93. 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 "192.168@1.1" 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:

class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> result = new ArrayList<>();
        List<String> subResult = new ArrayList<>();
        backtrack(0, 0, s, subResult, result);
        return result;
    }

    private void backtrack(int start, int parts, String s, List<String> subResult, List<String> result){
        // 如果已经有4段
        if (parts == 4){
            if (start == s.length()){
                result.add(String.join(".", subResult));
            }

            return;
        }

        // 剩余字符不够 or 太多, 剪枝
        int remaining = s.length() - start;
        if (remaining < (4 - parts) || remaining > ( 4 - parts) * 3){
            return;
        }


        // 每一段取 1~3 位
        for (int len = 1; len <= 3 && start + len <= s.length(); len++){
            String segment = s.substring(start, start + len);

            if (segment.length() > 1 && segment.charAt(0) == '0'){
                break;
            }

            int val = Integer.parseInt(segment);

            if (val > 255){
                break;
            }

            subResult.add(segment);
            backtrack(start + len, parts + 1, s, subResult, result);
            subResult.remove(subResult.size() - 1);
        }
    }
}

// TC: O(3^n)
// SC: O(n)

class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> result = new ArrayList<>();

        List<String> subResult = new ArrayList<>();

        backtrack(0, 0, s, subResult, result);
        return result;
    }

    private void backtrack(int start, int part, String s, List<String> subResult, List<String> result){
        if (part == 4){
            if (start == s.length()){
                result.add(String.join(".", subResult));
            }

            return;
        }
        // 0 1 2 3 4 5
        // 1 0 1 0 2 3      // s = 6 
        int remaining = s.length() - start;
        // 6 - 0
        // too small  6 
        if (remaining < 4 - part || remaining > (4 - part) * 3){
            return;
        }
        // [start 

        for (int len = 1; len <= 3 && (start + len <= s.length()); len++){
            String cur = s.substring(start, start + len); // 0 + 1 [0, 1)
            // 1
            if (cur.length() > 1 && cur.charAt(0) == '0'){
                return;
            }
            int val = Integer.parseInt(cur);
            if (val > 255){
                return;
            }

            subResult.add(cur);
            backtrack(start + len, part + 1, s, subResult, result);
            subResult.remove(subResult.size() - 1);
        }


    }
}