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package g3201_3300.s3280_convert_date_to_binary;

// #Easy #String #Math #2024_09_09_Time_3_ms_(100.00%)_Space_42.4_MB_(50.00%)

public class Solution {
private String helper(String str) {
return Integer.toBinaryString(Integer.parseInt(str));
}

public String convertDateToBinary(String date) {
StringBuilder sb = new StringBuilder();
for (String str : date.split("-")) {
sb.append(helper(str));
sb.append("-");
}
sb.deleteCharAt(sb.length() - 1);
return sb.toString();
}
}
35 changes: 35 additions & 0 deletions src/main/java/g3201_3300/s3280_convert_date_to_binary/readme.md
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3280\. Convert Date to Binary

Easy

You are given a string `date` representing a Gregorian calendar date in the `yyyy-mm-dd` format.

`date` can be written in its binary representation obtained by converting year, month, and day to their binary representations without any leading zeroes and writing them down in `year-month-day` format.

Return the **binary** representation of `date`.

**Example 1:**

**Input:** date = "2080-02-29"

**Output:** "100000100000-10-11101"

**Explanation:**

100000100000, 10, and 11101 are the binary representations of 2080, 02, and 29 respectively.

**Example 2:**

**Input:** date = "1900-01-01"

**Output:** "11101101100-1-1"

**Explanation:**

11101101100, 1, and 1 are the binary representations of 1900, 1, and 1 respectively.

**Constraints:**

* `date.length == 10`
* `date[4] == date[7] == '-'`, and all other `date[i]`'s are digits.
* The input is generated such that `date` represents a valid Gregorian calendar date between Jan 1<sup>st</sup>, 1900 and Dec 31<sup>st</sup>, 2100 (both inclusive).
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package g3201_3300.s3281_maximize_score_of_numbers_in_ranges;

// #Medium #Array #Sorting #Greedy #Binary_Search
// #2024_09_09_Time_47_ms_(100.00%)_Space_58.3_MB_(100.00%)

import java.util.Arrays;

public class Solution {
public int maxPossibleScore(int[] start, int d) {
Arrays.sort(start);
int n = start.length;
int l = 0;
int r = start[n - 1] - start[0] + d + 1;
while (l < r) {
int m = l + (r - l) / 2;
if (isPossible(start, d, m)) {
l = m + 1;
} else {
r = m;
}
}
return l - 1;
}

private boolean isPossible(int[] start, int d, int score) {
int pre = start[0];
for (int i = 1; i < start.length; i++) {
if (start[i] + d - pre < score) {
return false;
}
pre = Math.max(start[i], pre + score);
}
return true;
}
}
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3281\. Maximize Score of Numbers in Ranges

Medium

You are given an array of integers `start` and an integer `d`, representing `n` intervals `[start[i], start[i] + d]`.

You are asked to choose `n` integers where the <code>i<sup>th</sup></code> integer must belong to the <code>i<sup>th</sup></code> interval. The **score** of the chosen integers is defined as the **minimum** absolute difference between any two integers that have been chosen.

Return the **maximum** _possible score_ of the chosen integers.

**Example 1:**

**Input:** start = [6,0,3], d = 2

**Output:** 4

**Explanation:**

The maximum possible score can be obtained by choosing integers: 8, 0, and 4. The score of these chosen integers is `min(|8 - 0|, |8 - 4|, |0 - 4|)` which equals 4.

**Example 2:**

**Input:** start = [2,6,13,13], d = 5

**Output:** 5

**Explanation:**

The maximum possible score can be obtained by choosing integers: 2, 7, 13, and 18. The score of these chosen integers is `min(|2 - 7|, |2 - 13|, |2 - 18|, |7 - 13|, |7 - 18|, |13 - 18|)` which equals 5.

**Constraints:**

* <code>2 <= start.length <= 10<sup>5</sup></code>
* <code>0 <= start[i] <= 10<sup>9</sup></code>
* <code>0 <= d <= 10<sup>9</sup></code>
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package g3201_3300.s3282_reach_end_of_array_with_max_score;

// #Medium #Array #Greedy #2024_09_09_Time_9_ms_(100.00%)_Space_63.2_MB_(100.00%)

import java.util.List;

public class Solution {
public long findMaximumScore(List<Integer> nums) {
long res = 0;
long ma = 0;
for (int num : nums) {
res += ma;
ma = Math.max(ma, num);
}
return res;
}
}
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3282\. Reach End of Array With Max Score

Medium

You are given an integer array `nums` of length `n`.

