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Commit 893268c

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第442场周赛T1~T4 (4)
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public class Solution3492 {
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public int maxContainers(int n, int w, int maxWeight) {
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return Math.min(maxWeight / w, n * n);
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}
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}
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/*
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3492. 船上可以装载的最大集装箱数量
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https://leetcode.cn/problems/maximum-containers-on-a-ship/description/
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第 442 场周赛 T1。
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给你一个正整数 n,表示船上的一个 n x n 的货物甲板。甲板上的每个单元格可以装载一个重量 恰好 为 w 的集装箱。
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然而,如果将所有集装箱装载到甲板上,其总重量不能超过船的最大承载重量 maxWeight。
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请返回可以装载到船上的 最大 集装箱数量。
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提示:
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1 <= n <= 1000
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1 <= w <= 1000
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1 <= maxWeight <= 10^9
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数学。
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时间复杂度 O(1)。
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*/
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import java.util.HashSet;
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import java.util.Set;
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public class Solution3493 {
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public int numberOfComponents(int[][] properties, int k) {
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int n = properties.length;
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DSU dsu = new DSU(n);
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for (int i = 0; i < n; i++) {
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for (int j = i + 1; j < n; j++) {
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if (intersectGeK(properties[i], properties[j], k)) {
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dsu.union(i, j);
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}
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}
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}
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return dsu.sz;
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}
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// intersect >= k
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private boolean intersectGeK(int[] property1, int[] property2, int k) {
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Set<Integer> set = new HashSet<>();
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for (int v : property1) set.add(v);
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Set<Integer> set2 = new HashSet<>();
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for (int v : property2) set2.add(v);
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set.retainAll(set2);
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return set.size() >= k;
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}
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static class DSU {
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int[] fa;
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int sz;
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public DSU(int n) {
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fa = new int[n];
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for (int i = 0; i < n; i++) fa[i] = i;
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sz = n;
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}
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int find(int x) { // 查找
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return x == fa[x] ? fa[x] : (fa[x] = find(fa[x]));
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}
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void union(int p, int q) { // 合并
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p = find(p);
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q = find(q);
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if (p == q) return;
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fa[q] = p;
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sz--;
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}
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}
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}
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/*
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3493. 属性图
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https://leetcode.cn/problems/properties-graph/description/
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第 442 场周赛 T2。
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给你一个二维整数数组 properties,其维度为 n x m,以及一个整数 k。
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定义一个函数 intersect(a, b),它返回数组 a 和 b 中 共有的不同整数的数量 。
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构造一个 无向图,其中每个索引 i 对应 properties[i]。如果且仅当 intersect(properties[i], properties[j]) >= k(其中 i 和 j 的范围为 [0, n - 1] 且 i != j),节点 i 和节点 j 之间有一条边。
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返回结果图中 连通分量 的数量。
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提示:
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1 <= n == properties.length <= 100
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1 <= m == properties[i].length <= 100
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1 <= properties[i][j] <= 100
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1 <= k <= m
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并查集。
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时间复杂度 O(n^2 * m)。
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*/
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public class Solution3494 {
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public long minTime(int[] skill, int[] mana) {
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int n = skill.length;
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int m = mana.length;
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// f[i][j] 表示 药水 i 被巫师 j 处理完的时间,答案为 f[m][n]
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long[][] f = new long[m + 1][n + 1];
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for (int i = 1; i <= m; i++) {
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for (int j = 1; j <= n; j++) {
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f[i][j] = Math.max(f[i][j - 1], f[i - 1][j]) + (long) skill[j - 1] * mana[i - 1];
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}
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// 校准
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for (int j = n - 1; j >= 1; j--) {
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f[i][j] = f[i][j + 1] - (long) skill[j + 1 - 1] * mana[i - 1];
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}
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}
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return f[m][n];
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}
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}
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/*
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3494. 酿造药水需要的最少总时间
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https://leetcode.cn/problems/find-the-minimum-amount-of-time-to-brew-potions/description/
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第 442 场周赛 T3。
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给你两个长度分别为 n 和 m 的整数数组 skill 和 mana 。
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在一个实验室里,有 n 个巫师,他们必须按顺序酿造 m 个药水。每个药水的法力值为 mana[j],并且每个药水 必须 依次通过 所有 巫师处理,才能完成酿造。第 i 个巫师在第 j 个药水上处理需要的时间为 timeij = skill[i] * mana[j]。
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由于酿造过程非常精细,药水在当前巫师完成工作后 必须 立即传递给下一个巫师并开始处理。这意味着时间必须保持 同步,确保每个巫师在药水到达时 马上 开始工作。
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返回酿造所有药水所需的 最短 总时间。
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提示:
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n == skill.length
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m == mana.length
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1 <= n, m <= 5000
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1 <= mana[i], skill[i] <= 5000
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正向贪心 + 反向校准。
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时间复杂度 O(mn)。
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*/
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public class Solution3495 {
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public long minOperations(int[][] queries) {
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long ans = 0;
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for (int[] q : queries) {
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int l = q[0], r = q[1];
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long sum_level = 0;
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int k = 1;
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int prev = 1;
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int curr = 4;
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while (prev <= r) {
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int start_k = prev;
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int end_k = curr - 1;
