Design an algorithm to serialize and deserialize a binary tree. There is no restriction on how your serialization/deserialization algorithm should work. You just need to ensure that a binary tree can be serialized to a string and this string can be deserialized to the original tree structure.
-Clarification: The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.
+Clarification: The input/output format is the same as how LeetCode serializes a binary tree. You do not necessarily need to follow this format, so please be creative and come up with different approaches yourself.
Example 1:
diff --git a/solution/0500-0599/0593.Valid Square/README.md b/solution/0500-0599/0593.Valid Square/README.md index 1b818e3a203d9..ff04df42eea6f 100644 --- a/solution/0500-0599/0593.Valid Square/README.md +++ b/solution/0500-0599/0593.Valid Square/README.md @@ -29,7 +29,7 @@ tags:输入: p1 = [0,0], p2 = [1,1], p3 = [1,0], p4 = [0,1] -输出: True +输出: true
示例 2:
diff --git a/solution/0500-0599/0594.Longest Harmonious Subsequence/README.md b/solution/0500-0599/0594.Longest Harmonious Subsequence/README.md index cc09771a73c64..06b75b6a92e88 100644 --- a/solution/0500-0599/0594.Longest Harmonious Subsequence/README.md +++ b/solution/0500-0599/0594.Longest Harmonious Subsequence/README.md @@ -22,7 +22,7 @@ tags:和谐数组是指一个数组里元素的最大值和最小值之间的差别 正好是 1 。
给你一个整数数组 nums ，请你在所有可能的子序列中找到最长的和谐子序列的长度。
给你一个整数数组 nums ，请你在所有可能的 子序列 中找到最长的和谐子序列的长度。
数组的 子序列 是一个由数组派生出来的序列，它可以通过删除一些元素或不删除元素、且不改变其余元素的顺序而得到。
diff --git a/solution/0500-0599/0594.Longest Harmonious Subsequence/README_EN.md b/solution/0500-0599/0594.Longest Harmonious Subsequence/README_EN.md index cb21a5cbe29d1..606ae0f82eff8 100644 --- a/solution/0500-0599/0594.Longest Harmonious Subsequence/README_EN.md +++ b/solution/0500-0599/0594.Longest Harmonious Subsequence/README_EN.md @@ -22,9 +22,7 @@ tags:We define a harmonious array as an array where the difference between its maximum value and its minimum value is exactly 1.
Given an integer array nums, return the length of its longest harmonious subsequence among all its possible subsequences.
A subsequence of array is a sequence that can be derived from the array by deleting some or no elements without changing the order of the remaining elements.
+Given an integer array nums, return the length of its longest harmonious subsequence among all its possible subsequences.
Example 1:
diff --git a/solution/0900-0999/0975.Odd Even Jump/README.md b/solution/0900-0999/0975.Odd Even Jump/README.md index 71f1151615bcd..f5e96324a9b87 100644 --- a/solution/0900-0999/0975.Odd Even Jump/README.md +++ b/solution/0900-0999/0975.Odd Even Jump/README.md @@ -20,17 +20,17 @@ tags: -给定一个整数数组 A，你可以从某一起始索引出发，跳跃一定次数。在你跳跃的过程中，第 1、3、5... 次跳跃称为奇数跳跃，而第 2、4、6... 次跳跃称为偶数跳跃。
给定一个整数数组 arr，你可以从某一起始索引出发，跳跃一定次数。在你跳跃的过程中，第 1、3、5... 次跳跃称为奇数跳跃，而第 2、4、6... 次跳跃称为偶数跳跃。
你可以按以下方式从索引 i 向后跳转到索引 j（其中 i < j）：
j，使得 A[i] <= A[j]，A[j] 是可能的最小值。如果存在多个这样的索引 j，你只能跳到满足要求的最小索引 j 上。j，使得 A[i] >= A[j]，A[j] 是可能的最大值。如果存在多个这样的索引 j，你只能跳到满足要求的最小索引 j 上。j，使得 arr[i] <= arr[j]，且 arr[j] 的值尽可能小。如果存在多个这样的索引 j，你只能跳到满足要求的最小索引 j 上。j，使得 arr[i] >= arr[j]，且 arr[j] 的值尽可能大。如果存在多个这样的索引 j，你只能跳到满足要求的最小索引 j 上。i，可能无法进行合乎要求的跳跃。）如果从某一索引开始跳跃一定次数（可能是 0 次或多次），就可以到达数组的末尾（索引 A.length - 1），那么该索引就会被认为是好的起始索引。
如果从某一索引开始跳跃一定次数（可能是 0 次或多次），就可以到达数组的末尾（索引 arr.length - 1），那么该索引就会被认为是好的起始索引。
返回好的起始索引的数量。
@@ -38,10 +38,11 @@ tags:示例 1：
-输入：[10,13,12,14,15] ++输入：[10,13,12,14,15] 输出：2 解释： -从起始索引 i = 0 出发，我们可以跳到 i = 2，（因为 A[2] 是 A[1]，A[2]，A[3]，A[4] 中大于或等于 A[0] 的最小值），然后我们就无法继续跳下去了。 +从起始索引 i = 0 出发，我们可以跳到 i = 2，（因为 arr[2] 是 arr[1]，arr[2]，arr[3]，arr[4] 中大于或等于 arr[0] 的最小值），然后我们就无法继续跳下去了。 从起始索引 i = 1 和 i = 2 出发，我们可以跳到 i = 3，然后我们就无法继续跳下去了。 从起始索引 i = 3 出发，我们可以跳到 i = 4，到达数组末尾。 从起始索引 i = 4 出发，我们已经到达数组末尾。 @@ -50,16 +51,17 @@ tags:示例 2：
-输入：[2,3,1,1,4] ++输入：[2,3,1,1,4] 输出：3 解释： 从起始索引 i=0 出发，我们依次可以跳到 i = 1，i = 2，i = 3： -在我们的第一次跳跃（奇数）中，我们先跳到 i = 1，因为 A[1] 是（A[1]，A[2]，A[3]，A[4]）中大于或等于 A[0] 的最小值。 +在我们的第一次跳跃（奇数）中，我们先跳到 i = 1，因为 arr[1] 是（arr[1]，arr[2]，arr[3]，arr[4]）中大于或等于 arr[0] 的最小值。 -在我们的第二次跳跃（偶数）中，我们从 i = 1 跳到 i = 2，因为 A[2] 是（A[2]，A[3]，A[4]）中小于或等于 A[1] 的最大值。A[3] 也是最大的值，但 2 是一个较小的索引，所以我们只能跳到 i = 2，而不能跳到 i = 3。 +在我们的第二次跳跃（偶数）中，我们从 i = 1 跳到 i = 2，因为 arr[2] 是（arr[2]，arr[3]，arr[4]）中小于或等于 arr[1] 的最大值。arr[3] 也是最大的值，但 2 是一个较小的索引，所以我们只能跳到 i = 2，而不能跳到 i = 3。 -在我们的第三次跳跃（奇数）中，我们从 i = 2 跳到 i = 3，因为 A[3] 是（A[3]，A[4]）中大于或等于 A[2] 的最小值。 +在我们的第三次跳跃（奇数）中，我们从 i = 2 跳到 i = 3，因为 arr[3] 是（arr[3]，arr[4]）中大于或等于 arr[2] 的最小值。 我们不能从 i = 3 跳到 i = 4，所以起始索引 i = 0 不是好的起始索引。 @@ -73,7 +75,8 @@ tags:示例 3：
