diff --git a/solution/2200-2299/2226.Maximum Candies Allocated to K Children/README.md b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/README.md index eda1ef547b603..3f7b37b6d99be 100644 --- a/solution/2200-2299/2226.Maximum Candies Allocated to K Children/README.md +++ b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/README.md @@ -61,7 +61,11 @@ tags: ### 方法一：二分查找 -时间复杂度 $O(nlogn)$。 +我们注意到，如果每个小孩能分到糖果数 $v$，那么对于任意 $v' \lt v$，每个小孩也能分到 $v'$ 颗糖果。因此，我们可以使用二分查找的方法找到最大的 $v$，使得每个小孩能分到 $v$ 颗糖果。 + +我们定义二分查找的左边界 $l = 0$，右边界 $r = \max(\text{candies})$，其中 $\max(\text{candies})$ 表示数组 $\text{candies}$ 中的最大值。在二分查找的过程中，我们每次取 $v$ 的中间值 $v = \left\lfloor \frac{l + r + 1}{2} \right\rfloor$，然后计算每个小孩能分到的糖果数 $v$ 的总和，如果总和大于等于 $k$，则说明每个小孩能分到 $v$ 颗糖果，此时我们更新左边界 $l = v$，否则我们更新右边界 $r = v - 1$。最终，当 $l = r$ 时，我们找到了最大的 $v$。 + +时间复杂度 $O(n \times \log M)$，其中 $n$ 表示数组 $\text{candies}$ 的长度，而 $M$ 表示数组 $\text{candies}$ 中的最大值。空间复杂度 $O(1)$。 @@ -70,15 +74,14 @@ tags: ```python class Solution: def maximumCandies(self, candies: List[int], k: int) -> int: - left, right = 0, max(candies) - while left < right: - mid = (left + right + 1) >> 1 - cnt = sum(v // mid for v in candies) - if cnt >= k: - left = mid + l, r = 0, max(candies) + while l < r: + mid = (l + r + 1) >> 1 + if sum(x // mid for x in candies) >= k: + l = mid else: - right = mid - 1 - return left + r = mid - 1 + return l ``` #### Java @@ -86,20 +89,20 @@ class Solution: ```java class Solution { public int maximumCandies(int[] candies, long k) { - int left = 0, right = (int) 1e7; - while (left < right) { - int mid = (left + right + 1) >> 1; + int l = 0, r = Arrays.stream(candies).max().getAsInt(); + while (l < r) { + int mid = (l + r + 1) >> 1; long cnt = 0; - for (int v : candies) { - cnt += v / mid; + for (int x : candies) { + cnt += x / mid; } if (cnt >= k) { - left = mid; + l = mid; } else { - right = mid - 1; + r = mid - 1; } } - return left; + return l; } } ``` @@ -110,17 +113,20 @@ class Solution { class Solution { public: int maximumCandies(vector& candies, long long k) { - int left = 0, right = 1e7; - while (left < right) { - int mid = (left + right + 1) >> 1; + int l = 0, r = ranges::max(candies); + while (l < r) { + int mid = (l + r + 1) >> 1; long long cnt = 0; - for (int& v : candies) cnt += v / mid; - if (cnt >= k) - left = mid; - else - right = mid - 1; + for (int x : candies) { + cnt += x / mid; + } + if (cnt >= k) { + l = mid; + } else { + r = mid - 1; + } } - return left; + return l; } }; ``` @@ -129,20 +135,32 @@ public: ```go func maximumCandies(candies []int, k int64) int { - left, right := 0, int(1e7) - for left < right { - mid := (left + right + 1) >> 1 + return sort.Search(1e7, func(v int) bool { + v++ var cnt int64 - for _, v := range candies { - cnt += int64(v / mid) + for _, x := range candies { + cnt += int64(x / v) } - if cnt >= k { - left = mid - } else { - right = mid - 1 - } - } - return left + return cnt < k + }) +} +``` + +#### TypeScript + +```ts +function maximumCandies(candies: number[], k: number): number { + let [l, r] = [0, Math.max(...candies)]; + while (l < r) { + const mid = (l + r + 1) >> 1; + const cnt = candies.reduce((acc, cur) => acc + Math.floor(cur / mid), 0); + if (cnt >= k) { + l = mid; + } else { + r = mid - 1; + } + } + return l; } ``` diff --git a/solution/2200-2299/2226.Maximum Candies Allocated to K Children/README_EN.md b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/README_EN.md index 1e549fd12040b..ba3519f8c761b 100644 --- a/solution/2200-2299/2226.Maximum Candies Allocated to K Children/README_EN.md +++ b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/README_EN.md @@ -56,7 +56,13 @@ tags: -### Solution 1 +### Solution 1: Binary Search + +We notice that if each child can receive $v$ candies, then for any $v' \lt v$, each child can also receive $v'$ candies. Therefore, we can use binary search to find the maximum $v$ such that each child can receive $v$ candies. + +We define the left boundary of the binary search as $l = 0$ and the right boundary as $r = \max(\text{candies})$, where $\max(\text{candies})$ represents the maximum value in the array $\text{candies}$. During the binary search, we take the middle value $v = \left\lfloor \frac{l + r + 1}{2} \right\rfloor$ each time, and then calculate the total number of candies each child can receive. If the total is greater than or equal to $k$, it means each child can receive $v$ candies, so we update the left boundary $l = v$. Otherwise, we update the right boundary $r = v - 1$. Finally, when $l = r$, we have found the maximum $v$. + +The time complexity is $O(n \times \log M)$, where $n$ is the length of the array $\text{candies}$, and $M$ is the maximum value in the array $\text{candies}$. The space complexity is $O(1)$. @@ -65,15 +71,14 @@ tags: ```python class Solution: def maximumCandies(self, candies: List[int], k: int) -> int: - left, right = 0, max(candies) - while left < right: - mid = (left + right + 1) >> 1 - cnt = sum(v // mid for v in candies) - if cnt >= k: - left = mid + l, r = 0, max(candies) + while l < r: + mid = (l + r + 1) >> 1 + if sum(x // mid for x in candies) >= k: + l = mid else: - right = mid - 1 - return left + r = mid - 1 + return l ``` #### Java @@ -81,20 +86,20 @@ class Solution: ```java class Solution { public int maximumCandies(int[] candies, long k) { - int left = 0, right = (int) 1e7; - while (left < right) { - int mid = (left + right + 1) >> 1; + int l = 0, r = Arrays.stream(candies).max().getAsInt(); + while (l < r) { + int mid = (l + r + 1) >> 1; long cnt = 0; - for (int v : candies) { - cnt += v / mid; + for (int x : candies) { + cnt += x / mid; } if (cnt >= k) { - left = mid; + l = mid; } else { - right = mid - 1; + r = mid - 1; } } - return left; + return l; } } ``` @@ -105,17 +110,20 @@ class Solution { class Solution { public: int maximumCandies(vector& candies, long long k) { - int left = 0, right = 1e7; - while (left < right) { - int mid = (left + right + 1) >> 1; + int l = 0, r = ranges::max(candies); + while (l < r) { + int mid = (l + r + 1) >> 1; long long cnt = 0; - for (int& v : candies) cnt += v / mid; - if (cnt >= k) - left = mid; - else - right = mid - 1; + for (int x : candies) { + cnt += x / mid; + } + if (cnt >= k) { + l = mid; + } else { + r = mid - 1; + } } - return left; + return l; } }; ``` @@ -124,20 +132,32 @@ public: ```go func maximumCandies(candies []int, k int64) int { - left, right := 0, int(1e7) - for left < right { - mid := (left + right + 1) >> 1 + return sort.Search(1e7, func(v int) bool { + v++ var cnt int64 - for _, v := range candies { - cnt += int64(v / mid) + for _, x := range candies { + cnt += int64(x / v) } - if cnt >= k { - left = mid - } else { - right = mid - 1 - } - } - return left + return cnt < k + }) +} +``` + +#### TypeScript + +```ts +function maximumCandies(candies: number[], k: number): number { + let [l, r] = [0, Math.max(...candies)]; + while (l < r) { + const mid = (l + r + 1) >> 1; + const cnt = candies.reduce((acc, cur) => acc + Math.floor(cur / mid), 0); + if (cnt >= k) { + l = mid; + } else { + r = mid - 1; + } + } + return l; } ``` diff --git a/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.cpp b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.cpp index 