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feat: add solutions to lc problem: No.3477 (#4139)
No.3477.Fruits Into Baskets II
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‎solution/3400-3499/3477.Fruits Into Baskets II/README.md

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Original file line numberDiff line numberDiff line change
@@ -80,32 +80,118 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3477.Fr
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<!-- solution:start -->
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### 方法一
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### 方法一：模拟
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我们用一个长度为 $n$ 的布尔数组 $\textit{vis}$ 记录已经被使用的篮子，用一个答案变量 $\textit{ans}$ 记录所有未被放置的水果，初始时 $\textit{ans} = n$。
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接下来，我们遍历每一种水果 $x$，对于当前水果，我们遍历所有的篮子，找出第一个未被使用，且容量大于等于 $x$ 的篮子 $i$。如果找到了，那么答案 $\textit{ans}$ 减 $1$。
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遍历结束后，返回答案即可。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $\textit{fruits}$ 的长度。
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def numOfUnplacedFruits(self, fruits: List[int], baskets: List[int]) -> int:
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n = len(fruits)
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vis = [False] * n
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ans = n
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for x in fruits:
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for i, y in enumerate(baskets):
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if y >= x and not vis[i]:
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vis[i] = True
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ans -= 1
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break
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return ans
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```
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#### Java
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```java
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class Solution {
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public int numOfUnplacedFruits(int[] fruits, int[] baskets) {
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int n = fruits.length;
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boolean[] vis = new boolean[n];
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int ans = n;
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for (int x : fruits) {
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for (int i = 0; i < n; ++i) {
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if (baskets[i] >= x && !vis[i]) {
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vis[i] = true;
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--ans;
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break;
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}
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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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#### C++
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```cpp
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class Solution {
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public:
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int numOfUnplacedFruits(vector<int>& fruits, vector<int>& baskets) {
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int n = fruits.size();
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vector<bool> vis(n);
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int ans = n;
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for (int x : fruits) {
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for (int i = 0; i < n; ++i) {
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if (baskets[i] >= x && !vis[i]) {
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vis[i] = true;
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--ans;
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break;
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}
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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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#### Go
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```go
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func numOfUnplacedFruits(fruits []int, baskets []int) int {
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n := len(fruits)
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ans := n
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vis := make([]bool, n)
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for _, x := range fruits {
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for i, y := range baskets {
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if y >= x && !vis[i] {
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vis[i] = true
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ans--
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break
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}
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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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#### TypeScript
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```ts
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function numOfUnplacedFruits(fruits: number[], baskets: number[]): number {
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const n = fruits.length;
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const vis: boolean[] = Array(n).fill(false);
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let ans = n;
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for (const x of fruits) {
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for (let i = 0; i < n; ++i) {
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if (baskets[i] >= x && !vis[i]) {
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vis[i] = true;
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--ans;
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break;
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}
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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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<!-- tabs:end -->