-输入：[5,1,3,4,2] ++输入：[5,1,3,4,2] 输出：3 解释： 我们可以从起始索引 1，2，4 出发到达数组末尾。 @@ -84,8 +87,8 @@ tags:提示：
-
diff --git a/solution/1200-1299/1281.Subtract the Product and Sum of Digits of an Integer/README.md b/solution/1200-1299/1281.Subtract the Product and Sum of Digits of an Integer/README.md index 9fd1818df76b2..fb8cf597844ea 100644 --- a/solution/1200-1299/1281.Subtract the Product and Sum of Digits of an Integer/README.md +++ b/solution/1200-1299/1281.Subtract the Product and Sum of Digits of an Integer/README.md @@ -25,10 +25,10 @@ tags:- -
1 <= A.length <= 20000- +
0 <= A[i] < 100000- +
1 <= arr.length <= 200000 <= arr[i] < 100000示例 1：
输入：n = 234 -输出：15 +输出：15 解释： -各位数之积 = 2 * 3 * 4 = 24 -各位数之和 = 2 + 3 + 4 = 9 +各位数之积 = 2 * 3 * 4 = 24 +各位数之和 = 2 + 3 + 4 = 9 结果 = 24 - 9 = 15@@ -36,9 +36,9 @@ tags:输入：n = 4421 输出：21 -解释： -各位数之积 = 4 * 4 * 2 * 1 = 32 -各位数之和 = 4 + 4 + 2 + 1 = 11 +解释： +各位数之积 = 4 * 4 * 2 * 1 = 32 +各位数之和 = 4 + 4 + 2 + 1 = 11 结果 = 32 - 11 = 21diff --git a/solution/1200-1299/1281.Subtract the Product and Sum of Digits of an Integer/README_EN.md b/solution/1200-1299/1281.Subtract the Product and Sum of Digits of an Integer/README_EN.md index a6251a0511bcf..747d67bfc0a8b 100644 --- a/solution/1200-1299/1281.Subtract the Product and Sum of Digits of an Integer/README_EN.md +++ b/solution/1200-1299/1281.Subtract the Product and Sum of Digits of an Integer/README_EN.md @@ -25,10 +25,10 @@ Given an integer numbern, return the difference between the producInput: n = 234 -Output: 15 -Explanation: -Product of digits = 2 * 3 * 4 = 24 -Sum of digits = 2 + 3 + 4 = 9 +Output: 15 +Explanation: +Product of digits = 2 * 3 * 4 = 24 +Sum of digits = 2 + 3 + 4 = 9 Result = 24 - 9 = 15@@ -37,9 +37,9 @@ Result = 24 - 9 = 15Input: n = 4421 Output: 21 -Explanation: -Product of digits = 4 * 4 * 2 * 1 = 32 -Sum of digits = 4 + 4 + 2 + 1 = 11 +Explanation: +Product of digits = 4 * 4 * 2 * 1 = 32 +Sum of digits = 4 + 4 + 2 + 1 = 11 Result = 32 - 11 = 21diff --git a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/README.md b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/README.md index ebba6f60b91fa..6a05d1729477a 100644 --- a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/README.md +++ b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/README.md @@ -41,7 +41,7 @@ tags:输入：nums = [555,901,482,1771] -输出：1 +输出：1 解释： 只有 1771 是位数为偶数的数字。diff --git a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/README_EN.md b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/README_EN.md index 2a92c8c6d1e4b..85b2b1a18b99c 100644 --- a/solution/1200-1299/1295.Find Numbers with Even Number of Digits/README_EN.md +++ b/solution/1200-1299/1295.Find Numbers with Even Number of Digits/README_EN.md @@ -27,7 +27,7 @@ tags:Input: nums = [12,345,2,6,7896] Output: 2 -Explanation: +Explanation: 12 contains 2 digits (even number of digits). 345 contains 3 digits (odd number of digits). 2 contains 1 digit (odd number of digits). @@ -40,7 +40,7 @@ Therefore only 12 and 7896 contain an even number of digits.Input: nums = [555,901,482,1771] -Output: 1 +Output: 1 Explanation: Only 1771 contains an even number of digits.diff --git a/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/README_EN.md b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/README_EN.md index 5cd11d3f3f03c..daf591b44707a 100644 --- a/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/README_EN.md +++ b/solution/1200-1299/1299.Replace Elements with Greatest Element on Right Side/README_EN.md @@ -28,7 +28,7 @@ tags:Input: arr = [17,18,5,4,6,1] Output: [18,6,6,6,1,-1] -Explanation: +Explanation: - index 0 --> the greatest element to the right of index 0 is index 1 (18). - index 1 --> the greatest element to the right of index 1 is index 4 (6). - index 2 --> the greatest element to the right of index 2 is index 4 (6). diff --git a/solution/1400-1499/1470.Shuffle the Array/README.md b/solution/1400-1499/1470.Shuffle the Array/README.md index 376f0ef21ca39..0abf200137426 100644 --- a/solution/1400-1499/1470.Shuffle the Array/README.md +++ b/solution/1400-1499/1470.Shuffle the Array/README.md @@ -27,7 +27,7 @@ tags:diff --git a/solution/1400-1499/1475.Final Prices With a Special Discount in a Shop/README_EN.md b/solution/1400-1499/1475.Final Prices With a Special Discount in a Shop/README_EN.md index 72d7dfd45532f..7546d3beeda30 100644 --- a/solution/1400-1499/1475.Final Prices With a Special Discount in a Shop/README_EN.md +++ b/solution/1400-1499/1475.Final Prices With a Special Discount in a Shop/README_EN.md @@ -32,7 +32,7 @@ tags:示例 1：
