Restore Valid IPv4 Addresses

1. Step 1: Initialize Solution.restoreIpAddresses

   Replace the empty starter with the first real state owned by Solution.restoreIpAddresses. A small, named state is easier to verify than a complete algorithm. Establish it before adding the branch or loop that changes it.

2. Step 2: Assemble the primary transition

   Extend the initialized state with the next contiguous part of the popular solution. The transition explains how one input element or operation changes the state; boundaries are easier to reason about after this invariant is visible.

3. Step 3: Pass the Two Addresses case

   Complete the readable core algorithm for one representative Interview case. Build exactly four bounded segments and prune impossible remaining lengths.

4. Step 4: Harden the Leading Zero boundary

   Repair the reviewed boundary and pass the complete submission contract. Every valid address defines one sequence of four accepted segment lengths, which the search explores. Rejected segments violate an IPv4 rule, so emitted addresses are exactly the valid restorations.

• The segment 0 is valid, but 00 and 01 are not.
• recursion and backtracking

• LeetCode 93 — Restore IP Addresses
• Hello Interview — Backtracking Overview

• Complete One Backtracking Frame(opens in a new tab)

  Complete One Backtracking Frame isolates before expanding each choice, path equals the original caller-owned prefix; every emitted candidate contains exactly one additional element. That focused state discipline is required when implementing restore ip addresses as a complete Interview Problem.

Backtrack over segment lengths one through three, accepting non-leading-zero values at most 255 and emitting only after four segments consume all digits.

Why it works: Every valid address defines one sequence of four accepted segment lengths, which the search explores. Rejected segments violate an IPv4 rule, so emitted addresses are exactly the valid restorations.

class Solution:
    def restoreIpAddresses(self, text):
        """
        Checkpoint 1: initialize the state owned by this Interview contract.
        Checkpoint 2: assemble the primary transition without hiding the boundary.
        """
        result = []
        segments = []
        def backtrack(index):
            if len(segments) == 4:
                if index == len(text):
                    result.append('.'.join(segments))
                return
            remaining = len(text) - index
            slots = 4 - len(segments)
            if remaining < slots or remaining > 3 * slots:
                return
            for end in range(index + 1, min(index + 3, len(text)) + 1):
                segment = text[index:end]
                if len(segment) > 1 and segment[0] == '0':
                    break
                if int(segment) > 255:
                    continue
                segments.append(segment)
                backtrack(end)
                segments.pop()
        backtrack(0)
        return result