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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0093_restore_ip_addresses;

// #Medium #String #Backtracking #2022_06_21_Time_13_ms_(24.23%)_Space_42.8_MB_(71.26%)

import java.util.ArrayList;
import java.util.List;

/**
 * 93 - Restore IP Addresses\.
 *
 * Medium
 *
 * 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 = "1111"
 *
 * **Output:** ["1.1.1.1"] 
 *
 * **Example 4:**
 *
 * **Input:** s = "010010"
 *
 * **Output:** ["0.10.0.10","0.100.1.0"] 
 *
 * **Example 5:**
 *
 * **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:**
 *
 * *   `0 <= s.length <= 20`
 * *   `s` consists of digits only.
**/
public class Solution {
    public List restoreIpAddresses(String s) {
        List results = new ArrayList<>();
        step(s, 0, new int[4], 0, results);
        return results;
    }

    void step(String s, int pos, int[] octets, int count, List results) {
        if (count == 4 && pos == s.length()) {
            results.add(
                    String.valueOf(octets[0])
                            + '.'
                            + octets[1]
                            + '.'
                            + octets[2]
                            + '.'
                            + octets[3]);
        } else if (count < 4 && pos < 12) {
            int octet = 0;
            for (int i = 0; i < 3; i++) {
                if (pos + i < s.length()) {
                    int digit = s.charAt(pos + i) - '0';
                    octet = octet * 10 + digit;
                    if (octet < 256) {
                        octets[count] = octet;
                        step(s, pos + i + 1, octets, count + 1, results);
                    }
                    if (i == 0 && digit == 0) {
                        break;
                    }
                }
            }
        }
    }
}