‎solution/3400-3499/3477.Fruits Into Baskets II/README_EN.md

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@@ -78,32 +78,118 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3400-3499/3477.Fr
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Simulation
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We use a boolean array $\textit{vis}$ of length $n$ to record the baskets that have already been used, and a variable $\textit{ans}$ to record the number of fruits that have not been placed, initially $\textit{ans} = n$.
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Next, we traverse each fruit $x$. For the current fruit, we traverse all the baskets to find the first unused basket $i$ with a capacity greater than or equal to $x$. If found, we decrement $\textit{ans}$ by $1$.
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After traversing, we return the answer.
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The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{fruits}$.
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<!-- tabs:start -->
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#### Python3
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```python
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class Solution:
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def numOfUnplacedFruits(self, fruits: List[int], baskets: List[int]) -> int:
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n = len(fruits)
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vis = [False] * n
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ans = n
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for x in fruits:
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for i, y in enumerate(baskets):
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if y >= x and not vis[i]:
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vis[i] = True
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ans -= 1
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break
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return ans
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```
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#### Java
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```java
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class Solution {
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public int numOfUnplacedFruits(int[] fruits, int[] baskets) {
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int n = fruits.length;
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boolean[] vis = new boolean[n];
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int ans = n;
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for (int x : fruits) {
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for (int i = 0; i < n; ++i) {
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if (baskets[i] >= x && !vis[i]) {
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vis[i] = true;
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--ans;
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break;
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}
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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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#### C++
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```cpp
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class Solution {
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public:
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int numOfUnplacedFruits(vector<int>& fruits, vector<int>& baskets) {
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int n = fruits.size();
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vector<bool> vis(n);
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int ans = n;
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for (int x : fruits) {
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for (int i = 0; i < n; ++i) {
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if (baskets[i] >= x && !vis[i]) {
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vis[i] = true;
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--ans;
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break;
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}
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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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#### Go
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```go
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func numOfUnplacedFruits(fruits []int, baskets []int) int {
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n := len(fruits)
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ans := n
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vis := make([]bool, n)
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for _, x := range fruits {
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for i, y := range baskets {
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if y >= x && !vis[i] {
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vis[i] = true
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ans--
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break
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}
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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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#### TypeScript
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```ts
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function numOfUnplacedFruits(fruits: number[], baskets: number[]): number {
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const n = fruits.length;
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const vis: boolean[] = Array(n).fill(false);
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let ans = n;
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for (const x of fruits) {
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for (let i = 0; i < n; ++i) {
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if (baskets[i] >= x && !vis[i]) {
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vis[i] = true;
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--ans;
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break;
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}
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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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<!-- tabs:end -->

‎solution/3400-3499/3477.Fruits Into Baskets II/Solution.cpp

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class Solution {
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public:
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int numOfUnplacedFruits(vector<int>& fruits, vector<int>& baskets) {
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int n = fruits.size();
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vector<bool> vis(n);
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int ans = n;
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for (int x : fruits) {
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for (int i = 0; i < n; ++i) {
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if (baskets[i] >= x && !vis[i]) {
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vis[i] = true;
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--ans;
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break;
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}
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}
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}
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return ans;
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}
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};

‎solution/3400-3499/3477.Fruits Into Baskets II/Solution.go

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func numOfUnplacedFruits(fruits []int, baskets []int) int {
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n := len(fruits)
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ans := n
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vis := make([]bool, n)
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for _, x := range fruits {
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for i, y := range baskets {
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if y >= x && !vis[i] {
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vis[i] = true
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ans--
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break
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}
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}
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}
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return ans
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}

‎solution/3400-3499/3477.Fruits Into Baskets II/Solution.java

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class Solution {
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public int numOfUnplacedFruits(int[] fruits, int[] baskets) {
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int n = fruits.length;
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boolean[] vis = new boolean[n];
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int ans = n;
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for (int x : fruits) {
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for (int i = 0; i < n; ++i) {
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if (baskets[i] >= x && !vis[i]) {
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vis[i] = true;
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--ans;
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break;
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}
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}
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}
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return ans;
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}
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}

‎solution/3400-3499/3477.Fruits Into Baskets II/Solution.py

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class Solution:
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def numOfUnplacedFruits(self, fruits: List[int], baskets: List[int]) -> int:
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n = len(fruits)
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vis = [False] * n
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ans = n
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for x in fruits:
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for i, y in enumerate(baskets):
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if y >= x and not vis[i]:
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vis[i] = True
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ans -= 1
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break
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return ans

‎solution/3400-3499/3477.Fruits Into Baskets II/Solution.ts

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function numOfUnplacedFruits(fruits: number[], baskets: number[]): number {
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const n = fruits.length;
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const vis: boolean[] = Array(n).fill(false);
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let ans = n;
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for (const x of fruits) {
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for (let i = 0; i < n; ++i) {
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if (baskets[i] >= x && !vis[i]) {
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vis[i] = true;
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--ans;
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break;
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}
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}
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}
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return ans;
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}

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