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int overlap_start = Math.max(l, start_k);
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int overlap_end = Math.min(r, end_k);
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if (overlap_start <= overlap_end) {
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long count = overlap_end - overlap_start + 1;
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sum_level += count * k;
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}
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prev = curr;
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curr *= 4;
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k++;
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}
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ans += (sum_level + 1) / 2;
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}
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return ans;
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}
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}
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/*
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3495. 使数组元素都变为零的最少操作次数
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https://leetcode.cn/problems/minimum-operations-to-make-array-elements-zero/description/
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第 442 场周赛 T4。
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给你一个二维数组 queries,其中 queries[i] 形式为 [l, r]。每个 queries[i] 表示了一个元素范围从 l 到 r (包括 l 和 r )的整数数组 nums 。
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在一次操作中,你可以:
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选择一个查询数组中的两个整数 a 和 b。
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将它们替换为 floor(a / 4) 和 floor(b / 4)。
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你的任务是确定对于每个查询,将数组中的所有元素都变为零的 最少 操作次数。返回所有查询结果的总和。
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提示:
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1 <= queries.length <= 10^5
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queries[i].length == 2
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queries[i] == [l, r]
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1 <= l < r <= 10^9
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数学。
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时间复杂度 O(qlogR)。
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*/
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import org.junit.jupiter.api.Assertions;
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import org.junit.jupiter.api.Test;
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public class Solution3492Tests {
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private final Solution3492 solution3492 = new Solution3492();
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@Test
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public void example1() {
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int n = 2;
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int w = 3;
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int maxWeight = 15;
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int expected = 4;
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Assertions.assertEquals(expected, solution3492.maxContainers(n, w, maxWeight));
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}
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@Test
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public void example2() {
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int n = 3;
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int w = 5;
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int maxWeight = 20;
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int expected = 4;
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Assertions.assertEquals(expected, solution3492.maxContainers(n, w, maxWeight));
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}
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}
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import org.junit.jupiter.api.Assertions;
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import org.junit.jupiter.api.Test;
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public class Solution3493Tests {
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private final Solution3493 solution3493 = new Solution3493();
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@Test
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public void example1() {
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int[][] properties = UtUtils.stringToInts2("[[1,2],[1,1],[3,4],[4,5],[5,6],[7,7]]");
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int k = 1;
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int expected = 3;
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Assertions.assertEquals(expected, solution3493.numberOfComponents(properties, k));
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}
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@Test
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public void example2() {
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int[][] properties = UtUtils.stringToInts2("[[1,2,3],[2,3,4],[4,3,5]]");
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int k = 2;
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int expected = 1;
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Assertions.assertEquals(expected, solution3493.numberOfComponents(properties, k));
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}
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@Test
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public void example3() {
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int[][] properties = UtUtils.stringToInts2("[[1,1],[1,1]]");
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int k = 2;
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int expected = 2;
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Assertions.assertEquals(expected, solution3493.numberOfComponents(properties, k));
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}
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}
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import org.junit.jupiter.api.Assertions;
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import org.junit.jupiter.api.Test;
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public class Solution3494Tests {
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private final Solution3494 solution3494 = new Solution3494();
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@Test
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public void example1() {
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int[] skill = {1, 5, 2, 4};
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int[] mana = {5, 1, 4, 2};
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long expected = 110;
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Assertions.assertEquals(expected, solution3494.minTime(skill, mana));
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}
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@Test
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public void example2() {
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int[] skill = {1, 1, 1};
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int[] mana = {1, 1, 1};
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long expected = 5;
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Assertions.assertEquals(expected, solution3494.minTime(skill, mana));
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}
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@Test
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public void example3() {
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int[] skill = {1, 2, 3, 4};
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int[] mana = {1, 2};
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long expected = 21;
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Assertions.assertEquals(expected, solution3494.minTime(skill, mana));
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}
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}
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import org.junit.jupiter.api.Assertions;
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import org.junit.jupiter.api.Test;
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public class Solution3495Tests {
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private final Solution3495 solution3495 = new Solution3495();
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@Test
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public void example1() {
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int[][] queries = UtUtils.stringToInts2("[[1,2],[2,4]]");
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long expected = 3;
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Assertions.assertEquals(expected, solution3495.minOperations(queries));
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}
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@Test
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public void example2() {
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int[][] queries = UtUtils.stringToInts2("[[2,6]]");
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long expected = 4;
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Assertions.assertEquals(expected, solution3495.minOperations(queries));
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}
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}

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