097e6764f6518..86113d56f842e 100644 --- a/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.cpp +++ b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.cpp @@ -1,16 +1,19 @@ class Solution { public: int maximumCandies(vector& candies, long long k) { - int left = 0, right = 1e7; - while (left < right) { - int mid = (left + right + 1) >> 1; + int l = 0, r = ranges::max(candies); + while (l < r) { + int mid = (l + r + 1) >> 1; long long cnt = 0; - for (int& v : candies) cnt += v / mid; - if (cnt >= k) - left = mid; - else - right = mid - 1; + for (int x : candies) { + cnt += x / mid; + } + if (cnt >= k) { + l = mid; + } else { + r = mid - 1; + } } - return left; + return l; } -}; \ No newline at end of file +}; diff --git a/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.go b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.go index 5831b400a7166..800625e446274 100644 --- a/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.go +++ b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.go @@ -1,16 +1,10 @@ func maximumCandies(candies []int, k int64) int { - left, right := 0, int(1e7) - for left < right { - mid := (left + right + 1) >> 1 + return sort.Search(1e7, func(v int) bool { + v++ var cnt int64 - for _, v := range candies { - cnt += int64(v / mid) + for _, x := range candies { + cnt += int64(x / v) } - if cnt >= k { - left = mid - } else { - right = mid - 1 - } - } - return left -} \ No newline at end of file + return cnt < k + }) +} diff --git a/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.java b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.java index 02e14101bfe4b..c1f20f0e3aa57 100644 --- a/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.java +++ b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.java @@ -1,18 +1,18 @@ class Solution { public int maximumCandies(int[] candies, long k) { - int left = 0, right = (int) 1e7; - while (left < right) { - int mid = (left + right + 1) >> 1; + int l = 0, r = Arrays.stream(candies).max().getAsInt(); + while (l < r) { + int mid = (l + r + 1) >> 1; long cnt = 0; - for (int v : candies) { - cnt += v / mid; + for (int x : candies) { + cnt += x / mid; } if (cnt >= k) { - left = mid; + l = mid; } else { - right = mid - 1; + r = mid - 1; } } - return left; + return l; } -} \ No newline at end of file +} diff --git a/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.py b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.py index c526a09667a8f..efb97b20af52d 100644 --- a/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.py +++ b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.py @@ -1,11 +1,10 @@ class Solution: def maximumCandies(self, candies: List[int], k: int) -> int: - left, right = 0, max(candies) - while left < right: - mid = (left + right + 1) >> 1 - cnt = sum(v // mid for v in candies) - if cnt >= k: - left = mid + l, r = 0, max(candies) + while l < r: + mid = (l + r + 1) >> 1 + if sum(x // mid for x in candies) >= k: + l = mid else: - right = mid - 1 - return left + r = mid - 1 + return l diff --git a/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.ts b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.ts new file mode 100644 index 0000000000000..33000ce6584e2 --- /dev/null +++ b/solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.ts @@ -0,0 +1,13 @@ +function maximumCandies(candies: number[], k: number): number { + let [l, r] = [0, Math.max(...candies)]; + while (l < r) { + const mid = (l + r + 1) >> 1; + const cnt = candies.reduce((acc, cur) => acc + Math.floor(cur / mid), 0); + if (cnt >= k) { + l = mid; + } else { + r = mid - 1; + } + } + return l; +}