Added tasks 517, 518, 519

src/main/java/g0501_0600/s0517_super_washing_machines/Solution.java

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+package g0501_0600.s0517_super_washing_machines;
+
+// #Hard #Array #Greedy
+
+public class Solution {
+ // Reference: https://discuss.leetcode.com/topic/79938/super-short-easy-java-o-n-solution
+ public int findMinMoves(int[] machines) {
+ int total = 0;
+ for (int i : machines) {
+ total += i;
+ }
+ if (total % machines.length != 0) {
+ return -1;
+ }
+ int avg = total / machines.length;
+ int cnt = 0;
+ int max = 0;
+ for (int load : machines) {
+ cnt += load - avg;
+ max = Math.max(Math.max(max, Math.abs(cnt)), load - avg);
+ }
+ return max;
+ }
+}

src/main/java/g0501_0600/s0517_super_washing_machines/readme.md

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+517\. Super Washing Machines
+
+Hard
+
+You have `n` super washing machines on a line. Initially, each washing machine has some dresses or is empty.
+
+For each move, you could choose any `m` (`1 <= m <= n`) washing machines, and pass one dress of each washing machine to one of its adjacent washing machines at the same time.
+
+Given an integer array `machines` representing the number of dresses in each washing machine from left to right on the line, return _the minimum number of moves to make all the washing machines have the same number of dresses_. If it is not possible to do it, return `-1`.
+
+**Example 1:**
+
+**Input:** machines = [1,0,5]
+
+**Output:** 3
+
+**Explanation:** 1st move: 1 0 <-- 5 => 1 1 4 2nd move: 1 <-- 1 <-- 4 => 2 1 3 3rd move: 2 1 <-- 3 => 2 2 2
+
+**Example 2:**
+
+**Input:** machines = [0,3,0]
+
+**Output:** 2
+
+**Explanation:** 1st move: 0 <-- 3 0 => 1 2 0 2nd move: 1 2 --> 0 => 1 1 1
+
+**Example 3:**
+
+**Input:** machines = [0,2,0]
+
+**Output:** -1
+
+**Explanation:** It's impossible to make all three washing machines have the same number of dresses.
+
+**Constraints:**
+
+* `n == machines.length`
+* <code>1 <= n <= 10<sup>4</sup></code>
+* <code>0 <= machines[i] <= 10<sup>5</sup></code>

src/main/java/g0501_0600/s0518_coin_change_2/Solution.java

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+package g0501_0600.s0518_coin_change_2;
+
+// #Medium #Array #Dynamic_Programming
+
+public class Solution {
+ public int change(int amount, int[] coins) {
+ int[] dp = new int[amount + 1];
+ dp[0] = 1;
+ for (int coin : coins) {
+ for (int i = coin; i <= amount; i++) {
+ dp[i] += dp[i - coin];
+ }
+ }
+ return dp[amount];
+ }
+}

src/main/java/g0501_0600/s0518_coin_change_2/readme.md

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+518\. Coin Change 2
+
+Medium
+
+You are given an integer array `coins` representing coins of different denominations and an integer `amount` representing a total amount of money.
+
+Return _the number of combinations that make up that amount_. If that amount of money cannot be made up by any combination of the coins, return `0`.
+
+You may assume that you have an infinite number of each kind of coin.
+
+The answer is **guaranteed** to fit into a signed **32-bit** integer.
+
+**Example 1:**
+
+**Input:** amount = 5, coins = [1,2,5]
+
+**Output:** 4
+
+**Explanation:** there are four ways to make up the amount: 5=5 5=2+2+1 5=2+1+1+1 5=1+1+1+1+1
+
+**Example 2:**
+
+**Input:** amount = 3, coins = [2]
+
+**Output:** 0
+
+**Explanation:** the amount of 3 cannot be made up just with coins of 2.
+
+**Example 3:**
+
+**Input:** amount = 10, coins = [10]
+
+**Output:** 1
+
+**Constraints:**
+
+* `1 <= coins.length <= 300`
+* `1 <= coins[i] <= 5000`
+* All the values of `coins` are **unique**.
+* `0 <= amount <= 5000`

