Restore IP Addresses

Problem Statement

In the realm of networking, a valid IPv4 address is defined by four integers separated by dots. Each of these integers can range from 0 to 255. Importantly, these integers should not contain any leading zeros unless the number itself is zero (for instance, "0.1.2.3" is valid but "01.1.2.3" is not).

Given a straightforward input string s composed only of digits, the challenge is to determine all conceivable valid IP addresses that can be fashioned by inserting dots in appropriate places within s. The complications include ensuring that no integers exceed 255, no section starts with a zero if it has more than one digit, and modifying the original order of characters in s is not permitted. The task requires generating each potential configuration that meets the definition of a valid IP address and returning them in any sequence.

This problem tests the ability to manipulate strings, understand constraints thoroughly, and efficiently handle multiple nested conditions in programming.

Examples

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.

Approach and Intuition

To address the problem of generating all valid IP addresses from a string, a systematic approach involving the following steps can be employed:

1. Initial Verification:

   • Check if the string is too short or too long to potentially form an IP address, specifically considering that the shortest IP address string is 4 digits (all zeros) and the longest is 12 digits (all 255). If s length is outside this range, it can dramatically reduce the number of computations by early exit.

2. Backtracking Algorithm:

   • Utilize a recursive approach to try and fit each possible segment into the potential IP address. At each step, attempt to place a dot after 1, 2, or 3 digits and check if this forms a valid segment (0-255, no leading zeros unless the segment is '0').

3. Checking Conditions:

   • For each segment considered, validate the segment:
     • It must not exceed the value 255.
     • It must not have leading zeros unless the segment is exactly '0'.

4. Recursive Exploration:

   • On confirming a valid segment, recursively attempt to place the next dot, adjusting the position in the string and the part of the IP address being constructed.
   • If a segment is invalid, backtrack and try a different configuration.
   • Continue this until four valid segments are obtained or return to a previous step if the current path fails.

5. Gather Results:

   • Each time four valid segments are constructed, combine them into a dot-separated string and add this to the list of results.

Given the constraints and provided examples, this recursive method (often implemented with backtracking) efficiently explores all potential combinations and adheres to the conditions for constructing valid IP addresses. This approach ensures that all configurations are considered without repeating or missing potential valid answers.

Solutions

class Solution {
    bool isValidSegment(const string& str, int offset, int len) {
        return len == 1 || 
               (str[offset] != '0' && 
                (len < 3 || str.substr(offset, len) <= "255"));
    }
    
public:
    vector<string> generateIPAddresses(string str) {
        vector<string> result;
        int strLength = str.length();
        for (int a = max(1, strLength - 9); a <= 3 && a <= strLength - 3; ++a) {
            if (!isValidSegment(str, 0, a)) {
                continue;
            }
            for (int b = max(1, strLength - a - 6);
                 b <= 3 && b <= strLength - a - 2; ++b) {
                if (!isValidSegment(str, a, b)) {
                    continue;
                }
                for (int c = max(1, strLength - a - b - 3);
                     c <= 3 && c <= strLength - a - b - 1; ++c) {
                    if (isValidSegment(str, a + b, c) &&
                        isValidSegment(str, a + b + c,
                                       strLength - a - b - c)) {
                        result.push_back(str.substr(0, a) + "." +
                                         str.substr(a, b) + "." +
                                         str.substr(a + b, c) + "." +
                                         str.substr(a + b + c));
                    }
                }
            }
        }
        return result;
    }
};