93. Restore IP Addresses - Explanation

Description

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.

You are 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 = "101023"

Output: ["1.0.10.23","1.0.102.3","10.1.0.23","10.10.2.3","101.0.2.3"]

Example 2:

Input: s = "010010"

Output: ["0.10.0.10","0.100.1.0"]

Constraints:

1 <= s.length <= 20
s consists of digits only.

Topics

String Backtracking

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Prerequisites

Before attempting this problem, you should be comfortable with:

Backtracking - Exploring all possible combinations by making choices and undoing them when they lead to invalid solutions
Recursion - Breaking down the problem into smaller subproblems (placing one IP segment at a time)
String Manipulation - Extracting and validating substrings for IP segment constraints

1. Backtracking

Intuition

A valid IP address has exactly four segments, each containing 1 to 3 digits with a value between 0 and 255. Leading zeros are not allowed except for the segment "0" itself. Backtracking lets us explore all possible ways to place three dots in the string. At each step, we try taking 1, 2, or 3 characters for the current segment, validate it, and recurse for the remaining segments.

Algorithm

If the string length exceeds 12, return an empty list (maximum valid length is 12 digits).
Define a recursive function that tracks the current position i, number of segments placed dots, and the IP being built curIP.
Base case: if 4 segments are placed and we have consumed the entire string, add the IP to results.
For each call, try segment lengths of 1, 2, and 3 characters starting at the current position.
Skip segments with leading zeros (unless the segment is just "0") and values >= 256.
Recurse with the next position, incremented segment count, and updated IP string.
Return all valid IP addresses found.

class Solution:
    def restoreIpAddresses(self, s: str) -> List[str]:
        res = []
        if len(s) > 12:
            return res

        def backtrack(i, dots, curIP):
            if dots == 4 and i == len(s):
                res.append(curIP[:-1])
                return
            if dots > 4:
                return

            for j in range(i, min(i + 3, len(s))):
                if i != j and s[i] == "0":
                    continue
                if int(s[i: j + 1]) < 256:
                    backtrack(j + 1, dots + 1, curIP + s[i: j + 1] + ".")

        backtrack(0, 0, "")
        return res

public class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> res = new ArrayList<>();
        if (s.length() > 12) return res;

        backtrack(0, 0, "", s, res);
        return res;
    }

    private void backtrack(int i, int dots, String curIP, String s, List<String> res) {
        if (dots == 4 && i == s.length()) {
            res.add(curIP.substring(0, curIP.length() - 1));
            return;
        }
        if (dots > 4) return;

        for (int j = i; j < Math.min(i + 3, s.length()); j++) {
            if (i != j && s.charAt(i) == '0') continue;
            if (Integer.parseInt(s.substring(i, j + 1)) < 256) {
                backtrack(j + 1, dots + 1, curIP + s.substring(i, j + 1) + ".", s, res);
            }
        }
    }
}

class Solution {
        vector<string> res;

public:
    vector<string> restoreIpAddresses(string s) {
        if (s.length() > 12) return res;
        backtrack(s, 0, 0, "");
        return res;
    }

private:
    void backtrack(string& s, int i, int dots, string curIP) {
        if (dots == 4 && i == s.size()) {
            res.push_back(curIP.substr(0, curIP.size() - 1));
            return;
        }
        if (dots > 4) return;

        for (int j = i; j < min(i + 3, (int)s.size()); j++) {
            if (i != j && s[i] == '0') continue;
            if (stoi(s.substr(i, j - i + 1)) < 256) {
                backtrack(s, j + 1, dots + 1, curIP + s.substr(i, j - i + 1) + ".");
            }
        }
    }
};

class Solution {
    /**
     * @param {string} s
     * @return {string[]}
     */
    restoreIpAddresses(s) {
        const res = [];
        if (s.length > 12) return res;

        const backtrack = (i, dots, curIP) => {
            if (dots === 4 && i === s.length) {
                res.push(curIP.slice(0, -1));
                return;
            }
            if (dots > 4) return;

            for (let j = i; j < Math.min(i + 3, s.length); j++) {
                if (i !== j && s[i] === '0') continue;
                if (parseInt(s.slice(i, j + 1)) < 256) {
                    backtrack(j + 1, dots + 1, curIP + s.slice(i, j + 1) + '.');
                }
            }
        };

        backtrack(0, 0, '');
        return res;
    }
}

public class Solution {
    public List<string> RestoreIpAddresses(string s) {
        List<string> res = new List<string>();
        if (s.Length > 12) return res;

        void Backtrack(int i, int dots, string curIP) {
            if (dots == 4 && i == s.Length) {
                res.Add(curIP.Substring(0, curIP.Length - 1));
                return;
            }
            if (dots > 4) return;

            for (int j = i; j < Math.Min(i + 3, s.Length); j++) {
                if (i != j && s[i] == '0') continue;
                string part = s.Substring(i, j - i + 1);
                if (int.Parse(part) < 256) {
                    Backtrack(j + 1, dots + 1, curIP + part + ".");
                }
            }
        }

