LeetCode #93 — MEDIUM

Restore IP Addresses

Move from brute-force thinking to an efficient approach using backtracking strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Backtracking

Example 1

"25525511135"

Example 2

"0000"

Example 3

"101023"

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

import java.util.*;

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

    private void backtrack(String s, int idx, int parts, List<String> cur, List<String> ans) {
        if (parts == 4) {
            if (idx == s.length()) ans.add(String.join(".", cur));
            return;
        }

        for (int len = 1; len <= 3 && idx + len <= s.length(); len++) {
            String seg = s.substring(idx, idx + len);
            if (seg.length() > 1 && seg.charAt(0) == '0') break;
            int val = Integer.parseInt(seg);
            if (val > 255) continue;
            cur.add(seg);
            backtrack(s, idx + len, parts + 1, cur, ans);
            cur.remove(cur.size() - 1);
        }
    }
}

import "strconv"

func restoreIpAddresses(s string) []string {
    ans := []string{}
    cur := []string{}

    var backtrack func(idx, parts int)
    backtrack = func(idx, parts int) {
        if parts == 4 {
            if idx == len(s) {
                ans = append(ans, cur[0]+"."+cur[1]+"."+cur[2]+"."+cur[3])
            }
            return
        }

        for l := 1; l <= 3 && idx+l <= len(s); l++ {
            seg := s[idx : idx+l]
            if len(seg) > 1 && seg[0] == '0' {
                break
            }
            v, _ := strconv.Atoi(seg)
            if v > 255 {
                continue
            }
            cur = append(cur, seg)
            backtrack(idx+l, parts+1)
            cur = cur[:len(cur)-1]
        }
    }

    backtrack(0, 0)
    return ans
}

class Solution:
    def restoreIpAddresses(self, s: str) -> List[str]:
        ans: List[str] = []
        cur: List[str] = []

        def backtrack(idx: int, parts: int) -> None:
            if parts == 4:
                if idx == len(s):
                    ans.append('.'.join(cur))
                return

            for l in range(1, 4):
                if idx + l > len(s):
                    break
                seg = s[idx:idx + l]
                if len(seg) > 1 and seg[0] == '0':
                    break
                if int(seg) > 255:
                    continue
                cur.append(seg)
                backtrack(idx + l, parts + 1)
                cur.pop()

        backtrack(0, 0)
        return ans

impl Solution {
    pub fn restore_ip_addresses(s: String) -> Vec<String> {
        let mut ans = Vec::new();
        let mut cur: Vec<String> = Vec::new();

        fn backtrack(s: &str, idx: usize, parts: usize, cur: &mut Vec<String>, ans: &mut Vec<String>) {
            if parts == 4 {
                if idx == s.len() {
                    ans.push(cur.join("."));
                }
                return;
            }

            for l in 1..=3 {
                if idx + l > s.len() {
                    break;
                }
                let seg = &s[idx..idx + l];
                if seg.len() > 1 && seg.starts_with('0') {
                    break;
                }
                if seg.parse::<i32>().unwrap() > 255 {
                    continue;
                }
                cur.push(seg.to_string());
                backtrack(s, idx + l, parts + 1, cur, ans);
                cur.pop();
            }
        }

        backtrack(&s, 0, 0, &mut cur, &mut ans);
        ans
    }
}

function restoreIpAddresses(s: string): string[] {
  const ans: string[] = [];
  const cur: string[] = [];

  const backtrack = (idx: number, parts: number): void => {
    if (parts === 4) {
      if (idx === s.length) ans.push(cur.join('.'));
      return;
    }

    for (let len = 1; len <= 3 && idx + len <= s.length; len++) {
      const seg = s.slice(idx, idx + len);
      if (seg.length > 1 && seg[0] === '0') break;
      if (Number(seg) > 255) continue;
      cur.push(seg);
      backtrack(idx + len, parts + 1);
      cur.pop();
    }
  };

  backtrack(0, 0);
  return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × 3^4)

Space

O(n)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.