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‎src/main/java/g3401_3500/s3451_find_invalid_ip_addresses/readme.md

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3451\. Find Invalid IP Addresses
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Hard
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Table: `logs`
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+-------------+---------+
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| Column Name | Type |
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+-------------+---------+
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| log_id | int |
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| ip | varchar |
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| status_code | int |
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+-------------+---------+
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log_id is the unique key for this table.
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Each row contains server access log information including IP address and HTTP status code.
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Write a solution to find **invalid IP addresses**. An IPv4 address is invalid if it meets any of these conditions:
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* Contains numbers **greater than** `255` in any octet
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* Has **leading zeros** in any octet (like `01.02.03.04`)
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* Has **less or more** than `4` octets
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Return _the result table_ _ordered by_ `invalid_count`, `ip` _in **descending** order respectively_.
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The result format is in the following example.
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**Example:**
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**Input:**
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logs table:
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+--------+---------------+-------------+
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| log_id | ip | status_code |
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+--------+---------------+-------------+
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| 1 | 192.168.1.1 | 200 |
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| 2 | 256.1.2.3 | 404 |
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| 3 | 192.168.001.1 | 200 |
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| 4 | 192.168.1.1 | 200 |
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| 5 | 192.168.1 | 500 |
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| 6 | 256.1.2.3 | 404 |
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| 7 | 192.168.001.1 | 200 |
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+--------+---------------+-------------+
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**Output:**
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+---------------+--------------+
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| ip | invalid_count|
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+---------------+--------------+
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| 256.1.2.3 | 2 |
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| 192.168.001.1 | 2 |
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| 192.168.1 | 1 |
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+---------------+--------------+
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**Explanation:**
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* 256.1.2.3 is invalid because 256 > 255
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* 192.168.001.1 is invalid because of leading zeros
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* 192.168.1 is invalid because it has only 3 octets
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The output table is ordered by invalid\_count, ip in descending order respectively.

‎src/main/java/g3401_3500/s3451_find_invalid_ip_addresses/script.sql

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# Write your MySQL query statement below
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# #Hard #2025_02_16_Time_383_ms_(85.47%)_Space_0.0_MB_(100.00%)
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WITH cte_invalid_ip AS (
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SELECT log_id, ip
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FROM logs
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WHERE NOT regexp_like(ip, '^(?:[1-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])(?:[.](?:[1-9]|[1-9][0-9]|1[0-9][0-9]|2[0-4][0-9]|25[0-5])){3}$')
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),
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cte_invalid_ip_count AS (
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SELECT ip, count(log_id) as invalid_count
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FROM cte_invalid_ip
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GROUP BY ip
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)
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SELECT ip, invalid_count
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FROM cte_invalid_ip_count
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ORDER BY invalid_count DESC, ip DESC;

‎src/main/java/g3401_3500/s3452_sum_of_good_numbers/Solution.java

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package g3401_3500.s3452_sum_of_good_numbers;
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// #Easy #2025_02_16_Time_1_ms_(100.00%)_Space_44.30_MB_(100.00%)
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public class Solution {
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public int sumOfGoodNumbers(int[] nums, int k) {
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int totalSum = 0;
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int n = nums.length;
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for (int i = 0; i < n; i++) {
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boolean isGood = true;
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if (i - k >= 0) {
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if (nums[i] <= nums[i - k]) {
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isGood = false;
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}
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}
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if (i + k < n) {
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if (nums[i] <= nums[i + k]) {
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isGood = false;
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}
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}
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if (isGood) {
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totalSum += nums[i];
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}
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}
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return totalSum;
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}
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}

