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Commit d50384d

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Added tasks 3566-3569
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‎src/main/java/g0101_0200/s0178_rank_scores/script.sql

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Scores
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ORDER BY
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Rank ASC;
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‎src/main/java/g0101_0200/s0182_duplicate_emails/script.sql

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Email
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HAVING
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COUNT(Email) > 1;
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‎src/main/java/g3501_3600/s3558_number_of_ways_to_assign_edge_weights_i/Solution.java

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package g3501_3600.s3558_number_of_ways_to_assign_edge_weights_i;
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// #Medium #Math #Tree #Depth_First_Search #2025_05_27_Time_12_ms_(100.00%)_Space_106.62_MB_(76.01%)
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// #Medium #Math #Depth_First_Search #Tree #2025_05_27_Time_12_ms_(100.00%)_Space_106.62_MB_(76.01%)
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public class Solution {
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private static int mod = (int) 1e9 + 7;

‎src/main/java/g3501_3600/s3559_number_of_ways_to_assign_edge_weights_ii/Solution.java

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package g3501_3600.s3559_number_of_ways_to_assign_edge_weights_ii;
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// #Hard #Array #Dynamic_Programming #Math #Tree #Depth_First_Search
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// #Hard #Array #Dynamic_Programming #Math #Depth_First_Search #Tree
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// #2025_05_27_Time_138_ms_(64.66%)_Space_133.20_MB_(11.56%)
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import java.util.ArrayList;

‎src/main/java/g3501_3600/s3562_maximum_profit_from_trading_stocks_with_discounts/Solution.java

