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Diff for: ‎src/main/java/g3201_3300/s3254_find_the_power_of_k_size_subarrays_i/Solution.java

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package g3201_3300.s3254_find_the_power_of_k_size_subarrays_i;
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// #Medium #Array #Sliding_Window #2024_08_20_Time_1_ms_(100.00%)_Space_45.2_MB_(95.96%)
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public class Solution {
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public int[] resultsArray(int[] nums, int k) {
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int n = nums.length;
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int[] arr = new int[n - k + 1];
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int count = 0;
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for (int i = 1; i < k; i++) {
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if (nums[i] == nums[i - 1] + 1) {
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count++;
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}
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}
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arr[0] = (count == k - 1) ? nums[k - 1] : -1;
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for (int i = 1; i <= n - k; i++) {
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if (nums[i] == nums[i - 1] + 1) {
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count--;
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}
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if (nums[i + k - 1] == nums[i + k - 2] + 1) {
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count++;
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}
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arr[i] = (count == k - 1) ? nums[i + k - 1] : -1;
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}
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return arr;
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}
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}

Diff for: ‎src/main/java/g3201_3300/s3254_find_the_power_of_k_size_subarrays_i/readme.md

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3254\. Find the Power of K-Size Subarrays I
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Medium
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You are given an array of integers `nums` of length `n` and a _positive_ integer `k`.
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The **power** of an array is defined as:
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* Its **maximum** element if _all_ of its elements are **consecutive** and **sorted** in **ascending** order.
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* \-1 otherwise.
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You need to find the **power** of all subarrays of `nums` of size `k`.
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Return an integer array `results` of size `n - k + 1`, where `results[i]` is the _power_ of `nums[i..(i + k - 1)]`.
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**Example 1:**
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**Input:** nums = [1,2,3,4,3,2,5], k = 3
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**Output:** [3,4,-1,-1,-1]
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**Explanation:**
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There are 5 subarrays of `nums` of size 3:
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* `[1, 2, 3]` with the maximum element 3.
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* `[2, 3, 4]` with the maximum element 4.
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* `[3, 4, 3]` whose elements are **not** consecutive.
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* `[4, 3, 2]` whose elements are **not** sorted.
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* `[3, 2, 5]` whose elements are **not** consecutive.
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**Example 2:**
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**Input:** nums = [2,2,2,2,2], k = 4
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**Output:** [-1,-1]
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**Example 3:**
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**Input:** nums = [3,2,3,2,3,2], k = 2
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**Output:** [-1,3,-1,3,-1]
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**Constraints:**
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* `1 <= n == nums.length <= 500`
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* <code>1 <= nums[i] <= 10<sup>5</sup></code>
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* `1 <= k <= n`

Diff for: ‎src/main/java/g3201_3300/s3255_find_the_power_of_k_size_subarrays_ii/Solution.java

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package g3201_3300.s3255_find_the_power_of_k_size_subarrays_ii;
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// #Medium #Array #Sliding_Window #2024_08_20_Time_3_ms_(99.24%)_Space_64.1_MB_(67.44%)
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public class Solution {
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public int[] resultsArray(int[] nums, int k) {
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if (k == 1) {
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return nums;
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}
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int start = 0;
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int n = nums.length;
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int[] output = new int[n - k + 1];
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for (int i = 1; i < n; i++) {
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if (nums[i] != nums[i - 1] + 1) {
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start = i;
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}
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int index = i - k + 1;
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if (index >= 0) {
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if (start > index) {
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output[index] = -1;
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} else {
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output[index] = nums[i];
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}
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}
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}
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return output;
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}
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}

Diff for: ‎src/main/java/g3201_3300/s3255_find_the_power_of_k_size_subarrays_ii/readme.md

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3255\. Find the Power of K-Size Subarrays II
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Medium
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You are given an array of integers `nums` of length `n` and a _positive_ integer `k`.
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The **power** of an array is defined as:
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* Its **maximum** element if _all_ of its elements are **consecutive** and **sorted** in **ascending** order.
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* \-1 otherwise.
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You need to find the **power** of all subarrays of `nums` of size `k`.
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Return an integer array `results` of size `n - k + 1`, where `results[i]` is the _power_ of `nums[i..(i + k - 1)]`.
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**Example 1:**
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**Input:** nums = [1,2,3,4,3,2,5], k = 3
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**Output:** [3,4,-1,-1,-1]
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**Explanation:**
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There are 5 subarrays of `nums` of size 3:
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* `[1, 2, 3]` with the maximum element 3.
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* `[2, 3, 4]` with the maximum element 4.
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* `[3, 4, 3]` whose elements are **not** consecutive.
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* `[4, 3, 2]` whose elements are **not** sorted.
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* `[3, 2, 5]` whose elements are **not** consecutive.
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**Example 2:**
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**Input:** nums = [2,2,2,2,2], k = 4
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**Output:** [-1,-1]
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**Example 3:**
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**Input:** nums = [3,2,3,2,3,2], k = 2
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**Output:** [-1,3,-1,3,-1]
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**Constraints:**
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* <code>1 <= n == nums.length <= 10<sup>5</sup></code>
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* <code>1 <= nums[i] <= 10<sup>6</sup></code>
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* `1 <= k <= n`