输入：nums = [2,5,1,3,4,7], n = 3 -输出：[2,3,5,4,1,7] +输出：[2,3,5,4,1,7] 解释：由于 x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 ，所以答案为 [2,3,5,4,1,7]diff --git a/solution/1400-1499/1470.Shuffle the Array/README_EN.md b/solution/1400-1499/1470.Shuffle the Array/README_EN.md index efdc95874d1c1..8a806cd3fd73d 100644 --- a/solution/1400-1499/1470.Shuffle the Array/README_EN.md +++ b/solution/1400-1499/1470.Shuffle the Array/README_EN.md @@ -30,7 +30,7 @@ tags: Input: nums = [2,5,1,3,4,7], n = 3 -Output: [2,3,5,4,1,7] +Output: [2,3,5,4,1,7] Explanation: Since x1=2, x2=5, x3=1, y1=3, y2=4, y3=7 then the answer is [2,3,5,4,1,7]. diff --git a/solution/1400-1499/1473.Paint House III/README_EN.md b/solution/1400-1499/1473.Paint House III/README_EN.md index 804a5898db39a..0c13ec8adc04a 100644 --- a/solution/1400-1499/1473.Paint House III/README_EN.md +++ b/solution/1400-1499/1473.Paint House III/README_EN.md @@ -53,7 +53,7 @@ Cost of paint all houses (1 + 1 + 1 + 1 + 5) = 9. Input: houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3 Output: 11 Explanation: Some houses are already painted, Paint the houses of this way [2,2,1,2,2] -This array contains target = 3 neighborhoods, [{2,2}, {1}, {2,2}]. +This array contains target = 3 neighborhoods, [{2,2}, {1}, {2,2}]. Cost of paint the first and last house (10 + 1) = 11.Input: prices = [8,4,6,2,3] Output: [4,2,4,2,3] -Explanation: +Explanation: For item 0 with price[0]=8 you will receive a discount equivalent to prices[1]=4, therefore, the final price you will pay is 8 - 4 = 4. For item 1 with price[1]=4 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 4 - 2 = 2. For item 2 with price[2]=6 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 6 - 2 = 4. diff --git a/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/README.md b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/README.md index 366fe481c6cf9..af2911a1357a6 100644 --- a/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/README.md +++ b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/README.md @@ -90,7 +90,21 @@ tags: -### 方法一 +### 方法一：二分查找 + +根据题目描述，如果一个天数 $t$ 可以满足制作 $m$ 束花，那么对任意 $t' > t$，也一定可以满足制作 $m$ 束花。因此，我们可以使用二分查找的方法找到最小的满足制作 $m$ 束花的天数。 + +我们记花园中最大的开花天数为 $mx$，接下来，我们定义二分查找的左边界 $l = 1$，右边界 $r = mx + 1$。 + +然后，我们进行二分查找，对于每一个中间值 $\textit{mid} = \frac{l + r}{2}$，我们判断是否可以制作 $m$ 束花。如果可以，我们将右边界 $r$ 更新为 $\textit{mid}$，否则，我们将左边界 $l$ 更新为 $\textit{mid} + 1$。 + +最终，当 $l = r$ 时，结束二分查找。此时如果 $l > mx$，说明无法制作 $m$ 束花，返回 $-1$，否则返回 $l$。 + +因此，问题转换为判断一个天数 $\textit{days}$ 是否可以制作 $m$ 束花。 + +我们可以使用一个函数 $\text{check}(\textit{days})$ 来判断是否可以制作 $m$ 束花。具体地，我们从左到右遍历花园中的每一朵花，如果当前花开的天数小于等于 $\textit{days}$，我们将当前花加入到当前花束中，否则，我们将当前花束清空。当当前花束中的花的数量等于 $k$ 时，我们将当前花束的数量加一，并清空当前花束。最后，我们判断当前花束的数量是否大于等于 $m$，如果是，说明可以制作 $m$ 束花，否则，说明无法制作 $m$ 束花。 + +时间复杂度 $O(n \times \log M)$，其中 $n$ 和 $M$ 分别为花园中的花的数量和最大的开花天数，本题中 $M \leq 10^9$。空间复杂度 $O(1)$。 @@ -99,59 +113,50 @@ tags: ```python class Solution: def minDays(self, bloomDay: List[int], m: int, k: int) -> int: - if m * k > len(bloomDay): - return -1 - - def check(day: int) -> bool: + def check(days: int) -> int: cnt = cur = 0 - for bd in bloomDay: - cur = cur + 1 if bd <= day else 0 + for x in bloomDay: + cur = cur + 1 if x <= days else 0 if cur == k: cnt += 1 cur = 0 return cnt >= m - left, right = min(bloomDay), max(bloomDay) - while left < right: - mid = (left + right) >> 1 - if check(mid): - right = mid - else: - left = mid + 1 - return left + mx = max(bloomDay) + l = bisect_left(range(mx + 2), True, key=check) + return -1 if l > mx else l ``` #### Java ```java class Solution { + private int[] bloomDay; + private int m, k; + public int minDays(int[] bloomDay, int m, int k) { - if (m * k > bloomDay.length) { - return -1; - } - int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE; - for (int bd : bloomDay) { - min = Math.min(min, bd); - max = Math.max(max, bd); - } - int left = min, right = max; - while (left < right) { - int mid = (left + right) >>> 1; - if (check(bloomDay, m, k, mid)) { - right = mid; + this.bloomDay = bloomDay; + this.m = m; + this.k = k; + final int mx = (int) 1e9; + int l = 1, r = mx + 1; + while (l < r) { + int mid = (l + r) >> 1; + if (check(mid)) { + r = mid; } else { - left = mid + 1; + l = mid + 1; } } - return left; + return l > mx ? -1 : l; } - private boolean check(int[] bloomDay, int m, int k, int day) { + private boolean check(int days) { int cnt = 0, cur = 0; - for (int bd : bloomDay) { - cur = bd <= day ? cur + 1 : 0; + for (int x : bloomDay) { + cur = x <= days ? cur + 1 : 0; if (cur == k) { - cnt++; + ++cnt; cur = 0; } } @@ -166,36 +171,28 @@ class Solution { class Solution { public: int minDays(vector-& bloomDay, int m, int k) { - if (m * k > bloomDay.size()) { - return -1; - } - int mi = INT_MIN, mx = INT_MAX; - for (int& bd : bloomDay) { - mi = min(mi, bd); - mx = max(mx, bd); - } - int left = mi, right = mx; - while (left < right) { - int mid = left + right >> 1; - if (check(bloomDay, m, k, mid)) { - right = mid; - } else { - left = mid + 1; + int mx = ranges::max(bloomDay); + int l = 1, r = mx + 1; + auto check = [&](int days) { + int cnt = 0, cur = 0; + for (int x : bloomDay) { + cur = x <= days ? cur + 1 : 0; + if (cur == k) { + cnt++; + cur = 0; + } } - } - return left; - } - - bool check(vector & bloomDay, int m, int k, int day) { - int cnt = 0, cur = 0; - for (int& bd : bloomDay) { - cur = bd <= day ? cur + 1 : 0; - if (cur == k) { - ++cnt; - cur = 0; + return cnt >= m; + }; + while (l < r) { + int mid = (l + r) >> 1; + if (check(mid)) { + r = mid; + } else { + l = mid + 1; } } - return cnt >= m; + return l > mx ? -1 : l; } }; ``` @@ -204,40 +201,55 @@ public: ```go func minDays(bloomDay []int, m int, k int) int { - if m*k > len(bloomDay) { - return -1 - } - mi, mx := 0, 1000000000 - for _, bd := range bloomDay { - mi = min(mi, bd) - mx = max(mx, bd) - } - left, right := mi, mx - for left < right { - mid := (left + right) >> 1 - if check(bloomDay, m, k, mid) { - right = mid - } else { - left = mid + 1 + mx := slices.Max(bloomDay) + if l := sort.Search(mx+2, func(days int) bool { + cnt, cur := 0, 0 + for _, x := range bloomDay { + if x <= days { + cur++ + if cur == k { + cnt++ + cur = 0 + } + } else { + cur = 0 + } } + return cnt >= m + }); l <= mx { + return l } - return left + return -1 + } +``` -func check(bloomDay []int, m, k, day int) bool { - cnt, cur := 0, 0 - for _, bd := range bloomDay { - if bd <= day { - cur++ - } else { - cur = 0 - } - if cur == k { - cnt++ - cur = 0 - } - } - return cnt >= m +#### TypeScript + +```ts +function minDays(bloomDay: number[], m: number, k: number): number { + const mx = Math.max(...bloomDay); + let [l, r] = [1, mx + 1]; + const check = (days: number): boolean => { + let [cnt, cur] = [0, 0]; + for (const x of bloomDay) { + cur = x <= days ? cur + 1 : 0; + if (cur === k) { + cnt++; + cur = 0; + } + } + return cnt >= m; + }; + while (l < r) { + const mid = (l + r) >> 1; + if (check(mid)) { + r = mid; + } else { + l = mid + 1; + } + } + return l > mx ? -1 : l; } ``` diff --git a/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/README_EN.md b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/README_EN.md index 419d1cbfd33e4..31731f50e9996 100644 --- a/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/README_EN.md +++ b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/README_EN.md @@ -78,7 +78,21 @@ It is obvious that we can make two bouquets in different ways. -### Solution 1 +### Solution 1: Binary Search + +According to the problem description, if a day $t$ can satisfy making $m$ bouquets, then for any $t' > t$, it can also satisfy making $m$ bouquets. Therefore, we can use binary search to find the minimum day that satisfies making $m$ bouquets. + +Let $mx$ be the maximum blooming day in the garden. Next, we define the left boundary of the binary search as $l = 1$ and the right boundary as $r = mx + 1$. + +Then, we perform binary search. For each middle value $\textit{mid} = \frac{l + r}{2}$, we check if it is possible to make $m$ bouquets. If it is possible, we update the right boundary $r$ to $\textit{mid}$; otherwise, we update the left boundary $l$ to $\textit{mid} + 1$. + +Finally, when $l = r$, the binary search ends. At this point, if $l > mx$, it means it is not possible to make $m$ bouquets, and we return $-1$; otherwise, we return $l$. + +Therefore, the