‎src/main/java/g3401_3500/s3452_sum_of_good_numbers/readme.md

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3452\. Sum of Good Numbers
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Easy
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Given an array of integers `nums` and an integer `k`, an element `nums[i]` is considered **good** if it is **strictly** greater than the elements at indices `i - k` and `i + k` (if those indices exist). If neither of these indices _exists_, `nums[i]` is still considered **good**.
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Return the **sum** of all the **good** elements in the array.
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**Example 1:**
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**Input:** nums = [1,3,2,1,5,4], k = 2
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**Output:** 12
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**Explanation:**
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The good numbers are `nums[1] = 3`, `nums[4] = 5`, and `nums[5] = 4` because they are strictly greater than the numbers at indices `i - k` and `i + k`.
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**Example 2:**
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**Input:** nums = [2,1], k = 1
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**Output:** 2
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**Explanation:**
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The only good number is `nums[0] = 2` because it is strictly greater than `nums[1]`.
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**Constraints:**
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* `2 <= nums.length <= 100`
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* `1 <= nums[i] <= 1000`
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* `1 <= k <= floor(nums.length / 2)`

‎src/main/java/g3401_3500/s3453_separate_squares_i/Solution.java

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package g3401_3500.s3453_separate_squares_i;
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// #Medium #2025_02_16_Time_123_ms_(100.00%)_Space_87.97_MB_(_%)
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public class Solution {
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private double helper(double line, int[][] squares) {
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double aAbove = 0;
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double aBelow = 0;
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for (int[] square : squares) {
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int y = square[1];
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int l = square[2];
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double total = (double) l * l;
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if (line <= y) {
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aAbove += total;
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} else if (line >= y + l) {
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aBelow += total;
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} else {
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// The line intersects the square.
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double aboveHeight = (y + l) - line;
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double belowHeight = line - y;
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aAbove += l * aboveHeight;
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aBelow += l * belowHeight;
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}
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}
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return aAbove - aBelow;
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}
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public double separateSquares(int[][] squares) {
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double lo = 0;
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double hi = 2 * 1e9;
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for (int i = 0; i < 60; i++) {
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double mid = (lo + hi) / 2.0;
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double diff = helper(mid, squares);
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if (diff > 0) {
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lo = mid;
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} else {
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hi = mid;
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}
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}
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return hi;
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}
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}

‎src/main/java/g3401_3500/s3453_separate_squares_i/readme.md

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3453\. Separate Squares I
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Medium
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You are given a 2D integer array `squares`. Each <code>squares[i] = [x<sub>i</sub>, y<sub>i</sub>, l<sub>i</sub>]</code> represents the coordinates of the bottom-left point and the side length of a square parallel to the x-axis.
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Find the **minimum** y-coordinate value of a horizontal line such that the total area of the squares above the line _equals_ the total area of the squares below the line.
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Answers within <code>10<sup>-5</sup></code> of the actual answer will be accepted.
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**Note**: Squares **may** overlap. Overlapping areas should be counted **multiple times**.
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**Example 1:**
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**Input:** squares = [[0,0,1],[2,2,1]]
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**Output:** 1.00000
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**Explanation:**
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![](https://assets.leetcode.com/uploads/2025/01/06/4062example1drawio.png)
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Any horizontal line between `y = 1` and `y = 2` will have 1 square unit above it and 1 square unit below it. The lowest option is 1.
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**Example 2:**
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**Input:** squares = [[0,0,2],[1,1,1]]
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**Output:** 1.16667
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**Explanation:**
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![](https://assets.leetcode.com/uploads/2025/01/15/4062example2drawio.png)
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The areas are:
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* Below the line: `7/6 * 2 (Red) + 1/6 (Blue) = 15/6 = 2.5`.
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* Above the line: `5/6 * 2 (Red) + 5/6 (Blue) = 15/6 = 2.5`.
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Since the areas above and below the line are equal, the output is `7/6 = 1.16667`.
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**Constraints:**
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* <code>1 <= squares.length <= 5 * 10<sup>4</sup></code>
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* <code>squares[i] = [x<sub>i</sub>, y<sub>i</sub>, l<sub>i</sub>]</code>
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* `squares[i].length == 3`
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* <code>0 <= x<sub>i</sub>, y<sub>i</sub> <= 10<sup>9</sup></code>
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* <code>1 <= l<sub>i</sub> <= 10<sup>9</sup></code>

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