        Backtrack(0, 0, "");
        return res;
    }
}

func restoreIpAddresses(s string) []string {
    res := []string{}
    if len(s) > 12 {
        return res
    }

    var backtrack func(i, dots int, curIP string)
    backtrack = func(i, dots int, curIP string) {
        if dots == 4 && i == len(s) {
            res = append(res, curIP[:len(curIP)-1])
            return
        }
        if dots > 4 {
            return
        }

        for j := i; j < min(i+3, len(s)); j++ {
            if i != j && s[i] == '0' {
                continue
            }
            part := s[i : j+1]
            num, _ := strconv.Atoi(part)
            if num < 256 {
                backtrack(j+1, dots+1, curIP+part+".")
            }
        }
    }

    backtrack(0, 0, "")
    return res
}

func min(a, b int) int {
    if a < b {
        return a
    }
    return b
}

class Solution {
    fun restoreIpAddresses(s: String): List<String> {
        val res = mutableListOf<String>()
        if (s.length > 12) return res

        fun backtrack(i: Int, dots: Int, curIP: String) {
            if (dots == 4 && i == s.length) {
                res.add(curIP.dropLast(1))
                return
            }
            if (dots > 4) return

            for (j in i until minOf(i + 3, s.length)) {
                if (i != j && s[i] == '0') continue
                val part = s.substring(i, j + 1)
                if (part.toInt() < 256) {
                    backtrack(j + 1, dots + 1, curIP + part + ".")
                }
            }
        }

        backtrack(0, 0, "")
        return res
    }
}

class Solution {
    func restoreIpAddresses(_ s: String) -> [String] {
        var res = [String]()
        if s.count > 12 { return res }
        let chars = Array(s)

        func backtrack(_ i: Int, _ dots: Int, _ curIP: String) {
            if dots == 4 && i == chars.count {
                res.append(String(curIP.dropLast()))
                return
            }
            if dots > 4 { return }

            for j in i..<min(i + 3, chars.count) {
                if i != j && chars[i] == "0" { continue }
                let part = String(chars[i...j])
                if let num = Int(part), num < 256 {
                    backtrack(j + 1, dots + 1, curIP + part + ".")
                }
            }
        }

        backtrack(0, 0, "")
        return res
    }
}

Time & Space Complexity

Time complexity: $O(m ^ n * n)$
Space complexity: $O(m * n)$

Where $m$ is equals to $3$ as there are at most three digits in a valid segment and $n$ is equals to $4$ as there are four segments in a valid IP.

2. Iteration

Intuition

Since there are exactly four segments and each can have 1, 2, or 3 characters, we can enumerate all 81 combinations (3^4) of segment lengths directly. For each combination, we check if the total length matches the input string and whether each resulting segment is valid. This avoids recursion overhead while still exploring all possibilities.

Algorithm

If the string length exceeds 12, return an empty list.
Use four nested loops, each iterating from 1 to 3 (representing segment lengths seg1, seg2, seg3, seg4).
If the sum of all four segment lengths does not equal the string length, skip this combination.
Extract the four substrings based on the current segment lengths.
Validate each segment: no leading zeros (unless single digit) and value <= 255.
If all segments are valid, join them with dots and add to res.
Return all valid IP addresses found.

class Solution:
    def restoreIpAddresses(self, s: str) -> List[str]:
        res = []
        if len(s) > 12:
            return res

        def valid(num):
            return len(num) == 1 or (int(num) < 256 and num[0] != "0")

        def add(s1, s2, s3, s4):
            if s1 + s2 + s3 + s4 != len(s):
                return

            num1 = s[:s1]
            num2 = s[s1:s1+s2]
            num3 = s[s1+s2:s1+s2+s3]
            num4 = s[s1+s2+s3:]
            if valid(num1) and valid(num2) and valid(num3) and valid(num4):
                res.append(num1 + "." + num2 + "." + num3 + "." + num4)

        for seg1 in range(1, 4):
            for seg2 in range(1, 4):
                for seg3 in range(1, 4):
                    for seg4 in range(1, 4):
                        add(seg1, seg2, seg3, seg4)

        return res

public class Solution {
    public List<String> restoreIpAddresses(String s) {
        List<String> res = new ArrayList<>();
        if (s.length() > 12) return res;

        for (int seg1 = 1; seg1 < 4; seg1++) {
            for (int seg2 = 1; seg2 < 4; seg2++) {
                for (int seg3 = 1; seg3 < 4; seg3++) {
                    for (int seg4 = 1; seg4 < 4; seg4++) {
                        if (seg1 + seg2 + seg3 + seg4 != s.length()) continue;