problem is reduced to checking if a day $\textit{days}$ can make $m$ bouquets. + +We can use a function $\text{check}(\textit{days})$ to determine if it is possible to make $m$ bouquets. Specifically, we traverse each flower in the garden from left to right. If the blooming day of the current flower is less than or equal to $\textit{days}$, we add the current flower to the current bouquet; otherwise, we clear the current bouquet. When the number of flowers in the current bouquet equals $k$, we increment the bouquet count and clear the current bouquet. Finally, we check if the bouquet count is greater than or equal to $m$. If it is, it means it is possible to make $m$ bouquets; otherwise, it is not possible. + +The time complexity is $O(n \times \log M)$, where $n$ and $M$ are the number of flowers in the garden and the maximum blooming day, respectively. In this problem, $M \leq 10^9$. The space complexity is $O(1)$. @@ -87,59 +101,50 @@ It is obvious that we can make two bouquets in different ways. ```python class Solution: def minDays(self, bloomDay: List[int], m: int, k: int) -> int: - if m * k > len(bloomDay): - return -1 - - def check(day: int) -> bool: + def check(days: int) -> int: cnt = cur = 0 - for bd in bloomDay: - cur = cur + 1 if bd <= day else 0 + for x in bloomDay: + cur = cur + 1 if x <= days else 0 if cur == k: cnt += 1 cur = 0 return cnt >= m - left, right = min(bloomDay), max(bloomDay) - while left < right: - mid = (left + right) >> 1 - if check(mid): - right = mid - else: - left = mid + 1 - return left + mx = max(bloomDay) + l = bisect_left(range(mx + 2), True, key=check) + return -1 if l > mx else l ``` #### Java ```java class Solution { + private int[] bloomDay; + private int m, k; + public int minDays(int[] bloomDay, int m, int k) { - if (m * k > bloomDay.length) { - return -1; - } - int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE; - for (int bd : bloomDay) { - min = Math.min(min, bd); - max = Math.max(max, bd); - } - int left = min, right = max; - while (left < right) { - int mid = (left + right) >>> 1; - if (check(bloomDay, m, k, mid)) { - right = mid; + this.bloomDay = bloomDay; + this.m = m; + this.k = k; + final int mx = (int) 1e9; + int l = 1, r = mx + 1; + while (l < r) { + int mid = (l + r) >> 1; + if (check(mid)) { + r = mid; } else { - left = mid + 1; + l = mid + 1; } } - return left; + return l > mx ? -1 : l; } - private boolean check(int[] bloomDay, int m, int k, int day) { + private boolean check(int days) { int cnt = 0, cur = 0; - for (int bd : bloomDay) { - cur = bd <= day ? cur + 1 : 0; + for (int x : bloomDay) { + cur = x <= days ? cur + 1 : 0; if (cur == k) { - cnt++; + ++cnt; cur = 0; } } @@ -154,36 +159,28 @@ class Solution { class Solution { public: int minDays(vector & bloomDay, int m, int k) { - if (m * k > bloomDay.size()) { - return -1; - } - int mi = INT_MIN, mx = INT_MAX; - for (int& bd : bloomDay) { - mi = min(mi, bd); - mx = max(mx, bd); - } - int left = mi, right = mx; - while (left < right) { - int mid = left + right >> 1; - if (check(bloomDay, m, k, mid)) { - right = mid; - } else { - left = mid + 1; + int mx = ranges::max(bloomDay); + int l = 1, r = mx + 1; + auto check = [&](int days) { + int cnt = 0, cur = 0; + for (int x : bloomDay) { + cur = x <= days ? cur + 1 : 0; + if (cur == k) { + cnt++; + cur = 0; + } } - } - return left; - } - - bool check(vector & bloomDay, int m, int k, int day) { - int cnt = 0, cur = 0; - for (int& bd : bloomDay) { - cur = bd <= day ? cur + 1 : 0; - if (cur == k) { - ++cnt; - cur = 0; + return cnt >= m; + }; + while (l < r) { + int mid = (l + r) >> 1; + if (check(mid)) { + r = mid; + } else { + l = mid + 1; } } - return cnt >= m; + return l > mx ? -1 : l; } }; ``` @@ -192,40 +189,55 @@ public: ```go func minDays(bloomDay []int, m int, k int) int { - if m*k > len(bloomDay) { - return -1 - } - mi, mx := 0, 1000000000 - for _, bd := range bloomDay { - mi = min(mi, bd) - mx = max(mx, bd) - } - left, right := mi, mx - for left < right { - mid := (left + right) >> 1 - if check(bloomDay, m, k, mid) { - right = mid - } else { - left = mid + 1 + mx := slices.Max(bloomDay) + if l := sort.Search(mx+2, func(days int) bool { + cnt, cur := 0, 0 + for _, x := range bloomDay { + if x <= days { + cur++ + if cur == k { + cnt++ + cur = 0 + } + } else { + cur = 0 + } } + return cnt >= m + }); l <= mx { + return l } - return left + return -1 + } +``` -func check(bloomDay []int, m, k, day int) bool { - cnt, cur := 0, 0 - for _, bd := range bloomDay { - if bd <= day { - cur++ - } else { - cur = 0 - } - if cur == k { - cnt++ - cur = 0 - } - } - return cnt >= m +#### TypeScript + +```ts +function minDays(bloomDay: number[], m: number, k: number): number { + const mx = Math.max(...bloomDay); + let [l, r] = [1, mx + 1]; + const check = (days: number): boolean => { + let [cnt, cur] = [0, 0]; + for (const x of bloomDay) { + cur = x <= days ? cur + 1 : 0; + if (cur === k) { + cnt++; + cur = 0; + } + } + return cnt >= m; + }; + while (l < r) { + const mid = (l + r) >> 1; + if (check(mid)) { + r = mid; + } else { + l = mid + 1; + } + } + return l > mx ? -1 : l; } ``` diff --git a/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.cpp b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.cpp index de3424858b0ed..fe93831e60858 100644 --- a/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.cpp +++ b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.cpp @@ -1,35 +1,27 @@ class Solution { public: int minDays(vector & bloomDay, int m, int k) { - if (m * k > bloomDay.size()) { - return -1; - } - int mi = INT_MIN, mx = INT_MAX; - for (int& bd : bloomDay) { - mi = min(mi, bd); - mx = max(mx, bd); - } - int left = mi, right = mx; - while (left < right) { - int mid = left + right >> 1; - if (check(bloomDay, m, k, mid)) { - right = mid; - } else { - left = mid + 1; + int mx = ranges::max(bloomDay); + int l = 1, r = mx + 1; + auto check = [&](int days) { + int cnt = 0, cur = 0; + for (int x : bloomDay) { + cur = x <= days ? cur + 1 : 0; + if (cur == k) { + cnt++; + cur = 0; + } } - } - return left; - } - - bool check(vector & bloomDay, int m, int k, int day) { - int cnt = 0, cur = 0; - for (int& bd : bloomDay) { - cur = bd <= day ? cur + 1 : 0; - if (cur == k) { - ++cnt; - cur = 0; + return cnt >= m; + }; + while (l < r) { + int mid = (l + r) >> 1; + if (check(mid)) { + r = mid; + } else { + l = mid + 1; } } - return cnt >= m; + return l > mx ? -1 : l; } -}; \ No newline at end of file +}; diff --git a/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.go b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.go index c9986ab44b997..83d8734ed83db 100644 --- a/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.go +++ b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.go @@ -1,36 +1,22 @@ func minDays(bloomDay []int, m int, k int) int { - if m*k > len(bloomDay) { - return -1 - } - mi, mx := 0, 1000000000 - for _, bd := range bloomDay { - mi = min(mi, bd) - mx = max(mx, bd) - } - left, right := mi, mx - for left < right { - mid := (left + right) >> 1 - if check(bloomDay, m, k, mid) { - right = mid - } else { - left = mid + 1 + mx := slices.Max(bloomDay) + if l := sort.Search(mx+2, func(days int) bool { + cnt, cur := 0, 0 + for _, x := range bloomDay { + if x <= days { + cur++ + if cur == k { + cnt++ + cur = 0 + } + } else { + cur = 0 + } } + return cnt >= m + }); l <= mx { + return l } - return left -} + return -1 -func check(bloomDay []int, m, k, day int) bool { - cnt, cur := 0, 0 - for _, bd := range bloomDay { - if bd <= day { - cur++ - } else { - cur = 0 - } - if cur == k { - cnt++ - cur = 0 - } - } - return cnt >= m -} \ No newline at end of file +} diff --git a/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.java b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.java index ea962af7212e6..04a16e8fe01ee 100644 --- a/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.java +++ b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.java @@ -1,34 +1,33 @@ class Solution { + private int[] bloomDay; + private int m, k; + public int minDays(int[] bloomDay, int m, int k) { - if (m * k > bloomDay.length) { - return -1; - } - int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE; - for (int bd : bloomDay) { - min = Math.min(min, bd); - max = Math.max(max, bd); - } - int left = min, right = max; - while (left < right) { - int mid = (left + right) >>> 1; - if (check(bloomDay, m, k, mid)) { - right = mid; + this.bloomDay = bloomDay; + this.m = m; + this.k = k; + final int mx = (int) 1e9; + int l = 1, r = mx + 1; + while (l < r) { + int mid = (l + r) >> 1; + if (check(mid)) { + r = mid; } else { - left = mid + 1; + l = mid + 1; } } - return left; + return l > mx ? -1 : l; } - private boolean check(int[] bloomDay, int m, int k, int day) { + private boolean check(int days) { int cnt = 0, cur = 0; - for (int bd : bloomDay) { - cur = bd <= day ? cur + 1 : 0; + for (int x : bloomDay) { + cur = x <= days ? cur + 1 : 0; if (cur == k) { - cnt++; + ++cnt; cur = 0; } } return cnt >= m; } -} \ No newline at end of file +} diff --git a/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.py b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.py index dbfd1a53635a4..5475b146aa761 100644 --- a/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.py +++ b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.py @@ -1,22 +1,14 @@ class Solution: def minDays(self, bloomDay: List[int], m: int, k: int) -> int: - if m * k > len(bloomDay): - return -1 - - def check(day: int) -> bool: + def check(days: int) -> int: cnt = cur = 0 - for bd in bloomDay: - cur = cur + 1 if bd <= day else 0 + for x in bloomDay: + cur = cur + 1 if x <= days else 0 if cur == k: cnt += 1 cur = 0 return cnt >= m - left, right = min(bloomDay), max(bloomDay) - while left < right: - mid = (left + right) >> 1 - if check(mid): - right = mid - else: - left = mid + 1 - return left + mx = max(bloomDay) + l = bisect_left(range(mx + 2), True, key=check) + return -1 if l > mx else l diff --git a/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.ts b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.ts new file mode 100644 index 0000000000000..1ad67a6078e88 --- /dev/null +++ b/solution/1400-1499/1482.Minimum Number of Days to Make m Bouquets/Solution.ts @@ -0,0 +1,24 @@ +function minDays(bloomDay: number[], m: number, k: number): number { + const mx = Math.max(...bloomDay); + let [l, r] = [1, mx + 1]; + const check = (days: number): boolean => { + let [cnt, cur] = [0, 0]; + for (const x of bloomDay) { + cur = x <= days ? cur + 1 : 0; + if (cur === k) { + cnt++; + cur = 0; + } + } + return cnt >= m; + }; + while (l < r) { + const mid = (l + r) >> 1; + if (check(mid)) { + r = mid; + } else { + l = mid + 1; + } + } + return l > mx ? -1 : l; +} diff --git a/solution/1800-1899/1870.Minimum Speed to Arrive on Time/README.md b/solution/1800-1899/1870.Minimum Speed to Arrive on Time/README.md index f461cf8c542c6..2b8aaf021084b 100644 --- a/solution/1800-1899/1870.Minimum Speed to Arrive on Time/README.md +++ b/solution/1800-1899/1870.Minimum Speed to Arrive on Time/README.md @@ -27,11 +27,11 @@ tags: 例如，第 -1趟列车需要1.5小时，那你必须再等待0.5小时，搭乘在第 2 小时发车的第2趟列车。返回能满足你准时到达办公室所要求全部列车的 最小正整数 时速（单位：千米每小时），如果无法准时到达，则返回
+-1。返回能满足你在时限前到达办公室所要求全部列车的 最小正整数 时速（单位：千米每小时），如果无法准时到达，则返回
-1。生成的测试用例保证答案不超过
-107，且hour的 小数点后最多存在两位数字 。+
示例 1：
@@ -63,15 +63,15 @@ tags: 输出：-1 解释：不可能准时到达，因为第 3 趟列车最早是在第 2 小时发车。+
提示：
diff --git a/solution/2600-2699/2694.Event Emitter/README_EN.md b/solution/2600-2699/2694.Event Emitter/README_EN.md index e7c28c1bf15bf..6759594507da2 100644 --- a/solution/2600-2699/2694.Event Emitter/README_EN.md +++ b/solution/2600-2699/2694.Event Emitter/README_EN.md @@ -33,7 +33,7 @@ tags:
- -
n == dist.length- -
1 <= n <= 105- -
1 <= dist[i] <= 105- +
1 <= hour <= 109- +
1 <= n <= 105- +
1 <= dist[i] <= 1051 <= hour <= 109hours中，小数点后最多存在两位数字Input: actions = ["EventEmitter", "emit", "subscribe", "subscribe", "emit"], -values = [[], ["firstEvent", "function cb1() { return 5; }"], ["firstEvent", "function cb1() { return 6; }"], ["firstEvent"]] +values = [[], ["firstEvent"], ["firstEvent", "function cb1() { return 5; }"], ["firstEvent", "function cb1() { return 6; }"], ["firstEvent"]] Output: [[],["emitted",[]],["subscribed"],["subscribed"],["emitted",[5,6]]] Explanation: const emitter = new EventEmitter(); diff --git a/solution/3100-3199/3126.Server Utilization Time/README.md b/solution/3100-3199/3126.Server Utilization Time/README.md index 814b0577b4e33..1fe601412550e 100644 --- a/solution/3100-3199/3126.Server Utilization Time/README.md +++ b/solution/3100-3199/3126.Server Utilization Time/README.md @@ -31,7 +31,7 @@ session_status 是 ('start', 'stop') 的 ENUM (category)。 这张表的每一行包含 server_id, status_time 和 session_status。-编写一个解决方案来查找服务器 运行 的 总时间。输出应四舍五入为最接近的 整天数。
+编写一个解决方案来查找服务器 运行 的 总时间。输出应向下舍入为最接近的 整天数。
以 任意 顺序返回结果表。
