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Commitd50384d

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Added tasks 3566-3569
1 parent2ee3575 commitd50384d

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17 files changed

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17 files changed

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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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RankASC;
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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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packageg3501_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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publicclassSolution {
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privatestaticintmod = (int)1e9 +7;

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

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packageg3501_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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importjava.util.ArrayList;

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

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packageg3501_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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importjava.util.ArrayList;
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packageg3501_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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publicclassSolution {
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publicbooleancheckEqualPartitions(int[]nums,longtarget) {
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for (intnum :nums) {
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if (target %num !=0) {
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returnfalse;
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}
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}
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longpro =1;
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for (longn :nums) {
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pro *=n;
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}
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returnpro ==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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packageg3501_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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importjava.util.Arrays;
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publicclassSolution {
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publicint[][]minAbsDiff(int[][]grid,intk) {
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introws =grid.length;
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intcols =grid[0].length;
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int[][]result =newint[rows -k +1][cols -k +1];
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for (intx =0;x <=rows -k;x++) {
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for (inty =0;y <=cols -k;y++) {
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intsize =k *k;
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int[]elements =newint[size];
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intidx =0;
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for (inti =x;i <x +k;i++) {
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for (intj =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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intminDiff =Integer.MAX_VALUE;
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for (inti =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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returnresult;
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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 × 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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packageg3501_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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importjava.util.ArrayDeque;
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importjava.util.ArrayList;
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importjava.util.Arrays;
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importjava.util.List;
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importjava.util.Queue;
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@SuppressWarnings({"java:S135","java:S6541"})
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publicclassSolution {
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privatestaticclassState {
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intx;
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inty;
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intenergy;
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intmask;
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intsteps;
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State(intx,inty,intenergy,intmask,intsteps) {
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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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publicintminMoves(String[]classroom,intenergy) {
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intm =classroom.length;
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intn =classroom[0].length();
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char[][]grid =newchar[m][n];
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for (inti =0;i <m;i++) {
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grid[i] =classroom[i].toCharArray();
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}
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intstartX = -1;
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intstartY = -1;
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List<int[]>lumetarkon =newArrayList<>();
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for (inti =0;i <m;i++) {
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for (intj =0;j <n;j++) {
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charc =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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}elseif (c =='L') {
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lumetarkon.add(newint[] {i,j});
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}
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}
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}
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inttotalLitter =lumetarkon.size();
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intallMask = (1 <<totalLitter) -1;
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int[][][]visited =newint[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 =newArrayDeque<>();
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queue.offer(newState(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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Statecurr =queue.poll();
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if (curr.mask ==allMask) {
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returncurr.steps;
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}
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for (int[]dir :dirs) {
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intnx =curr.x +dir[0];
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intny =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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intnextEnergy =curr.energy -1;
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if (nextEnergy <0) {
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continue;
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}
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charcell =grid[nx][ny];
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if (cell =='R') {
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nextEnergy =energy;
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}
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intnextMask =curr.mask;
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if (cell =='L') {
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for (inti =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(newState(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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3568\. Minimum Moves to Clean the Classroom
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Medium
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You are given an`m x n` grid`classroom` where a student volunteer is tasked with cleaning up litter scattered around the room. Each cell in the grid is one of the following:
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*`'S'`: Starting position of the student
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*`'L'`: Litter that must be collected (once collected, the cell becomes empty)
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*`'R'`: Reset area that restores the student's energy to full capacity, regardless of their current energy level (can be used multiple times)
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*`'X'`: Obstacle the student cannot pass through
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*`'.'`: Empty space
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You are also given an integer`energy`, representing the student's maximum energy capacity. The student starts with this energy from the starting position`'S'`.
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Each move to an adjacent cell (up, down, left, or right) costs 1 unit of energy. If the energy reaches 0, the student can only continue if they are on a reset area`'R'`, which resets the energy to its**maximum** capacity`energy`.
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Return the**minimum** number of moves required to collect all litter items, or`-1` if it's impossible.
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**Example 1:**
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**Input:** classroom =["S.", "XL"], energy = 2
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**Output:** 2
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**Explanation:**
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* The student starts at cell`(0, 0)` with 2 units of energy.
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* Since cell`(1, 0)` contains an obstacle 'X', the student cannot move directly downward.
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* A valid sequence of moves to collect all litter is as follows:
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* Move 1: From`(0, 0)` →`(0, 1)` with 1 unit of energy and 1 unit remaining.
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* Move 2: From`(0, 1)` →`(1, 1)` to collect the litter`'L'`.
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* The student collects all the litter using 2 moves. Thus, the output is 2.
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**Example 2:**
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**Input:** classroom =["LS", "RL"], energy = 4
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**Output:** 3
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**Explanation:**
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* The student starts at cell`(0, 1)` with 4 units of energy.
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* A valid sequence of moves to collect all litter is as follows:
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* Move 1: From`(0, 1)` →`(0, 0)` to collect the first litter`'L'` with 1 unit of energy used and 3 units remaining.
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* Move 2: From`(0, 0)` →`(1, 0)` to`'R'` to reset and restore energy back to 4.
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* Move 3: From`(1, 0)` →`(1, 1)` to collect the second litter`'L'`.
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* The student collects all the litter using 3 moves. Thus, the output is 3.
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**Example 3:**
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**Input:** classroom =["L.S", "RXL"], energy = 3
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**Output:**\-1
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**Explanation:**
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No valid path collects all`'L'`.
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**Constraints:**
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*`1 <= m == classroom.length <= 20`
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*`1 <= n == classroom[i].length <= 20`
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*`classroom[i][j]` is one of`'S'`,`'L'`,`'R'`,`'X'`, or`'.'`
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*`1 <= energy <= 50`
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* There is exactly**one**`'S'` in the grid.
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* There are**at most** 10`'L'` cells in the grid.

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