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feat: add solutions to lc problem: No.2226 #3759

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Nov 13, 2024
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feat: add solutions to lc problem: No.2226 #3759

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94 changes: 56 additions & 38 deletions solution/2200-2299/2226.Maximum Candies Allocated to K Children/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -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 $O(1)$。

<!-- tabs:start -->

Expand All @@ -70,36 +74,35 @@ 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

```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;
}
}
```
Expand All @@ -110,17 +113,20 @@ class Solution {
class Solution {
public:
int maximumCandies(vector<int>& 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;
}
};
```
Expand All @@ -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;
}
```

Expand Down
96 changes: 58 additions & 38 deletions solution/2200-2299/2226.Maximum Candies Allocated to K Children/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -56,7 +56,13 @@ tags:

<!-- solution:start -->

### 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 $O(1)$.

<!-- tabs:start -->

Expand All @@ -65,36 +71,35 @@ 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

```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;
}
}
```
Expand All @@ -105,17 +110,20 @@ class Solution {
class Solution {
public:
int maximumCandies(vector<int>& 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;
}
};
```
Expand All @@ -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;
}
```

Expand Down
23 changes: 13 additions & 10 deletions solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,16 +1,19 @@
class Solution {
public:
int maximumCandies(vector<int>& 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;
}
};
};
20 changes: 7 additions & 13 deletions solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.go
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Original file line number Diff line number Diff line change
@@ -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
}
return cnt < k
})
}
18 changes: 9 additions & 9 deletions solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.java
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Original file line number Diff line number Diff line change
@@ -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;
}
}
}
15 changes: 7 additions & 8 deletions solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.py
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Original file line number Diff line number Diff line change
@@ -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
13 changes: 13 additions & 0 deletions solution/2200-2299/2226.Maximum Candies Allocated to K Children/Solution.ts
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Original file line number Diff line number Diff line change
@@ -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;
}
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