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Commit 76b3609

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第466场周赛T1~T4 (4)
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public class Solution3674 {
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public int minOperations(int[] nums) {
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for (int i = 1; i < nums.length; i++) {
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if (nums[i - 1] != nums[i]) return 1;
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}
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return 0;
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}
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}
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/*
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3674. 数组元素相等的最小操作次数
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https://leetcode.cn/problems/minimum-operations-to-equalize-array/description/
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第 466 场周赛 T1。
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给你一个长度为 n 的整数数组 nums。
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在一次操作中,可以选择任意子数组 nums[l...r] (0 <= l <= r < n),并将该子数组中的每个元素 替换 为所有元素的 按位与(bitwise AND)结果。
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返回使数组 nums 中所有元素相等所需的最小操作次数。
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子数组 是数组中连续的、非空的元素序列。
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提示:
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1 <= n == nums.length <= 100
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1 <= nums[i] <= 10^5
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脑筋急转弯。
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如果 nums 所有元素都相等,无需操作,返回 0。否则返回 1。
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时间复杂度 O(n)。
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*/
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public class Solution3675 {
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public int minOperations(String s) {
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int[] cnt = new int[26];
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for (char c : s.toCharArray()) {
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cnt[c - 'a']++;
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}
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for (int i = 1; i < 26; i++) {
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if (cnt[i] > 0) return 26 - i;
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}
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return 0;
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}
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}
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/*
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3675. 转换字符串的最小操作次数
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https://leetcode.cn/problems/minimum-operations-to-transform-string/description/
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第 466 场周赛 T2。
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给你一个仅由小写英文字母组成的字符串 s。
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你可以执行以下操作任意次(包括零次):
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- 选择字符串中出现的一个字符 c,并将 每个 出现的 c 替换为英文字母表中 下一个 小写字母。
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返回将 s 转换为仅由 'a' 组成的字符串所需的最小操作次数。
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注意:字母表是循环的,因此 'z' 的下一个字母是 'a'。
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提示:
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1 <= s.length <= 5 * 10^5
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s 仅由小写英文字母组成。
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统计出现过的字符。答案就是最小的非 a 字母到 z+1 的距离。
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时间复杂度 O(n)。
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*/
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import java.util.ArrayDeque;
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import java.util.Deque;
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public class Solution3676 {
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public long bowlSubarrays(int[] nums) {
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int n = nums.length;
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int[] left = new int[n];
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int[] right = new int[n];
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Deque<Integer> st = new ArrayDeque<>();
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for (int i = 0; i < n; i++) {
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while (!st.isEmpty() && nums[st.peek()] <= nums[i]) {
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st.pop();
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}
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left[i] = st.isEmpty() ? -1 : st.peek();
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st.push(i);
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}
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st.clear();
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for (int i = n - 1; i >= 0; i--) {
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while (!st.isEmpty() && nums[st.peek()] <= nums[i]) {
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st.pop();
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}
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right[i] = st.isEmpty() ? n : st.peek();
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st.push(i);
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}
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long ans = 0;
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for (int i = 0; i < n; i++) {
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if (left[i] != -1 && right[i] != n) {
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ans++;
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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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3676. 碗子数组的数目
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https://leetcode.cn/problems/count-bowl-subarrays/description/
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第 466 场周赛 T3。
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给你一个整数数组 nums,包含 互不相同 的元素。
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nums 的一个子数组 nums[l...r] 被称为 碗(bowl),如果它满足以下条件:
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- 子数组的长度至少为 3。也就是说,r - l + 1 >= 3。
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- 其两端元素的 最小值 严格大于 中间所有元素的 最大值。也就是说,min(nums[l], nums[r]) > max(nums[l + 1], ..., nums[r - 1])。
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返回 nums 中 碗 子数组的数量。
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子数组 是数组中连续的元素序列。
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提示:
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3 <= nums.length <= 10^5
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1 <= nums[i] <= 10^9
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nums 由不同的元素组成。
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单调栈。统计合法下标。返回值 long 是迷惑性质。
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时间复杂度 O(n)。
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https://yuanbao.tencent.com/chat/naQivTmsDa/0f153432-97c4-4f12-972f-bae0f4f6d805
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*/
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public class Solution3677 {
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public int countBinaryPalindromes(long n) {
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if (n == 0) {
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return 1;
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}
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String binaryStr = Long.toBinaryString(n);
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int totalLen = binaryStr.length();
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long count = 1;
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for (int L = 1; L < totalLen; L++) {
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int halfLen = (L + 1) / 2;
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count += (1L << (halfLen - 1));
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}
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int halfLen = (totalLen + 1) / 2;
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long low = 1L << (halfLen - 1);
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long high = (1L << halfLen) - 1;
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long left = low, right = high;
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long lastValid = low - 1;
