = "(100)"
Output: [(10, 0)]
Explanation: 
1.0 is not allowed.

Constraints

  • 4 <= s.length <= 12.
  • s[0] = “(“, s[s.length - 1] = “)”, and the other elements in s are digits.

Solution

It is a sort of permutation problem where we need to find the possibilities of place “,” in the digit. The position of comma(‘,’) separates the string into two parts. For example, with a string like "123", we could separate it into fragments "1" and "23""12" and "3", or "123".

After splitting the string into segments, we can evaluate the combinations of placing decimals for every segment.  for each fragment, we have a choice of where to put the period, to create a list of choices. For example, "123" could be made into "1.23""12.3", or "123".

We also need to ignore extraneous zeroes on the left and right side of the decimal.

Code

Java

class Solution { //aw
public List<String> ambiguousCoordinates(String S) {
List<String> ans = new ArrayList();
for (int i = 2; i < S.length()1; ++i)
for (String left: make(S, 1, i))
for (String right: make(S, i, S.length()1))
ans.add("(" + left + ", " + right + ")");
return ans;
}
public List<String> make(String S, int i, int j) {
// Make on S.substring(i, j)
List<String> ans = new ArrayList();
for (int d = 1; d <= ji; ++d) {
String left = S.substring(i, i+d);
String right = S.substring(i+d, j);
if ((!left.startsWith("0") || left.equals("0"))
&& !right.endsWith("0"))
ans.add(left + (d < ji ? "." : "") + right);
}
return ans;
}
}

Python

class Solution(object):
def ambiguousCoordinates(self, S):
def make(frag):
N = len(frag)
for d in xrange(1, N+1):
left = frag[:d]
right = frag[d:]
if ((not left.startswith('0') or left == '0')
and (not right.endswith('0'))):
yield left + ('.' if d != N else '') + right
S = S[1:1]
return ["({}, {})".format(*cand)
for i in xrange(1, len(S))
for cand in itertools.product(make(S[:i]), make(S[i:]))]

Complexity Analysis

Time Complexity = O(N3), where N is the length S.

Space Complexity = O(N3), to store the answer

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