                        String num1 = s.substring(0, seg1);
                        String num2 = s.substring(seg1, seg1 + seg2);
                        String num3 = s.substring(seg1 + seg2, seg1 + seg2 + seg3);
                        String num4 = s.substring(seg1 + seg2 + seg3);

                        if (isValid(num1) && isValid(num2) && isValid(num3) && isValid(num4)) {
                            res.add(num1 + "." + num2 + "." + num3 + "." + num4);
                        }
                    }
                }
            }
        }
        return res;
    }

    private boolean isValid(String num) {
        if (num.length() > 1 && num.charAt(0) == '0') return false;
        int value = Integer.parseInt(num);
        return value <= 255;
    }
}

class Solution {
public:
    vector<string> restoreIpAddresses(string s) {
        vector<string> res;
        if (s.size() > 12) return res;

        auto valid = [&](string& num) {
            if (num.size() > 1 && num[0] == '0') return false;
            int value = stoi(num);
            return value <= 255;
        };

        for (int seg1 = 1; seg1 < 4; ++seg1) {
            for (int seg2 = 1; seg2 < 4; ++seg2) {
                for (int seg3 = 1; seg3 < 4; ++seg3) {
                    for (int seg4 = 1; seg4 < 4; ++seg4) {
                        if (seg1 + seg2 + seg3 + seg4 != s.size()) continue;

                        string num1 = s.substr(0, seg1);
                        string num2 = s.substr(seg1, seg2);
                        string num3 = s.substr(seg1 + seg2, seg3);
                        string num4 = s.substr(seg1 + seg2 + seg3);

                        if (valid(num1) && valid(num2) && valid(num3) && valid(num4)) {
                            res.push_back(num1 + "." + num2 + "." + num3 + "." + num4);
                        }
                    }
                }
            }
        }
        return res;
    }
};

class Solution {
    /**
     * @param {string} s
     * @return {string[]}
     */
    restoreIpAddresses(s) {
        const res = [];
        if (s.length > 12) return res;

        const isValid = (num) => {
            if (num.length > 1 && num[0] === '0') return false;
            const value = parseInt(num, 10);
            return value <= 255;
        };

        for (let seg1 = 1; seg1 < 4; seg1++) {
            for (let seg2 = 1; seg2 < 4; seg2++) {
                for (let seg3 = 1; seg3 < 4; seg3++) {
                    for (let seg4 = 1; seg4 < 4; seg4++) {
                        if (seg1 + seg2 + seg3 + seg4 !== s.length) continue;

                        const num1 = s.substring(0, seg1);
                        const num2 = s.substring(seg1, seg1 + seg2);
                        const num3 = s.substring(
                            seg1 + seg2,
                            seg1 + seg2 + seg3,
                        );
                        const num4 = s.substring(seg1 + seg2 + seg3);

                        if (
                            isValid(num1) &&
                            isValid(num2) &&
                            isValid(num3) &&
                            isValid(num4)
                        ) {
                            res.push(`${num1}.${num2}.${num3}.${num4}`);
                        }
                    }
                }
            }
        }
        return res;
    }
}

public class Solution {
    public List<string> RestoreIpAddresses(string s) {
        List<string> res = new List<string>();
        if (s.Length > 12) return res;

        bool Valid(string num) {
            return num.Length == 1 || (int.Parse(num) < 256 && num[0] != '0');
        }

        void Add(int s1, int s2, int s3, int s4) {
            if (s1 + s2 + s3 + s4 != s.Length) return;

            string num1 = s.Substring(0, s1);
            string num2 = s.Substring(s1, s2);
            string num3 = s.Substring(s1 + s2, s3);
            string num4 = s.Substring(s1 + s2 + s3);

            if (Valid(num1) && Valid(num2) && Valid(num3) && Valid(num4)) {
                res.Add(num1 + "." + num2 + "." + num3 + "." + num4);
            }
        }