diff --git a/solution/3200-3299/3279.Maximum Total Area Occupied by Pistons/README.md b/solution/3200-3299/3279.Maximum Total Area Occupied by Pistons/README.md index 6d82924e44957..b0de44bc3caf5 100644 --- a/solution/3200-3299/3279.Maximum Total Area Occupied by Pistons/README.md +++ b/solution/3200-3299/3279.Maximum Total Area Occupied by Pistons/README.md @@ -13,7 +13,7 @@ tags: -# [3279. 活塞占据的最大总面积 🔒](https://leetcode.cn/problems/maximum-total-area-occupied-by-pistons) +# [3279. 活塞占据的最大总区域 🔒](https://leetcode.cn/problems/maximum-total-area-occupied-by-pistons) [English Version](/solution/3200-3299/3279.Maximum%20Total%20Area%20Occupied%20by%20Pistons/README_EN.md) @@ -21,13 +21,13 @@ tags: -一台旧车的引擎中有一些活塞，我们想要计算活塞 下方 的 最大 面积。
+一台旧车的引擎中有一些活塞，我们想要计算活塞 下方 的 最大 区域。
给定：
@@ -38,7 +38,7 @@ tags:
- 一个整数
-height，表示活塞 最大 可到达的高度。- 一个整数数组
+positions，其中positions[i]是活塞i的当前位置，等于其 下方 的当前面积。- 一个整数数组
positions，其中positions[i]是活塞i的当前位置，等于其 下方 的当前区域。- 一个字符串
directions，其中directions[i]是活塞i的当前移动方向，'U'表示向上，'D'表示向下。如果一个活塞到达了其中一个终点，即 -positions[i] == 0或positions[i] == height，它的方向将会改变。返回所有活塞下方的最大可能面积。
+返回所有活塞下方的最大可能区域。
@@ -51,7 +51,7 @@ tags:
解释：
-当前活塞的位置下方面积最大。
+当前活塞的位置下方区域最大。
示例 2：
@@ -63,7 +63,7 @@ tags:解释：
-三秒后，活塞将会位于
+[3, 3, 3, 6]，此时下方面积最大。三秒后，活塞将会位于
[3, 3, 3, 6]，此时下方区域最大。diff --git a/solution/3200-3299/3284.Sum of Consecutive Subarrays/README.md b/solution/3200-3299/3284.Sum of Consecutive Subarrays/README.md index 102998410f492..70a4882133bec 100644 --- a/solution/3200-3299/3284.Sum of Consecutive Subarrays/README.md +++ b/solution/3200-3299/3284.Sum of Consecutive Subarrays/README.md @@ -18,7 +18,7 @@ tags: -
如果一个长度为
+n的数组arr符合下面条件，可以称它 连续：如果一个长度为
n的数组arr符合下面其中一个条件，可以称它 连续：
- 对于所有的
@@ -27,7 +27,7 @@ tags:1 <= i < n，arr[i] - arr[i - 1] == 1。数组的 值 是其元素的和。
-例如，
+[3, 4, 5]是一个值为 12 的连续数组，并且[9, 8]是另一个值为 17 的子数组。而[3, 4, 3]和[8, 6]都不连续。例如，
[3, 4, 5]是一个值为 12 的连续数组，并且[9, 8]是另一个值为 17 的连续数组。而[3, 4, 3]和[8, 6]都不连续。给定一个整数数组
diff --git a/solution/3200-3299/3299.Sum of Consecutive Subsequences/README.md b/solution/3200-3299/3299.Sum of Consecutive Subsequences/README.md index ef1a94f109d51..c49a65dd8e684 100644 --- a/solution/3200-3299/3299.Sum of Consecutive Subsequences/README.md +++ b/solution/3200-3299/3299.Sum of Consecutive Subsequences/README.md @@ -18,7 +18,7 @@ tags: -nums，返回所有 连续 子数组 的 值 之和。如果一个长度为
+n的数组arr符合下面条件，可以称它 连续：如果一个长度为
n的数组arr符合下面其中一个条件，可以称它 连续：
- 对于所有的
@@ -27,7 +27,7 @@ tags:1 <= i < n，arr[i] - arr[i - 1] == 1。数组的 值 是其元素的和。
-例如，
+[3, 4, 5]是一个值为 12 的连续数组，并且[9, 8]是另一个值为 17 的子数组。而[3, 4, 3]和[8, 6]都不连续。例如，
[3, 4, 5]是一个值为 12 的连续数组，并且[9, 8]是另一个值为 17 的连续数组。而[3, 4, 3]和[8, 6]都不连续。给定一个整数数组
diff --git a/solution/3300-3399/3303.Find the Occurrence of First Almost Equal Substring/README.md b/solution/3300-3399/3303.Find the Occurrence of First Almost Equal Substring/README.md index 062013d915c69..65a8e2ca1843f 100644 --- a/solution/3300-3399/3303.Find the Occurrence of First Almost Equal Substring/README.md +++ b/solution/3300-3399/3303.Find the Occurrence of First Almost Equal Substring/README.md @@ -73,7 +73,7 @@ tags:nums，返回所有 连续 非空 子序列 的 值 之和。提示：
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diff --git a/solution/README.md b/solution/README.md index 56eec3a709d7f..860caf1e6e16f 100644 --- a/solution/README.md +++ b/solution/README.md @@ -3289,7 +3289,7 @@ | 3276 | [选择矩阵中单元格的最大得分](/solution/3200-3299/3276.Select%20Cells%20in%20Grid%20With%20Maximum%20Score/README.md) | `位运算`,`数组`,`动态规划`,`状态压缩`,`矩阵` | 困难 | 第 413 场周赛 | | 3277 | [查询子数组最大异或值](/solution/3200-3299/3277.Maximum%20XOR%20Score%20Subarray%20Queries/README.md) | `数组`,`动态规划` | 困难 | 第 413 场周赛 | | 3278 | [寻找数据科学家职位的候选人 II](/solution/3200-3299/3278.Find%20Candidates%20for%20Data%20Scientist%20Position%20II/README.md) | `数据库` | 中等 | 🔒 | -| 3279 | [活塞占据的最大总面积](/solution/3200-3299/3279.Maximum%20Total%20Area%20Occupied%20by%20Pistons/README.md) | `数组`,`哈希表`,`字符串`,`计数`,`前缀和`,`模拟` | 困难 | 🔒 | +| 3279 | [活塞占据的最大总区域](/solution/3200-3299/3279.Maximum%20Total%20Area%20Occupied%20by%20Pistons/README.md) | `数组`,`哈希表`,`字符串`,`计数`,`前缀和`,`模拟` | 困难 | 🔒 | | 3280 | [将日期转换为二进制表示](/solution/3200-3299/3280.Convert%20Date%20to%20Binary/README.md) | `数学`,`字符串` | 简单 | 第 414 场周赛 | | 3281 | [范围内整数的最大得分](/solution/3200-3299/3281.Maximize%20Score%20of%20Numbers%20in%20Ranges/README.md) | `贪心`,`数组`,`二分查找`,`排序` | 中等 | 第 414 场周赛 | | 3282 | [到达数组末尾的最大得分](/solution/3200-3299/3282.Reach%20End%20of%20Array%20With%20Max%20Score/README.md) | `贪心`,`数组` | 中等 | 第 414 场周赛 |- +
1 <= pattern.length < s.length <= 3 * 1051 <= pattern.length < s.length <= 105s和pattern都只包含小写英文字母。