src/main/java/g0501_0600/s0519_random_flip_matrix/Solution.java

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+package g0501_0600.s0519_random_flip_matrix;
+
+// #Medium #Hash_Table #Math #Randomized #Reservoir_Sampling
+
+import java.security.SecureRandom;
+import java.util.HashSet;
+import java.util.Set;
+
+public class Solution {
+
+ private final int cols;
+ private final int total;
+ private final SecureRandom rand;
+ private final Set<Integer> available;
+
+ public Solution(int nRows, int nCols) {
+ this.cols = nCols;
+ this.rand = new SecureRandom();
+ this.available = new HashSet<>();
+ this.total = nRows * this.cols;
+ }
+
+ public int[] flip() {
+ int x = rand.nextInt(this.total);
+ while (available.contains(x)) {
+ x = rand.nextInt(this.total);
+ }
+
+ this.available.add(x);
+ return new int[] {x / this.cols, x % this.cols};
+ }
+
+ public void reset() {
+ this.available.clear();
+ }
+}

src/main/java/g0501_0600/s0519_random_flip_matrix/readme.md

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+519\. Random Flip Matrix
+
+Medium
+
+There is an `m x n` binary grid `matrix` with all the values set `0` initially. Design an algorithm to randomly pick an index `(i, j)` where `matrix[i][j] == 0` and flips it to `1`. All the indices `(i, j)` where `matrix[i][j] == 0` should be equally likely to be returned.
+
+Optimize your algorithm to minimize the number of calls made to the **built-in** random function of your language and optimize the time and space complexity.
+
+Implement the `Solution` class:
+
+* `Solution(int m, int n)` Initializes the object with the size of the binary matrix `m` and `n`.
+* `int[] flip()` Returns a random index `[i, j]` of the matrix where `matrix[i][j] == 0` and flips it to `1`.
+* `void reset()` Resets all the values of the matrix to be `0`.
+
+**Example 1:**
+
+**Input** ["Solution", "flip", "flip", "flip", "reset", "flip"] [[3, 1], [], [], [], [], []]
+
+**Output:** [null, [1, 0], [2, 0], [0, 0], null, [2, 0]]
+
+**Explanation:** Solution solution = new Solution(3, 1); solution.flip(); // return [1, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned. solution.flip(); // return [2, 0], Since [1,0] was returned, [2,0] and [0,0] solution.flip(); // return [0, 0], Based on the previously returned indices, only [0,0] can be returned. solution.reset(); // All the values are reset to 0 and can be returned. solution.flip(); // return [2, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned.
+
+**Constraints:**
+
+* <code>1 <= m, n <= 10<sup>4</sup></code>
+* There will be at least one free cell for each call to `flip`.
+* At most `1000` calls will be made to `flip` and `reset`.

src/test/java/g0501_0600/s0517_super_washing_machines/SolutionTest.java

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+package g0501_0600.s0517_super_washing_machines;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+ @Test
+ void findMinMoves() {
+ assertThat(new Solution().findMinMoves(new int[] {1, 0, 5}), equalTo(3));
+ }
+
+ @Test
+ void findMinMoves2() {
+ assertThat(new Solution().findMinMoves(new int[] {0, 3, 0}), equalTo(2));
+ }
+}

src/test/java/g0501_0600/s0518_coin_change_2/SolutionTest.java

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+package g0501_0600.s0518_coin_change_2;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+ @Test
+ void change() {
+ assertThat(new Solution().change(5, new int[] {1, 2, 5}), equalTo(4));
+ }
+
+ @Test
+ void change2() {
+ assertThat(new Solution().change(3, new int[] {2}), equalTo(0));
+ }
+
+ @Test
+ void change3() {
+ assertThat(new Solution().change(10, new int[] {10}), equalTo(1));
+ }
+}

src/test/java/g0501_0600/s0519_random_flip_matrix/SolutionTest.java

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+package g0501_0600.s0519_random_flip_matrix;
+
+import static org.hamcrest.CoreMatchers.equalTo;
+import static org.hamcrest.MatcherAssert.assertThat;
+
+import com_github_leetcode.CommonUtils;
+import org.junit.jupiter.api.Test;
+
+class SolutionTest {
+ @Test
+ void solutionTest() {
+ Solution solution = new Solution(3, 1);
+ CommonUtils.printArray(solution.flip());
+ CommonUtils.printArray(solution.flip());
+ CommonUtils.printArray(solution.flip());
+ solution.reset();
+ CommonUtils.printArray(solution.flip());
+ assertThat(true, equalTo(true));
+ }
+}