Diff for: ‎src/main/java/g3201_3300/s3256_maximum_value_sum_by_placing_three_rooks_i/Solution.java

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package g3201_3300.s3256_maximum_value_sum_by_placing_three_rooks_i;
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// #Hard #Array #Dynamic_Programming #Matrix #Enumeration
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// #2024_08_20_Time_7_ms_(100.00%)_Space_45.1_MB_(90.97%)
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import java.util.Arrays;
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public class Solution {
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public long maximumValueSum(int[][] board) {
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int n = board.length;
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int m = board[0].length;
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int[][] tb = new int[n][m];
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tb[0] = Arrays.copyOf(board[0], m);
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for (int i = 1; i < n; i++) {
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for (int j = 0; j < m; j++) {
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tb[i][j] = Math.max(tb[i - 1][j], board[i][j]);
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}
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}
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int[][] bt = new int[n][m];
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bt[n - 1] = Arrays.copyOf(board[n - 1], m);
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for (int i = n - 2; i >= 0; i--) {
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for (int j = 0; j < m; j++) {
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bt[i][j] = Math.max(bt[i + 1][j], board[i][j]);
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}
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}
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long ans = Long.MIN_VALUE;
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for (int i = 1; i < n - 1; i++) {
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int[][] max3Top = getMax3(tb[i - 1]);
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int[][] max3Cur = getMax3(board[i]);
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int[][] max3Bottom = getMax3(bt[i + 1]);
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for (int[] topCand : max3Top) {
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for (int[] curCand : max3Cur) {
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for (int[] bottomCand : max3Bottom) {
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if (topCand[1] != curCand[1]
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&& topCand[1] != bottomCand[1]
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&& curCand[1] != bottomCand[1]) {
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long cand = (long) topCand[0] + curCand[0] + bottomCand[0];
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ans = Math.max(ans, cand);
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}
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}
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}
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}
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}
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return ans;
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}
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private int[][] getMax3(int[] row) {
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int m = row.length;
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int[][] ans = new int[3][2];
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Arrays.fill(ans, new int[] {Integer.MIN_VALUE, -1});
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for (int j = 0; j < m; j++) {
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if (row[j] >= ans[0][0]) {
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ans[2] = ans[1];
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ans[1] = ans[0];
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ans[0] = new int[] {row[j], j};
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} else if (row[j] >= ans[1][0]) {
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ans[2] = ans[1];
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ans[1] = new int[] {row[j], j};
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} else if (row[j] > ans[2][0]) {
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ans[2] = new int[] {row[j], j};
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}
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}
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return ans;
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}
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}

Diff for: ‎src/main/java/g3201_3300/s3256_maximum_value_sum_by_placing_three_rooks_i/readme.md

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3256\. Maximum Value Sum by Placing Three Rooks I
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Hard
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You are given a `m x n` 2D array `board` representing a chessboard, where `board[i][j]` represents the **value** of the cell `(i, j)`.
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Rooks in the **same** row or column **attack** each other. You need to place _three_ rooks on the chessboard such that the rooks **do not** **attack** each other.
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Return the **maximum** sum of the cell **values** on which the rooks are placed.
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**Example 1:**
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**Input:** board = [[-3,1,1,1],[-3,1,-3,1],[-3,2,1,1]]
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**Output:** 4
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**Explanation:**
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![](https://assets.leetcode.com/uploads/2024/08/08/rooks2.png)
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We can place the rooks in the cells `(0, 2)`, `(1, 3)`, and `(2, 1)` for a sum of `1 + 1 + 2 = 4`.
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**Example 2:**
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**Input:** board = [[1,2,3],[4,5,6],[7,8,9]]
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**Output:** 15
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**Explanation:**
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We can place the rooks in the cells `(0, 0)`, `(1, 1)`, and `(2, 2)` for a sum of `1 + 5 + 9 = 15`.
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**Example 3:**
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**Input:** board = [[1,1,1],[1,1,1],[1,1,1]]
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**Output:** 3
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**Explanation:**
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We can place the rooks in the cells `(0, 2)`, `(1, 1)`, and `(2, 0)` for a sum of `1 + 1 + 1 = 3`.
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**Constraints:**
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* `3 <= m == board.length <= 100`
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* `3 <= n == board[i].length <= 100`
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* <code>-10<sup>9</sup> <= board[i][j] <= 10<sup>9</sup></code>

