LeetCode #3451 — HARD

Find Invalid IP Addresses

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Table: logs

+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| log_id      | int     |
| ip          | varchar |
| status_code | int     |
+-------------+---------+
log_id is the unique key for this table.
Each row contains server access log information including IP address and HTTP status code.

Write a solution to find invalid IP addresses. An IPv4 address is invalid if it meets any of these conditions:

Contains numbers greater than 255 in any octet
Has leading zeros in any octet (like 01.02.03.04)
Has less or more than 4 octets

Return the result table ordered by invalid_count, ip in descending order respectively.

The result format is in the following example.

Example:

Input:

logs table:

+--------+---------------+-------------+
| log_id | ip            | status_code | 
+--------+---------------+-------------+
| 1      | 192.168.1.1   | 200         | 
| 2      | 256.1.2.3     | 404         | 
| 3      | 192.168.001.1 | 200         | 
| 4      | 192.168.1.1   | 200         | 
| 5      | 192.168.1     | 500         | 
| 6      | 256.1.2.3     | 404         | 
| 7      | 192.168.001.1 | 200         | 
+--------+---------------+-------------+

Output:

+---------------+--------------+
| ip            | invalid_count|
+---------------+--------------+
| 256.1.2.3     | 2            |
| 192.168.001.1 | 2            |
| 192.168.1     | 1            |
+---------------+--------------+

Explanation:

256.1.2.3 is invalid because 256 > 255
192.168.001.1 is invalid because of leading zeros
192.168.1 is invalid because it has only 3 octets

The output table is ordered by invalid_count, ip in descending order respectively.

Step 01

Brute Force Baseline

Problem summary: Table: logs +-------------+---------+ | Column Name | Type | +-------------+---------+ | log_id | int | | ip | varchar | | status_code | int | +-------------+---------+ log_id is the unique key for this table. Each row contains server access log information including IP address and HTTP status code. Write a solution to find invalid IP addresses. An IPv4 address is invalid if it meets any of these conditions: Contains numbers greater than 255 in any octet Has leading zeros in any octet (like 01.02.03.04) Has less or more than 4 octets Return the result table ordered by invalid_count, ip in descending order respectively. The result format is in the following example.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

{"headers":{"logs":["log_id","ip","status_code"]},"rows":{"logs":[[1,"192.168.1.1",200],[2,"256.1.2.3",404],[3,"192.168.001.1",200],[4,"192.168.1.1",200],[5,"192.168.1",500],[6,"256.1.2.3",404],[7,"192.168.001.1",200]]}}

Step 02

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

Largest constraint values

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

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3451: Find Invalid IP Addresses
// Auto-generated Java example from py.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (py):
// # Accepted solution for LeetCode #3451: Find Invalid IP Addresses
// import pandas as pd
// 
// 
// def find_invalid_ips(logs: pd.DataFrame) -> pd.DataFrame:
//     def is_valid_ip(ip: str) -> bool:
//         octets = ip.split(".")
//         if len(octets) != 4:
//             return False
//         for octet in octets:
//             if not octet.isdigit():
//                 return False
//             value = int(octet)
//             if not 0 <= value <= 255 or octet != str(value):
//                 return False
//         return True
// 
//     logs["is_valid"] = logs["ip"].apply(is_valid_ip)
//     invalid_ips = logs[~logs["is_valid"]]
//     invalid_count = invalid_ips["ip"].value_counts().reset_index()
//     invalid_count.columns = ["ip", "invalid_count"]
//     result = invalid_count.sort_values(
//         by=["invalid_count", "ip"], ascending=[False, False]
//     )
//     return result

// Accepted solution for LeetCode #3451: Find Invalid IP Addresses
// Auto-generated Go example from py.
func exampleSolution() {
}
// Reference (py):
// # Accepted solution for LeetCode #3451: Find Invalid IP Addresses
// import pandas as pd
// 
// 
// def find_invalid_ips(logs: pd.DataFrame) -> pd.DataFrame:
//     def is_valid_ip(ip: str) -> bool:
//         octets = ip.split(".")
//         if len(octets) != 4:
//             return False
//         for octet in octets:
//             if not octet.isdigit():
//                 return False
//             value = int(octet)
//             if not 0 <= value <= 255 or octet != str(value):
//                 return False
//         return True
// 
//     logs["is_valid"] = logs["ip"].apply(is_valid_ip)
//     invalid_ips = logs[~logs["is_valid"]]
//     invalid_count = invalid_ips["ip"].value_counts().reset_index()
//     invalid_count.columns = ["ip", "invalid_count"]
//     result = invalid_count.sort_values(
//         by=["invalid_count", "ip"], ascending=[False, False]
//     )
//     return result

# Accepted solution for LeetCode #3451: Find Invalid IP Addresses
import pandas as pd


def find_invalid_ips(logs: pd.DataFrame) -> pd.DataFrame:
    def is_valid_ip(ip: str) -> bool:
        octets = ip.split(".")
        if len(octets) != 4:
            return False
        for octet in octets:
            if not octet.isdigit():
                return False
            value = int(octet)
            if not 0 <= value <= 255 or octet != str(value):
                return False
        return True

    logs["is_valid"] = logs["ip"].apply(is_valid_ip)
    invalid_ips = logs[~logs["is_valid"]]
    invalid_count = invalid_ips["ip"].value_counts().reset_index()
    invalid_count.columns = ["ip", "invalid_count"]
    result = invalid_count.sort_values(
        by=["invalid_count", "ip"], ascending=[False, False]
    )
    return result

// Accepted solution for LeetCode #3451: Find Invalid IP Addresses
// Rust example auto-generated from py reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (py):
// # Accepted solution for LeetCode #3451: Find Invalid IP Addresses
// import pandas as pd
// 
// 
// def find_invalid_ips(logs: pd.DataFrame) -> pd.DataFrame:
//     def is_valid_ip(ip: str) -> bool:
//         octets = ip.split(".")
//         if len(octets) != 4:
//             return False
//         for octet in octets:
//             if not octet.isdigit():
//                 return False
//             value = int(octet)
//             if not 0 <= value <= 255 or octet != str(value):
//                 return False
//         return True
// 
//     logs["is_valid"] = logs["ip"].apply(is_valid_ip)
//     invalid_ips = logs[~logs["is_valid"]]
//     invalid_count = invalid_ips["ip"].value_counts().reset_index()
//     invalid_count.columns = ["ip", "invalid_count"]
//     result = invalid_count.sort_values(
//         by=["invalid_count", "ip"], ascending=[False, False]
//     )
//     return result

// Accepted solution for LeetCode #3451: Find Invalid IP Addresses
// Auto-generated TypeScript example from py.
function exampleSolution(): void {
}
// Reference (py):
// # Accepted solution for LeetCode #3451: Find Invalid IP Addresses
// import pandas as pd
// 
// 
// def find_invalid_ips(logs: pd.DataFrame) -> pd.DataFrame:
//     def is_valid_ip(ip: str) -> bool:
//         octets = ip.split(".")
//         if len(octets) != 4:
//             return False
//         for octet in octets:
//             if not octet.isdigit():
//                 return False
//             value = int(octet)
//             if not 0 <= value <= 255 or octet != str(value):
//                 return False
//         return True
// 
//     logs["is_valid"] = logs["ip"].apply(is_valid_ip)
//     invalid_ips = logs[~logs["is_valid"]]
//     invalid_count = invalid_ips["ip"].value_counts().reset_index()
//     invalid_count.columns = ["ip", "invalid_count"]
//     result = invalid_count.sort_values(
//         by=["invalid_count", "ip"], ascending=[False, False]
//     )
//     return result

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)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.