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Commit 93c7dc1

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第151场双周赛T1~T4 (4)
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import java.util.Arrays;
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public class Solution3467 {
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public int[] transformArray(int[] nums) {
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int n = nums.length;
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int even = 0;
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for (int v : nums) {
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if (v % 2 == 0) even++;
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}
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int[] ans = new int[n];
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Arrays.fill(ans, 1);
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for (int i = 0; i < even; i++) {
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ans[i] = 0;
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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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3467. 将数组按照奇偶性转化
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https://leetcode.cn/problems/transform-array-by-parity/description/
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第 151 场双周赛 T1。
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给你一个整数数组 nums。请你按照以下顺序 依次 执行操作,转换 nums:
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1.将每个偶数替换为 0。
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2.将每个奇数替换为 1。
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3.按 非递减 顺序排序修改后的数组。
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执行完这些操作后,返回结果数组。
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提示:
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1 <= nums.length <= 100
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1 <= nums[i] <= 1000
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遍历计数。
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时间复杂度 O(n)。
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*/
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public class Solution3468 {
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public int countArrays(int[] original, int[][] bounds) {
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int n = original.length;
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int minUp = Integer.MAX_VALUE;
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int minDown = Integer.MAX_VALUE;
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for (int i = 0; i < n; i++) {
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int ui = bounds[i][0], vi = bounds[i][1];
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int up = vi - original[i]; // 上移多少
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int down = original[i] - ui; // 下移多少
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minUp = Math.min(minUp, up);
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minDown = Math.min(minDown, down);
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}
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return Math.max(0, minUp + minDown + 1);
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}
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}
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/*
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3468. 可行数组的数目
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https://leetcode.cn/problems/find-the-number-of-copy-arrays/description/
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第 151 场双周赛 T2。
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给你一个长度为 n 的数组 original 和一个长度为 n x 2 的二维数组 bounds,其中 bounds[i] = [ui, vi]。
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你需要找到长度为 n 且满足以下条件的 可能的 数组 copy 的数量:
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1.对于 1 <= i <= n - 1 ,都有 (copy[i] - copy[i - 1]) == (original[i] - original[i - 1]) 。
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2.对于 0 <= i <= n - 1 ,都有 ui <= copy[i] <= vi 。
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返回满足这些条件的数组数目。
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提示:
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2 <= n == original.length <= 10^5
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1 <= original[i] <= 10^9
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bounds.length == n
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bounds[i].length == 2
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1 <= bounds[i][0] <= bounds[i][1] <= 10^9
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移项得 copy[i]=copy[0]+original[i]−original[0]
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求这 n 个 copy[0] 的交集。
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时间复杂度 O(n)。
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*/
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import java.util.Arrays;
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public class Solution3469 {
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private int n;
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private int[] nums;
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private int[][] memo;
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public int minCost(int[] nums) {
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n = nums.length;
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this.nums = nums;
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memo = new int[n][n];
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for (int i = 0; i < n; i++) {
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Arrays.fill(memo[i], -1);
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}
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return dfs(1, 0);
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}
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private int dfs(int i, int j) {
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if (i == n) return nums[j];
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if (i == n - 1) return Math.max(nums[j], nums[i]);
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if (memo[i][j] != -1) return memo[i][j];
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int a = nums[j], b = nums[i], c = nums[i + 1];
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// 移除 a,b
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int res = dfs(i + 2, i + 1) + Math.max(a, b);
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// 移除 a,c
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res = Math.min(res, dfs(i + 2, i) + Math.max(a, c));
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// 移除 b,c
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res = Math.min(res, dfs(i + 2, j) + Math.max(b, c));
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return memo[i][j] = res;
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}
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}
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/*
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3469. 移除所有数组元素的最小代价
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https://leetcode.cn/problems/find-minimum-cost-to-remove-array-elements/description/
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第 151 场双周赛 T3。
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给你一个整数数组 nums。你的任务是在每一步中执行以下操作之一,直到 nums 为空,从而移除 所有元素 :
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- 从 nums 的前三个元素中选择任意两个元素并移除它们。此操作的成本为移除的两个元素中的 最大值 。
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- 如果 nums 中剩下的元素少于三个,则一次性移除所有剩余元素。此操作的成本为剩余元素中的 最大值 。
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返回移除所有元素所需的最小成本。
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提示:
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1 <= nums.length <= 1000
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1 <= nums[i] <= 10^6
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记忆化搜索,边界挺难写的。
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时间复杂度 O(n^2)。
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rating 2115 (clist.by)
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*/
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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 Solution3470 {
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// 预处理交替排列的方案数
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private static final List<Long> F = new ArrayList<>();
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static {
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F.add(1L);
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for (int i = 1; F.getLast() < 1e15; i++) {
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F.add(F.getLast() * i);
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F.add(F.getLast() * i);
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}
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}
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public int[] permute(int n, long k) {
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// k 改成从 0 开始,方便计算
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k--;
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if (n < F.size() && k >= F.get(n) * (2 - n % 2)) { // n 是偶数的时候,方案数乘以 2
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return new int[0];
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}
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// cand 表示剩余未填入 ans 的数字
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// cand[0] 保存偶数,cand[1] 保存奇数