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while (left <= right) {
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long mid = left + (right - left) / 2;
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long palin = constructPalindrome(mid, totalLen);
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if (palin <= n) {
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lastValid = mid;
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left = mid + 1;
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} else {
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right = mid - 1;
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}
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}
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if (lastValid >= low) {
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count += (lastValid - low + 1);
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}
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return (int) count;
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}
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private long constructPalindrome(long p, int len) {
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int halfLen = (len + 1) / 2;
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if (len % 2 == 0) {
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long reversed = reverseBits(p, halfLen);
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return (p << halfLen) | reversed;
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} else {
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long reversed = reverseBits(p >> 1, halfLen - 1);
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return (p << (halfLen - 1)) | reversed;
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}
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}
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private long reverseBits(long num, int bits) {
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long rev = 0;
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for (int i = 0; i < bits; i++) {
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rev = (rev << 1) | (num & 1);
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num >>= 1;
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}
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return rev;
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}
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}
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/*
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3677. 统计二进制回文数字的数目
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https://leetcode.cn/problems/count-binary-palindromic-numbers/description/
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第 466 场周赛 T4。
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给你一个 非负 整数 n。
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如果一个 非负 整数的二进制表示(不含前导零)正着读和倒着读都一样,则称该数为 二进制回文数。
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返回满足 0 <= k <= n 且 k 的二进制表示是回文数的整数 k 的数量。
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注意: 数字 0 被认为是二进制回文数,其表示为 "0"。
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提示:
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0 <= n <= 10^15
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方法思路
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1.特殊情况处理:如果n为0,直接返回1,因为0是二进制回文数。
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2.计算二进制长度:将n转换为二进制字符串,并计算其长度len。
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3.统计较短长度的回文数:对于每个长度L从1到len-1,计算该长度的二进制回文数的数量。这些回文数的数量由前半部分的位数决定,即2^(halfLen-1),其中halfLen是(L+1)/2。
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4.统计相同长度的回文数:对于长度为len的二进制回文数,使用二分查找来高效地确定满足条件的回文数数量,而不是逐个生成和检查。
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https://chat.deepseek.com/a/chat/s/f16c1854-1de7-4750-b0c3-19b280f76ae2
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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 Solution3674Tests {
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private final Solution3674 solution3674 = new Solution3674();
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@Test
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public void example1() {
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int[] nums = {1, 2};
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int expected = 1;
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Assertions.assertEquals(expected, solution3674.minOperations(nums));
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}
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@Test
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public void example2() {
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int[] nums = {5, 5, 5};
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int expected = 0;
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Assertions.assertEquals(expected, solution3674.minOperations(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 Solution3675Tests {
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private final Solution3675 solution3675 = new Solution3675();
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@Test
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public void example1() {
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String s = "yz";
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int expected = 2;
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Assertions.assertEquals(expected, solution3675.minOperations(s));
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}
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@Test
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public void example2() {
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String s = "a";
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int expected = 0;
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Assertions.assertEquals(expected, solution3675.minOperations(s));
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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 Solution3676Tests {
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private final Solution3676 solution3676 = new Solution3676();
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@Test
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public void example1() {
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int[] nums = {2, 5, 3, 1, 4};
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long expected = 2;
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Assertions.assertEquals(expected, solution3676.bowlSubarrays(nums));
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}
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@Test
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public void example2() {
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int[] nums = {5, 1, 2, 3, 4};
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long expected = 3;
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Assertions.assertEquals(expected, solution3676.bowlSubarrays(nums));
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}
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@Test
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public void example3() {
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int[] nums = {1000000000, 999999999, 999999998};
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long expected = 0;
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Assertions.assertEquals(expected, solution3676.bowlSubarrays(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 Solution3677Tests {
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private final Solution3677 solution3677 = new Solution3677();
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@Test
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public void example1() {
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long n = 9;
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int expected = 6;
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Assertions.assertEquals(expected, solution3677.countBinaryPalindromes(n));
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}
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@Test
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public void example2() {
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long n = 0;
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int expected = 1;
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Assertions.assertEquals(expected, solution3677.countBinaryPalindromes(n));
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}
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}

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