        for (int seg1 = 1; seg1 < 4; seg1++) {
            for (int seg2 = 1; seg2 < 4; seg2++) {
                for (int seg3 = 1; seg3 < 4; seg3++) {
                    for (int seg4 = 1; seg4 < 4; seg4++) {
                        Add(seg1, seg2, seg3, seg4);
                    }
                }
            }
        }

        return res;
    }
}

func restoreIpAddresses(s string) []string {
    res := []string{}
    if len(s) > 12 {
        return res
    }

    valid := func(num string) bool {
        if len(num) > 1 && num[0] == '0' {
            return false
        }
        val, _ := strconv.Atoi(num)
        return val <= 255
    }

    for seg1 := 1; seg1 < 4; seg1++ {
        for seg2 := 1; seg2 < 4; seg2++ {
            for seg3 := 1; seg3 < 4; seg3++ {
                for seg4 := 1; seg4 < 4; seg4++ {
                    if seg1+seg2+seg3+seg4 != len(s) {
                        continue
                    }

                    num1 := s[:seg1]
                    num2 := s[seg1 : seg1+seg2]
                    num3 := s[seg1+seg2 : seg1+seg2+seg3]
                    num4 := s[seg1+seg2+seg3:]

                    if valid(num1) && valid(num2) && valid(num3) && valid(num4) {
                        res = append(res, num1+"."+num2+"."+num3+"."+num4)
                    }
                }
            }
        }
    }
    return res
}

class Solution {
    fun restoreIpAddresses(s: String): List<String> {
        val res = mutableListOf<String>()
        if (s.length > 12) return res

        fun valid(num: String): Boolean {
            if (num.length > 1 && num[0] == '0') return false
            return num.toInt() <= 255
        }

        for (seg1 in 1 until 4) {
            for (seg2 in 1 until 4) {
                for (seg3 in 1 until 4) {
                    for (seg4 in 1 until 4) {
                        if (seg1 + seg2 + seg3 + seg4 != s.length) continue

                        val num1 = s.substring(0, seg1)
                        val num2 = s.substring(seg1, seg1 + seg2)
                        val num3 = s.substring(seg1 + seg2, seg1 + seg2 + seg3)
                        val num4 = s.substring(seg1 + seg2 + seg3)

                        if (valid(num1) && valid(num2) && valid(num3) && valid(num4)) {
                            res.add("$num1.$num2.$num3.$num4")
                        }
                    }
                }
            }
        }
        return res
    }
}

class Solution {
    func restoreIpAddresses(_ s: String) -> [String] {
        var res = [String]()
        if s.count > 12 { return res }
        let chars = Array(s)

        func valid(_ num: String) -> Bool {
            if num.count > 1 && num.first == "0" { return false }
            guard let val = Int(num) else { return false }
            return val <= 255
        }

        for seg1 in 1..<4 {
            for seg2 in 1..<4 {
                for seg3 in 1..<4 {
                    for seg4 in 1..<4 {
                        if seg1 + seg2 + seg3 + seg4 != chars.count { continue }

                        let num1 = String(chars[0..<seg1])
                        let num2 = String(chars[seg1..<(seg1 + seg2)])
                        let num3 = String(chars[(seg1 + seg2)..<(seg1 + seg2 + seg3)])
                        let num4 = String(chars[(seg1 + seg2 + seg3)...])

                        if valid(num1) && valid(num2) && valid(num3) && valid(num4) {
                            res.append("\(num1).\(num2).\(num3).\(num4)")
                        }
                    }
                }
            }
        }
        return res
    }
}

Time & Space Complexity

Time complexity: $O(m ^ n * n)$
Space complexity: $O(m * n)$

Where $m$ is equals to $3$ as there are at most three digits in a valid segment and $n$ is equals to $4$ as there are four segments in a valid IP.

Common Pitfalls

Allowing Leading Zeros in Multi-Digit Segments

Segments like "01" or "001" are invalid in IP addresses, but "0" alone is valid. The validation must specifically reject multi-digit segments that start with zero while accepting single-digit zeros.

Missing the Upper Bound Check for Segment Values

Each segment must be at most 255. Forgetting to check this constraint or using < 256 instead of <= 255 can lead to subtle bugs, especially for three-digit segments like "256" which appear valid at first glance.

Not Validating Total String Length Early

An IP address can have at most 12 digits (four segments of 3 digits each) and at least 4 digits (four segments of 1 digit each). Failing to check these bounds upfront leads to unnecessary computation on inputs that cannot possibly form valid IPs.