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Commit 4b81f52

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第468场周赛T1~T4 (4)
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public class Solution3688 {
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public int evenNumberBitwiseORs(int[] nums) {
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int ans = 0;
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for (int v : nums) {
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if (v % 2 == 0) {
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ans |= v;
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}
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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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3688. 偶数的按位或运算
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https://leetcode.cn/problems/bitwise-or-of-even-numbers-in-an-array/description/
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第 468 场周赛 T1。
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给你一个整数数组 nums。
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返回数组中所有 偶数 的按位 或 运算结果。
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如果 nums 中没有偶数,返回 0。
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提示:
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1 <= nums.length <= 100
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1 <= nums[i] <= 100
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中国时间 2025年09月21日 周日 10:30
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佛山。
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模拟。
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*/
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import java.util.Arrays;
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public class Solution3689 {
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public long maxTotalValue(int[] nums, int k) {
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int mx = Arrays.stream(nums).max().orElseThrow();
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int mn = Arrays.stream(nums).min().orElseThrow();
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return (long) (mx - mn) * k;
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}
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}
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/*
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3689. 最大子数组总值 I
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https://leetcode.cn/problems/maximum-total-subarray-value-i/description/
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第 468 场周赛 T2。
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给定一个长度为 n 的整数数组 nums 和一个整数 k。
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你必须从 nums 中选择 恰好 k 个非空子数组 nums[l..r]。子数组可以重叠,同一个子数组(相同的 l 和 r)可以 被选择超过一次。
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子数组 nums[l..r] 的 值 定义为:max(nums[l..r]) - min(nums[l..r])。
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总值 是所有被选子数组的 值 之和。
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返回你能实现的 最大 可能总值。
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子数组 是数组中连续的 非空 元素序列。
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提示:
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1 <= n == nums.length <= 5 * 10^4
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0 <= nums[i] <= 10^9
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1 <= k <= 10^5
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脑筋急转弯。最大值 - 最小值 的值乘以 k 即可。
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*/
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import java.util.*;
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import java.util.stream.Collectors;
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public class Solution3690 {
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public int minSplitMerge(int[] nums1, int[] nums2) {
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int n = nums1.length;
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List<Integer> nums2List = toList(nums2);
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Set<List<Integer>> vis = new HashSet<>();
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vis.add(toList(nums1));
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List<List<Integer>> q = List.of(toList(nums1));
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for (int ans = 0; ; ans++) {
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List<List<Integer>> tmp = q;
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q = new ArrayList<>();
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for (List<Integer> a : tmp) {
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if (a.equals(nums2List)) {
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return ans;
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}
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for (int l = 0; l < n; l++) {
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for (int r = l + 1; r <= n; r++) {
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List<Integer> sub = a.subList(l, r);
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List<Integer> b = new ArrayList<>(a);
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b.subList(l, r).clear(); // 从 b 中移除 sub
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for (int i = 0; i <= b.size(); i++) {
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List<Integer> c = new ArrayList<>(b);
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c.addAll(i, sub);
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if (vis.add(c)) { // c 不在 vis 中
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q.add(c);
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}
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}
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}
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}
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}
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}
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}
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private List<Integer> toList(int[] nums) {
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return Arrays.stream(nums).boxed().collect(Collectors.toList());
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}
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}
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/*
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3690. 拆分合并数组
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https://leetcode.cn/problems/split-and-merge-array-transformation/description/
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第 468 场周赛 T3。
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给你两个长度为 n 的整数数组 nums1 和 nums2。你可以对 nums1 执行任意次下述的 拆分合并操作:
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1.选择一个子数组 nums1[L..R]。
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2.移除该子数组,留下前缀 nums1[0..L-1](如果 L = 0 则为空)和后缀 nums1[R+1..n-1](如果 R = n - 1 则为空)。
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3.将移除的子数组(按原顺序)重新插入到剩余数组的 任意 位置(即,在任意两个元素之间、最开始或最后面)。
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返回将 nums1 转换为 nums2 所需的 最少拆分合并操作 次数。
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提示:
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2 <= n == nums1.length == nums2.length <= 6
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-10^5 <= nums1[i], nums2[i] <= 10^5
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nums2 是 nums1 的一个 排列。
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根据数据范围,至多有 6!=720 个不同的排列,这很小,考虑暴力。
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https://leetcode.cn/problems/split-and-merge-array-transformation/solutions/3787914/bfspythonjavacgo-by-endlesscheng-86ya/
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时间复杂度 O(n! * n^4)。
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*/
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import java.util.Comparator;
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import java.util.PriorityQueue;
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public class Solution3691 {
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public long maxTotalValue(int[] nums, int k) {
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int n = nums.length;
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SparseTable st = new SparseTable(nums);
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PriorityQueue<int[]> pq = new PriorityQueue<>(Comparator.comparingInt(o -> -o[0])); // 最大堆
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for (int i = 0; i < n; i++) {
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pq.add(new int[]{st.query(i, n - 1), i, n - 1}); // 子数组值,左端点,右端点加一
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}
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long ans = 0;
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while (k-- > 0 && pq.peek()[0] > 0) {
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int[] top = pq.poll();
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ans += top[0];
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top[2]--;
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top[0] = st.query(top[1], top[2]);
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pq.add(top);
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}
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return ans;
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}
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static class SparseTable {
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long[][] mx, mi;
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int[] logTable;
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public SparseTable(int[] arr) {