Your goal is to start at index `0` and reach index `n - 1`. You can only jump to indices **greater** than your current index.

The score for a jump from index `i` to index `j` is calculated as `(j - i) * nums[i]`.

Return the **maximum** possible **total score** by the time you reach the last index.

**Example 1:**

**Input:** nums = [1,3,1,5]

**Output:** 7

**Explanation:**

First, jump to index 1 and then jump to the last index. The final score is `1 * 1 + 2 * 3 = 7`.

**Example 2:**

**Input:** nums = [4,3,1,3,2]

**Output:** 16

**Explanation:**

Jump directly to the last index. The final score is `4 * 4 = 16`.

**Constraints:**

* <code>1 <= nums.length <= 10<sup>5</sup></code>
* <code>1 <= nums[i] <= 10<sup>5</sup></code>
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package g3201_3300.s3283_maximum_number_of_moves_to_kill_all_pawns;

// #Hard #Array #Math #Breadth_First_Search #Bit_Manipulation #Bitmask #Game_Theory
// #2024_09_09_Time_250_ms_(98.43%)_Space_50.1_MB_(66.27%)

import java.util.LinkedList;
import java.util.Queue;

public class Solution {
private static final int[][] KNIGHT_MOVES = {
{-2, -1}, {-2, 1}, {-1, -2}, {-1, 2},
{1, -2}, {1, 2}, {2, -1}, {2, 1}
};
private int[][] distances;
private Integer[][] memo;

public int maxMoves(int kx, int ky, int[][] positions) {
int n = positions.length;
distances = new int[n + 1][n + 1];
memo = new Integer[n + 1][1 << n];
// Calculate distances between all pairs of positions (including knight's initial position)
for (int i = 0; i < n; i++) {
distances[n][i] = calculateMoves(kx, ky, positions[i][0], positions[i][1]);
for (int j = i + 1; j < n; j++) {
int dist =
calculateMoves(
positions[i][0], positions[i][1], positions[j][0], positions[j][1]);
distances[i][j] = distances[j][i] = dist;
}
}
return minimax(n, (1 << n) - 1, true);
}

private int minimax(int lastPos, int remainingPawns, boolean isAlice) {
if (remainingPawns == 0) {
return 0;
}
if (memo[lastPos][remainingPawns] != null) {
return memo[lastPos][remainingPawns];
}
int result = isAlice ? 0 : Integer.MAX_VALUE;
for (int i = 0; i < distances.length - 1; i++) {
if ((remainingPawns & (1 << i)) != 0) {
int newRemainingPawns = remainingPawns & ~(1 << i);
int moveValue = distances[lastPos][i] + minimax(i, newRemainingPawns, !isAlice);

if (isAlice) {
result = Math.max(result, moveValue);
} else {
result = Math.min(result, moveValue);
}
}
}
memo[lastPos][remainingPawns] = result;
return result;
}

private int calculateMoves(int x1, int y1, int x2, int y2) {
if (x1 == x2 && y1 == y2) {
return 0;
}
boolean[][] visited = new boolean[50][50];
Queue<int[]> queue = new LinkedList<>();
queue.offer(new int[] {x1, y1, 0});
visited[x1][y1] = true;
while (!queue.isEmpty()) {
int[] current = queue.poll();
int x = current[0];
int y = current[1];
int moves = current[2];
for (int[] move : KNIGHT_MOVES) {
int nx = x + move[0];
int ny = y + move[1];
if (nx == x2 && ny == y2) {
return moves + 1;
}
if (nx >= 0 && nx < 50 && ny >= 0 && ny < 50 && !visited[nx][ny]) {
queue.offer(new int[] {nx, ny, moves + 1});
visited[nx][ny] = true;
}
}
}
// Should never reach here if input is valid
return -1;
}
}
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3283\. Maximum Number of Moves to Kill All Pawns

Hard

There is a `50 x 50` chessboard with **one** knight and some pawns on it. You are given two integers `kx` and `ky` where `(kx, ky)` denotes the position of the knight, and a 2D array `positions` where <code>positions[i] = [x<sub>i</sub>, y<sub>i</sub>]</code> denotes the position of the pawns on the chessboard.