Diff for: ‎src/main/java/g3201_3300/s3257_maximum_value_sum_by_placing_three_rooks_ii/Solution.java

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package g3201_3300.s3257_maximum_value_sum_by_placing_three_rooks_ii;
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// #Hard #Array #Dynamic_Programming #Matrix #Enumeration
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// #2024_08_20_Time_18_ms_(99.59%)_Space_74.2_MB_(66.81%)
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import java.util.Arrays;
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public class Solution {
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public long maximumValueSum(int[][] board) {
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int n = board.length;
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int m = board[0].length;
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int[][] tb = new int[n][m];
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tb[0] = Arrays.copyOf(board[0], m);
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for (int i = 1; i < n; i++) {
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for (int j = 0; j < m; j++) {
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tb[i][j] = Math.max(tb[i - 1][j], board[i][j]);
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}
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}
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int[][] bt = new int[n][m];
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bt[n - 1] = Arrays.copyOf(board[n - 1], m);
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for (int i = n - 2; i >= 0; i--) {
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for (int j = 0; j < m; j++) {
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bt[i][j] = Math.max(bt[i + 1][j], board[i][j]);
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}
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}
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long ans = Long.MIN_VALUE;
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for (int i = 1; i < n - 1; i++) {
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int[][] max3Top = getMax3(tb[i - 1]);
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int[][] max3Cur = getMax3(board[i]);
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int[][] max3Bottom = getMax3(bt[i + 1]);
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for (int[] topCand : max3Top) {
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for (int[] curCand : max3Cur) {
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for (int[] bottomCand : max3Bottom) {
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if (topCand[1] != curCand[1]
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&& topCand[1] != bottomCand[1]
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&& curCand[1] != bottomCand[1]) {
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long cand = (long) topCand[0] + curCand[0] + bottomCand[0];
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ans = Math.max(ans, cand);
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}
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}
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}
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}
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}
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return ans;
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}
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private int[][] getMax3(int[] row) {
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int m = row.length;
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int[][] ans = new int[3][2];
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Arrays.fill(ans, new int[] {Integer.MIN_VALUE, -1});
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for (int j = 0; j < m; j++) {
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if (row[j] >= ans[0][0]) {
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ans[2] = ans[1];
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ans[1] = ans[0];
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ans[0] = new int[] {row[j], j};
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} else if (row[j] >= ans[1][0]) {
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ans[2] = ans[1];
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ans[1] = new int[] {row[j], j};
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} else if (row[j] > ans[2][0]) {
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ans[2] = new int[] {row[j], j};
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}
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}
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return ans;
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}
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}

Diff for: ‎src/main/java/g3201_3300/s3257_maximum_value_sum_by_placing_three_rooks_ii/readme.md

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3257\. Maximum Value Sum by Placing Three Rooks II
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Hard
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You are given a `m x n` 2D array `board` representing a chessboard, where `board[i][j]` represents the **value** of the cell `(i, j)`.
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Rooks in the **same** row or column **attack** each other. You need to place _three_ rooks on the chessboard such that the rooks **do not** **attack** each other.
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Return the **maximum** sum of the cell **values** on which the rooks are placed.
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**Example 1:**
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**Input:** board = [[-3,1,1,1],[-3,1,-3,1],[-3,2,1,1]]
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**Output:** 4
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**Explanation:**
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![](https://assets.leetcode.com/uploads/2024/08/08/rooks2.png)
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We can place the rooks in the cells `(0, 2)`, `(1, 3)`, and `(2, 1)` for a sum of `1 + 1 + 2 = 4`.
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**Example 2:**
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**Input:** board = [[1,2,3],[4,5,6],[7,8,9]]
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**Output:** 15
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**Explanation:**
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We can place the rooks in the cells `(0, 0)`, `(1, 1)`, and `(2, 2)` for a sum of `1 + 5 + 9 = 15`.
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**Example 3:**
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**Input:** board = [[1,1,1],[1,1,1],[1,1,1]]
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**Output:** 3
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**Explanation:**
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We can place the rooks in the cells `(0, 2)`, `(1, 1)`, and `(2, 0)` for a sum of `1 + 1 + 1 = 3`.
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**Constraints:**
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* `3 <= m == board.length <= 500`
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* `3 <= n == board[i].length <= 500`
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* <code>-10<sup>9</sup> <= board[i][j] <= 10<sup>9</sup></code>

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