Restore IP Addresses
A valid IP address consists of exactly four integers separated by single dots. Each integer is between 0
and 255
(inclusive) and cannot have leading zeros.
For example, "0.1.2.201"
and "192.168.1.1"
are valid IP addresses, but "0.011.255.245"
, "192.168.1.312"
and "[email protected]"
are invalid IP addresses.
Given a string s
containing only digits, return all possible valid IP addresses that can be formed by inserting dots into s
. You are not allowed to reorder or remove any digits in s
. You may return the valid IP addresses in any order.
Example 1:
Input: s = "25525511135"
Output: ["255.255.11.135","255.255.111.35"]
Example 2:
Input: s = "0000"
Output: ["0.0.0.0"]
Example 3:
Input: s = "101023"
Output: ["1.0.10.23","1.0.102.3","10.1.0.23","10.10.2.3","101.0.2.3"]
Constraints:
1 <= s.length <= 20
s
consists of digits only.
Solution
We want to apply our template backtracking1 to this problem. To fill in the logic:
is_leaf
:start_index == len(s) and
len(path) == 4`, when all of the digits are used and there are exactly 4 segments.get_edges
: the edges are the potential segments (of size 1-3) that starts atstart_index
.is_valid
: is the edge (integer) between 0 to 255? and does it not have a leading zero?
Implementation
1def restoreIpAddresses(self, s: str) -> List[str]:
2 def to_ip_address(path):
3 address = path[0]
4 for i in range(1, 4): address += "." + path[i]
5 return address
6
7 def get_edges(start_index):
8 segments = []
9 for i in range(start_index, start_index + 3):
10 if i < len(s) : # if not out of bound
11 segments.append(s[start_index:i+1]) # up to and including s[i]
12 return segments
13
14 def is_valid(num):
15 if num == "0": return True
16 elif num[0] == "0": return False # leading zero
17 elif int(num) > 255: return False # out of range
18 else: return True
19
20 def dfs(start_index, path):
21 if len(path) > 4: return
22 if start_index == len(s): # if all digits are used
23 if len(path) == 4: # and there are exactly four segments
24 ans.append(to_ip_address(path)) # add address to the result
25 return
26 for edge in get_edges(start_index):
27 if is_valid(edge):
28 path.append(edge)
29 dfs(start_index + len(edge), path)
30 path.pop()
31 ans = []
32 print(s[0:len(s)+1])
33 dfs(0, [])
34 return ans
How would you design a stack which has a function min
that returns the minimum element in the stack, in addition to push
and pop
? All push
, pop
, min
should have running time O(1)
.
Solution Implementation
Problem: Given a list of tasks and a list of requirements, compute a sequence of tasks that can be performed, such that we complete every task once while satisfying all the requirements.
Which of the following method should we use to solve this problem?
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