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List<Integer>[] cand = new ArrayList[2];
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Arrays.setAll(cand, e -> new ArrayList<>());
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for (int i = 1; i <= n; i++) {
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cand[i % 2].add(i);
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}
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int[] ans = new int[n];
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int parity = 1; // 当前要填入 ans[i] 的数的奇偶性
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for (int i = 0; i < n; i++) {
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int j = 0;
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if (n - 1 - i < F.size()) {
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// 比如示例 1,按照第一个数分组,每一组的大小都是 size=2
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// 知道 k 和 size 就知道我们要去哪一组
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long size = F.get(n - 1 - i);
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j = (int) (k / size); // 去第 j 组
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k %= size;
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// n 是偶数的情况,第一个数既可以填奇数又可以填偶数,要特殊处理
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if (n % 2 == 0 && i == 0) {
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parity = 1 - j % 2;
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j /= 2;
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}
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} // else j=0,在 n 很大的情况下,只能按照 1,2,3,... 的顺序填
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ans[i] = cand[parity].remove(j);
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parity ^= 1; // 下一个数的奇偶性
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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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3470. 全排列 IV
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https://leetcode.cn/problems/permutations-iv/description/
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第 151 场双周赛 T4。
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给你两个整数 n 和 k,一个 交替排列 是前 n 个正整数的排列,且任意相邻 两个 元素不都为奇数或都为偶数。
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返回第 k 个 交替排列 ,并按 字典序 排序。如果有效的 交替排列 少于 k 个,则返回一个空列表。
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提示:
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1 <= n <= 100
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1 <= k <= 10^15
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从左往右构造。
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时间复杂度 O(n^2)。
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相似题目: 60. 排列序列
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https://leetcode.cn/problems/permutation-sequence/
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rating 2496 (clist.by)
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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 Solution3467Tests {
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private final Solution3467 solution3467 = new Solution3467();
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@Test
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public void example1() {
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int[] nums = {4, 3, 2, 1};
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int[] expected = {0, 0, 1, 1};
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Assertions.assertArrayEquals(expected, solution3467.transformArray(nums));
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}
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@Test
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public void example2() {
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int[] nums = {1, 5, 1, 4, 2};
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int[] expected = {0, 0, 1, 1, 1};
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Assertions.assertArrayEquals(expected, solution3467.transformArray(nums));
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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 Solution3468Tests {
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private final Solution3468 solution3468 = new Solution3468();
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@Test
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public void example1() {
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int[] original = {1, 2, 3, 4};
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int[][] bounds = UtUtils.stringToInts2("[[1,2],[2,3],[3,4],[4,5]]");
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int expected = 2;
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Assertions.assertEquals(expected, solution3468.countArrays(original, bounds));
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}
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@Test
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public void example2() {
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int[] original = {1, 2, 3, 4};
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int[][] bounds = UtUtils.stringToInts2("[[1,10],[2,9],[3,8],[4,7]]");
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int expected = 4;
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Assertions.assertEquals(expected, solution3468.countArrays(original, bounds));
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}
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@Test
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public void example3() {
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int[] original = {1, 2, 1, 2};
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int[][] bounds = UtUtils.stringToInts2("[[1,1],[2,3],[3,3],[2,3]]");
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int expected = 0;
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Assertions.assertEquals(expected, solution3468.countArrays(original, bounds));
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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 Solution3469Tests {
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private final Solution3469 solution3469 = new Solution3469();
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@Test
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public void example1() {
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int[] nums = {6, 2, 8, 4};
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int expected = 12;
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Assertions.assertEquals(expected, solution3469.minCost(nums));
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}
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@Test
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public void example2() {
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int[] nums = {2, 1, 3, 3};
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int expected = 5;
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Assertions.assertEquals(expected, solution3469.minCost(nums));
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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 Solution3470Tests {
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private final Solution3470 solution3470 = new Solution3470();
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@Test
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public void example1() {
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int n = 4;
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long k = 6;
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int[] expected = {3, 4, 1, 2};
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Assertions.assertArrayEquals(expected, solution3470.permute(n, k));
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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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long k = 2;
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int[] expected = {3, 2, 1};
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Assertions.assertArrayEquals(expected, solution3470.permute(n, k));
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}
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@Test
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public void example3() {
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int n = 2;
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long k = 3;
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int[] expected = {};
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Assertions.assertArrayEquals(expected, solution3470.permute(n, k));
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}
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// 补充用例
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@Test
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public void example4() {
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// https://leetcode.cn/problems/permutations-iv/submissions/608196751/
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int n = 41;
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long k = 872502217664402L;
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int[] expected = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 32, 21, 24, 33, 38, 31, 28, 37, 26, 27, 34, 23, 22, 29, 30, 25, 20, 35, 36, 41, 40, 39};
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Assertions.assertArrayEquals(expected, solution3470.permute(n, k));
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}
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}

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