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package g3501_3600.s3562_maximum_profit_from_trading_stocks_with_discounts;
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// #Hard #Array #Dynamic_Programming #Tree #Depth_First_Search
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// #Hard #Array #Dynamic_Programming #Depth_First_Search #Tree
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// #2025_05_27_Time_27_ms_(100.00%)_Space_45.29_MB_(82.12%)
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import java.util.ArrayList;
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package g3501_3600.s3566_partition_array_into_two_equal_product_subsets;
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// #Medium #Array #Bit_Manipulation #Recursion #Enumeration
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// #2025_06_03_Time_0_ms_(100.00%)_Space_42.45_MB_(36.66%)
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public class Solution {
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public boolean checkEqualPartitions(int[] nums, long target) {
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for (int num : nums) {
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if (target % num != 0) {
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return false;
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}
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}
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long pro = 1;
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for (long n : nums) {
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pro *= n;
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}
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return pro == target * target;
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}
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}
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3566\. Partition Array into Two Equal Product Subsets
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Medium
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You are given an integer array `nums` containing **distinct** positive integers and an integer `target`.
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Determine if you can partition `nums` into two **non-empty** **disjoint** **subsets**, with each element belonging to **exactly one** subset, such that the product of the elements in each subset is equal to `target`.
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Return `true` if such a partition exists and `false` otherwise.
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A **subset** of an array is a selection of elements of the array.
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**Example 1:**
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**Input:** nums = [3,1,6,8,4], target = 24
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**Output:** true
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**Explanation:** The subsets `[3, 8]` and `[1, 6, 4]` each have a product of 24. Hence, the output is true.
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**Example 2:**
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**Input:** nums = [2,5,3,7], target = 15
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**Output:** false
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**Explanation:** There is no way to partition `nums` into two non-empty disjoint subsets such that both subsets have a product of 15. Hence, the output is false.
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**Constraints:**
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* `3 <= nums.length <= 12`
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* <code>1 <= target <= 10<sup>15</sup></code>
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* `1 <= nums[i] <= 100`
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* All elements of `nums` are **distinct**.
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package g3501_3600.s3567_minimum_absolute_difference_in_sliding_submatrix;
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// #Medium #Array #Sorting #Matrix #2025_06_03_Time_7_ms_(99.69%)_Space_45.24_MB_(99.03%)
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import java.util.Arrays;
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public class Solution {
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public int[][] minAbsDiff(int[][] grid, int k) {
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int rows = grid.length;
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int cols = grid[0].length;
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int[][] result = new int[rows - k + 1][cols - k + 1];
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for (int x = 0; x <= rows - k; x++) {
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for (int y = 0; y <= cols - k; y++) {
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int size = k * k;
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int[] elements = new int[size];
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int idx = 0;
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for (int i = x; i < x + k; i++) {
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for (int j = y; j < y + k; j++) {
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elements[idx++] = grid[i][j];
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}
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}
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Arrays.sort(elements);
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int minDiff = Integer.MAX_VALUE;
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for (int i = 1; i < size; i++) {
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if (elements[i] != elements[i - 1]) {
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minDiff = Math.min(minDiff, elements[i] - elements[i - 1]);
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if (minDiff == 1) {
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break;
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}
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}
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}
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result[x][y] = minDiff == Integer.MAX_VALUE ? 0 : minDiff;
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}
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}
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return result;
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}
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}
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3567\. Minimum Absolute Difference in Sliding Submatrix
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Medium
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You are given an `m x n` integer matrix `grid` and an integer `k`.
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For every contiguous `k x k` **submatrix** of `grid`, compute the **minimum absolute** difference between any two **distinct** values within that **submatrix**.
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Return a 2D array `ans` of size `(m - k + 1) x (n - k + 1)`, where `ans[i][j]` is the minimum absolute difference in the submatrix whose top-left corner is `(i, j)` in `grid`.
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**Note**: If all elements in the submatrix have the same value, the answer will be 0.
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A submatrix `(x1, y1, x2, y2)` is a matrix that is formed by choosing all cells `matrix[x][y]` where `x1 <= x <= x2` and `y1 <= y <= y2`.
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**Example 1:**
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**Input:** grid = [[1,8],[3,-2]], k = 2
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**Output:** [[2]]
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**Explanation:**
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* There is only one possible `k x k` submatrix: `[[1, 8], [3, -2]]`.
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* Distinct values in the submatrix are `[1, 8, 3, -2]`.
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* The minimum absolute difference in the submatrix is `|1 - 3| = 2`. Thus, the answer is `[[2]]`.
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**Example 2:**
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**Input:** grid = [[3,-1]], k = 1
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**Output:** [[0,0]]
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**Explanation:**
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* Both `k x k` submatrix has only one distinct element.
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* Thus, the answer is `[[0, 0]]`.
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**Example 3:**
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**Input:** grid = [[1,-2,3],[2,3,5]], k = 2
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**Output:** [[1,2]]
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**Explanation:**
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* There are two possible `k ×ばつ k` submatrix:
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* Starting at `(0, 0)`: `[[1, -2], [2, 3]]`.
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* Distinct values in the submatrix are `[1, -2, 2, 3]`.
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* The minimum absolute difference in the submatrix is `|1 - 2| = 1`.
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* Starting at `(0, 1)`: `[[-2, 3], [3, 5]]`.
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* Distinct values in the submatrix are `[-2, 3, 5]`.
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* The minimum absolute difference in the submatrix is `|3 - 5| = 2`.
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* Thus, the answer is `[[1, 2]]`.
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**Constraints:**
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* `1 <= m == grid.length <= 30`
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* `1 <= n == grid[i].length <= 30`
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* <code>-10<sup>5</sup> <= grid[i][j] <= 10<sup>5</sup></code>
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* `1 <= k <= min(m, n)`
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package g3501_3600.s3568_minimum_moves_to_clean_the_classroom;
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// #Medium #Array #Hash_Table #Breadth_First_Search #Matrix #Bit_Manipulation
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// #2025_06_03_Time_94_ms_(99.86%)_Space_53.76_MB_(99.86%)
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import java.util.ArrayDeque;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.List;
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import java.util.Queue;
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@SuppressWarnings({"java:S135", "java:S6541"})
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public class Solution {
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private static class State {
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int x;
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int y;
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int energy;
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int mask;
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int steps;
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State(int x, int y, int energy, int mask, int steps) {
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this.x = x;
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this.y = y;
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this.energy = energy;
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this.mask = mask;
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this.steps = steps;
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}
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}
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public int minMoves(String[] classroom, int energy) {
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int m = classroom.length;
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int n = classroom[0].length();
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char[][] grid = new char[m][n];
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for (int i = 0; i < m; i++) {
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grid[i] = classroom[i].toCharArray();
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}
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int startX = -1;
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int startY = -1;
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List<int[]> lumetarkon = new ArrayList<>();
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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char c = grid[i][j];
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if (c == 'S') {
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startX = i;
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startY = j;
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} else if (c == 'L') {
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lumetarkon.add(new int[] {i, j});
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}
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}
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}
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int totalLitter = lumetarkon.size();
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int allMask = (1 << totalLitter) - 1;
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int[][][] visited = new int[m][n][1 << totalLitter];
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for (int[][] layer : visited) {
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for (int[] row : layer) {
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Arrays.fill(row, -1);
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}
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}
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Queue<State> queue = new ArrayDeque<>();
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queue.offer(new State(startX, startY, energy, 0, 0));
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visited[startX][startY][0] = energy;
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int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
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while (!queue.isEmpty()) {
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State curr = queue.poll();
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if (curr.mask == allMask) {
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return curr.steps;
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}
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for (int[] dir : dirs) {
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int nx = curr.x + dir[0];
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int ny = curr.y + dir[1];
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if (nx < 0 || ny < 0 || nx >= m || ny >= n || grid[nx][ny] == 'X') {
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continue;
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}
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int nextEnergy = curr.energy - 1;
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if (nextEnergy < 0) {
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continue;
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}
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char cell = grid[nx][ny];
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if (cell == 'R') {
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nextEnergy = energy;
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}
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int nextMask = curr.mask;
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if (cell == 'L') {
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for (int i = 0; i < lumetarkon.size(); i++) {
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int[] pos = lumetarkon.get(i);
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if (pos[0] == nx && pos[1] == ny) {
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nextMask |= (1 << i);
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break;
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}
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}
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}
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if (visited[nx][ny][nextMask] < nextEnergy) {
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visited[nx][ny][nextMask] = nextEnergy;
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queue.offer(new State(nx, ny, nextEnergy, nextMask, curr.steps + 1));
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}
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}
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}
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return -1;
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}
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}

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