Alice and Bob play a _turn-based_ game, where Alice goes first. In each player's turn:

* The player _selects_ a pawn that still exists on the board and captures it with the knight in the **fewest** possible **moves**. **Note** that the player can select **any** pawn, it **might not** be one that can be captured in the **least** number of moves.
* In the process of capturing the _selected_ pawn, the knight **may** pass other pawns **without** capturing them. **Only** the _selected_ pawn can be captured in _this_ turn.

Alice is trying to **maximize** the **sum** of the number of moves made by _both_ players until there are no more pawns on the board, whereas Bob tries to **minimize** them.

Return the **maximum** _total_ number of moves made during the game that Alice can achieve, assuming both players play **optimally**.

Note that in one **move,** a chess knight has eight possible positions it can move to, as illustrated below. Each move is two cells in a cardinal direction, then one cell in an orthogonal direction.

![](https://assets.leetcode.com/uploads/2024/08/01/chess_knight.jpg)

**Example 1:**

**Input:** kx = 1, ky = 1, positions = [[0,0]]

**Output:** 4

**Explanation:**

![](https://assets.leetcode.com/uploads/2024/08/16/gif3.gif)

The knight takes 4 moves to reach the pawn at `(0, 0)`.

**Example 2:**

**Input:** kx = 0, ky = 2, positions = [[1,1],[2,2],[3,3]]

**Output:** 8

**Explanation:**

**![](https://assets.leetcode.com/uploads/2024/08/16/gif4.gif)**

* Alice picks the pawn at `(2, 2)` and captures it in two moves: `(0, 2) -> (1, 4) -> (2, 2)`.
* Bob picks the pawn at `(3, 3)` and captures it in two moves: `(2, 2) -> (4, 1) -> (3, 3)`.
* Alice picks the pawn at `(1, 1)` and captures it in four moves: `(3, 3) -> (4, 1) -> (2, 2) -> (0, 3) -> (1, 1)`.

**Example 3:**

**Input:** kx = 0, ky = 0, positions = [[1,2],[2,4]]

**Output:** 3

**Explanation:**

* Alice picks the pawn at `(2, 4)` and captures it in two moves: `(0, 0) -> (1, 2) -> (2, 4)`. Note that the pawn at `(1, 2)` is not captured.
* Bob picks the pawn at `(1, 2)` and captures it in one move: `(2, 4) -> (1, 2)`.

**Constraints:**

* `0 <= kx, ky <= 49`
* `1 <= positions.length <= 15`
* `positions[i].length == 2`
* `0 <= positions[i][0], positions[i][1] <= 49`
* All `positions[i]` are unique.
* The input is generated such that `positions[i] != [kx, ky]` for all `0 <= i < positions.length`.
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package g3201_3300.s3280_convert_date_to_binary;

import static org.hamcrest.CoreMatchers.equalTo;
import static org.hamcrest.MatcherAssert.assertThat;

import org.junit.jupiter.api.Test;

class SolutionTest {
@Test
void convertDateToBinary() {
assertThat(
new Solution().convertDateToBinary("2080-02-29"), equalTo("100000100000-10-11101"));
}

@Test
void convertDateToBinary2() {
assertThat(new Solution().convertDateToBinary("1900-01-01"), equalTo("11101101100-1-1"));
}
}
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package g3201_3300.s3281_maximize_score_of_numbers_in_ranges;

import static org.hamcrest.CoreMatchers.equalTo;
import static org.hamcrest.MatcherAssert.assertThat;

import org.junit.jupiter.api.Test;

class SolutionTest {
@Test
void maxPossibleScore() {
assertThat(new Solution().maxPossibleScore(new int[] {6, 0, 3}, 2), equalTo(4));
}

@Test
void maxPossibleScore2() {
assertThat(new Solution().maxPossibleScore(new int[] {2, 6, 13, 13}, 5), equalTo(5));
}
}
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package g3201_3300.s3282_reach_end_of_array_with_max_score;

import static org.hamcrest.CoreMatchers.equalTo;
import static org.hamcrest.MatcherAssert.assertThat;

import java.util.List;
import org.junit.jupiter.api.Test;

class SolutionTest {
@Test
void findMaximumScore() {
assertThat(new Solution().findMaximumScore(List.of(1, 3, 1, 5)), equalTo(7L));
}

@Test
void findMaximumScore2() {
assertThat(new Solution().findMaximumScore(List.of(4, 3, 1, 3, 2)), equalTo(16L));
}
}
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