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int n = arr.length;
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int maxLog = log2(n) + 1;
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mx = new long[n][maxLog];
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mi = new long[n][maxLog];
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logTable = new int[n + 1];
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// 预处理对数表
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for (int i = 2; i <= n; ++i) {
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logTable[i] = logTable[i / 2] + 1;
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}
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// 初始化 ST 表
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for (int i = 0; i < n; ++i) {
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mx[i][0] = arr[i];
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mi[i][0] = arr[i];
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}
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// 动态规划填充表
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for (int j = 1; (1 << j) <= n; ++j) {
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for (int i = 0; i + (1 << j) <= n; ++i) {
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mx[i][j] = Math.max(mx[i][j - 1], mx[i + (1 << (j - 1))][j - 1]);
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mi[i][j] = Math.min(mi[i][j - 1], mi[i + (1 << (j - 1))][j - 1]);
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}
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}
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}
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// ceil(log2(x)) = 32 - numberOfLeadingZeros(x - 1)
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int log2(long x) {
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return 64 - Long.numberOfLeadingZeros(x - 1);
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}
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long query_max(int l, int r) {
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int j = logTable[r - l + 1];
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return Math.max(mx[l][j], mx[r - (1 << j) + 1][j]);
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}
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long query_min(int l, int r) {
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int j = logTable[r - l + 1];
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return Math.min(mi[l][j], mi[r - (1 << j) + 1][j]);
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}
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int query(int l, int r) {
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return (int) (query_max(l, r) - query_min(l, r));
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}
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}
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}
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/*
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3691. 最大子数组总值 II
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https://leetcode.cn/problems/maximum-total-subarray-value-ii/description/
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第 468 场周赛 T4。
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给你一个长度为 n 的整数数组 nums 和一个整数 k。
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你必须从 nums 中选择 恰好 k 个 不同 的非空子数组 nums[l..r]。子数组可以重叠,但同一个子数组(相同的 l 和 r)不能 被选择超过一次。
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子数组 nums[l..r] 的 值 定义为:max(nums[l..r]) - min(nums[l..r])。
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总值 是所有被选子数组的 值 之和。
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返回你能实现的 最大 可能总值。
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子数组 是数组中连续的 非空 元素序列。
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提示:
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1 <= n == nums.length <= 5 * 10^4
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0 <= nums[i] <= 10^9
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1 <= k <= min(10^5, n * (n + 1) / 2)
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ST 表 + 最大堆
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https://leetcode.cn/problems/maximum-total-subarray-value-ii/solutions/3787892/st-biao-zui-da-dui-pythonjavacgo-by-endl-igja/
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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 Solution3688Tests {
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private final Solution3688 solution3688 = new Solution3688();
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@Test
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public void example1() {
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int[] nums = {1, 2, 3, 4, 5, 6};
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int expected = 6;
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Assertions.assertEquals(expected, solution3688.evenNumberBitwiseORs(nums));
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}
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@Test
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public void example2() {
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int[] nums = {7, 9, 11};
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int expected = 0;
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Assertions.assertEquals(expected, solution3688.evenNumberBitwiseORs(nums));
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}
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@Test
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public void example3() {
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int[] nums = {1, 8, 16};
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int expected = 24;
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Assertions.assertEquals(expected, solution3688.evenNumberBitwiseORs(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 Solution3689Tests {
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private final Solution3689 solution3689 = new Solution3689();
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@Test
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public void example1() {
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int[] nums = {1, 3, 2};
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int k = 2;
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long expected = 4;
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Assertions.assertEquals(expected, solution3689.maxTotalValue(nums, k));
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}
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@Test
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public void example2() {
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int[] nums = {4, 2, 5, 1};
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int k = 3;
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long expected = 12;
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Assertions.assertEquals(expected, solution3689.maxTotalValue(nums, 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 Solution3690Tests {
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private final Solution3690 solution3690 = new Solution3690();
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@Test
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public void example1() {
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int[] nums1 = {3, 1, 2};
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int[] nums2 = {1, 2, 3};
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int expected = 1;
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Assertions.assertEquals(expected, solution3690.minSplitMerge(nums1, nums2));
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}
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@Test
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public void example2() {
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int[] nums1 = {1, 1, 2, 3, 4, 5};
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int[] nums2 = {5, 4, 3, 2, 1, 1};
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int expected = 3;
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Assertions.assertEquals(expected, solution3690.minSplitMerge(nums1, nums2));
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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 Solution3691Tests {
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private final Solution3691 solution3691 = new Solution3691();
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@Test
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public void example1() {
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int[] nums = {1, 3, 2};
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int k = 2;
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long expected = 4;
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Assertions.assertEquals(expected, solution3691.maxTotalValue(nums, k));
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}
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@Test
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public void example2() {
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int[] nums = {4, 2, 5, 1};
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int k = 3;
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long expected = 12;
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Assertions.assertEquals(expected, solution3691.maxTotalValue(nums, k));
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}
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}

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