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Commit 6dd3ade

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Added tasks 365, 367, 368.
1 parent f014f4a commit 6dd3ade

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package g0301_0400.s0365_water_and_jug_problem;
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// #Medium #Math #Depth_First_Search #Breadth_First_Search
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public class Solution {
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private int gcd(int n1, int n2) {
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if (n2 == 0) {
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return n1;
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}
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return gcd(n2, n1 % n2);
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}
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public boolean canMeasureWater(int jug1, int jug2, int target) {
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if (jug1 + jug2 < target) {
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return false;
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}
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int gcd = gcd(jug1, jug2);
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return target % gcd == 0;
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}
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}
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365\. Water and Jug Problem
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Medium
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You are given two jugs with capacities `jug1Capacity` and `jug2Capacity` liters. There is an infinite amount of water supply available. Determine whether it is possible to measure exactly `targetCapacity` liters using these two jugs.
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If `targetCapacity` liters of water are measurable, you must have `targetCapacity` liters of water contained **within one or both buckets** by the end.
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Operations allowed:
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* Fill any of the jugs with water.
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* Empty any of the jugs.
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* Pour water from one jug into another till the other jug is completely full, or the first jug itself is empty.
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**Example 1:**
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**Input:** jug1Capacity = 3, jug2Capacity = 5, targetCapacity = 4
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**Output:** true
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**Explanation:** The famous [Die Hard](https://www.youtube.com/watch?v=BVtQNK_ZUJg&ab_channel=notnek01) example
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**Example 2:**
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**Input:** jug1Capacity = 2, jug2Capacity = 6, targetCapacity = 5
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**Output:** false
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**Example 3:**
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**Input:** jug1Capacity = 1, jug2Capacity = 2, targetCapacity = 3
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**Output:** true
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**Constraints:**
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* <code>1 <= jug1Capacity, jug2Capacity, targetCapacity <= 10<sup>6</sup></code>
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package g0301_0400.s0367_valid_perfect_square;
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// #Easy #Math #Binary_Search
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public class Solution {
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public boolean isPerfectSquare(int num) {
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if (num == 0) {
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// If num is 0 return false
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return false;
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}
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// long datatype can holds huge number.
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long start = 0;
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long end = num;
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long mid = 0;
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while (start <= end) {
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// untill start is lesser or equal to end do this
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// Finding middle value
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mid = start + (end - start) / 2;
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if (mid * mid == num) {
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// if mid*mid == num return true
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return true;
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} else if (mid * mid < num) {
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// if num is greater than mid*mid then make start = mid + 1
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start = mid + 1;
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} else if (mid * mid > num) {
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// if num is lesser than mid*mid then make end = mid - 1
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end = mid - 1;
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}
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}
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return false;
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}
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}
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367\. Valid Perfect Square
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Easy
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Given a **positive** integer _num_, write a function which returns True if _num_ is a perfect square else False.
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**Follow up:** **Do not** use any built-in library function such as `sqrt`.
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**Example 1:**
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**Input:** num = 16
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**Output:** true
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**Example 2:**
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**Input:** num = 14
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**Output:** false
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**Constraints:**
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* `1 <= num <= 2^31 - 1`
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package g0301_0400.s0368_largest_divisible_subset;
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// #Medium #Array #Dynamic_Programming #Math #Sorting
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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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public class Solution {
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// Helper class containing value and an arraylist
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public class Helper {
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int val;
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List<Integer> al;
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Helper(int val) {
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this.val = val;
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al = new ArrayList<>();
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}
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}
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public List<Integer> largestDivisibleSubset(int[] nums) {
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// Initializing Helper type array
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Helper[] dp = new Helper[nums.length];
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// Sorting given array
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Arrays.sort(nums);
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// max variable will keep track size of Longest Divisible Subset
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int max = 0;
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// index variable will keep track the index at which dp contains Longest Divisible Subset
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int index = 0;
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dp[0] = new Helper(1);
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dp[0].al.add(nums[0]);
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// Operation similar to LIS
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for (int i = 1; i < nums.length; i++) {
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int max2 = 0;
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int pos = i;
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for (int j = 0; j < i; j++) {
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if (nums[i] % nums[j] == 0) {
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if (max2 < dp[j].val) {
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max2 = dp[j].val;
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pos = j;
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}
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}
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}
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// max2 = 0, means no element exists which can divide the present element
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if (max2 == 0) {
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dp[i] = new Helper(max2 + 1);
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dp[i].al.add(nums[i]);
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} else {
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dp[i] = new Helper(max2 + 1);
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for (int val : dp[pos].al) {
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dp[i].al.add(val);
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}
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dp[i].al.add(nums[i]);
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}
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if (max2 > max) {
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max = max2;
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index = i;
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}
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}
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// Returning Largest Divisible Subset
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return dp[index].al;
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}
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}
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368\. Largest Divisible Subset
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Medium
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Given a set of **distinct** positive integers `nums`, return the largest subset `answer` such that every pair `(answer[i], answer[j])` of elements in this subset satisfies:
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* `answer[i] % answer[j] == 0`, or
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* `answer[j] % answer[i] == 0`
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If there are multiple solutions, return any of them.
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**Example 1:**
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**Input:** nums = \[1,2,3\]
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**Output:** \[1,2\]
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**Explanation:** \[1,3\] is also accepted.
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**Example 2:**
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**Input:** nums = \[1,2,4,8\]
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**Output:** \[1,2,4,8\]
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**Constraints:**
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* `1 <= nums.length <= 1000`
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* <code>1 <= nums[i] <= 2 * 10<sup>9</sup></code>
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* All the integers in `nums` are **unique**.
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package g0301_0400.s0365_water_and_jug_problem;
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import static org.hamcrest.CoreMatchers.equalTo;
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import static org.hamcrest.MatcherAssert.assertThat;
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import org.junit.jupiter.api.Test;
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class SolutionTest {
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@Test
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void canMeasureWater() {
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assertThat(new Solution().canMeasureWater(3, 5, 4), equalTo(true));
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}
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@Test
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void canMeasureWater2() {
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assertThat(new Solution().canMeasureWater(2, 6, 5), equalTo(false));
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}
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@Test
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void canMeasureWater3() {
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assertThat(new Solution().canMeasureWater(1, 2, 3), equalTo(true));
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}
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}
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package g0301_0400.s0367_valid_perfect_square;
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import static org.hamcrest.CoreMatchers.equalTo;
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import static org.hamcrest.MatcherAssert.assertThat;
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import org.junit.jupiter.api.Test;
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class SolutionTest {
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@Test
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void isPerfectSquare() {
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assertThat(new Solution().isPerfectSquare(16), equalTo(true));
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}
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@Test
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void isPerfectSquare2() {
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assertThat(new Solution().isPerfectSquare(14), equalTo(false));
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}
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}
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package g0301_0400.s0368_largest_divisible_subset;
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import static org.hamcrest.CoreMatchers.equalTo;
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import static org.hamcrest.MatcherAssert.assertThat;
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import java.util.ArrayList;
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import java.util.Arrays;
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import org.junit.jupiter.api.Test;
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class SolutionTest {
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@Test
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void largestDivisibleSubset() {
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assertThat(
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new Solution().largestDivisibleSubset(new int[] {1, 2, 3}),
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equalTo(new ArrayList<Integer>(Arrays.asList(1, 2))));
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}
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@Test
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void largestDivisibleSubset2() {
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assertThat(
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new Solution().largestDivisibleSubset(new int[] {1, 2, 4, 8}),
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equalTo(new ArrayList<Integer>(Arrays.asList(1, 2, 4, 8))));
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}
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}

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