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feat: add solutions to lc problem: No.2706 (#2158)
No.2706.Buy Two Chocolates
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‎solution/2700-2799/2706.Buy Two Chocolates/README.md

+115-14
Original file line numberDiff line numberDiff line change
@@ -42,6 +42,18 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：排序**
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我们可以将巧克力的价格从小到大排序，然后取前两个价格相加，就是我们购买两块巧克力的最小花费 $cost$。如果这个花费大于我们拥有的钱数，那么我们就返回 `money`，否则返回 `money - cost`。
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时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 是数组 `prices` 的长度。
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**方法一：一次遍历**
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我们可以在一次遍历中找到最小的两个价格，然后计算花费。
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时间复杂度 $O(n)$，其中 $n$ 是数组 `prices` 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
@@ -56,6 +68,19 @@ class Solution:
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return money if money < cost else money - cost
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```
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```python
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class Solution:
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def buyChoco(self, prices: List[int], money: int) -> int:
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a = b = inf
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for x in prices:
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if x < a:
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a, b = x, a
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elif x < b:
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b = x
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cost = a + b
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return money if money < cost else money - cost
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -70,6 +95,24 @@ class Solution {
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}
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```
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```java
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class Solution {
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public int buyChoco(int[] prices, int money) {
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int a = 1000, b = 1000;
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for (int x : prices) {
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if (x < a) {
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b = a;
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a = x;
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} else if (x < b) {
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b = x;
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}
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}
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int cost = a + b;
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return money < cost ? money : money - cost;
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}
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}
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```
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### **C++**
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```cpp
@@ -83,6 +126,25 @@ public:
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};
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```
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```cpp
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class Solution {
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public:
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int buyChoco(vector<int>& prices, int money) {
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int a = 1000, b = 1000;
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for (int x : prices) {
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if (x < a) {
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b = a;
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a = x;
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} else if (x < b) {
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b = x;
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}
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}
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int cost = a + b;
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return money < cost ? money : money - cost;
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}
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};
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```
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### **Go**
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```go
@@ -96,6 +158,24 @@ func buyChoco(prices []int, money int) int {
96158
}
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```
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```go
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func buyChoco(prices []int, money int) int {
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a, b := 1001, 1001
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for _, x := range prices {
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if x < a {
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a, b = x, a
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} else if x < b {
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b = x
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}
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}
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cost := a + b
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if money < cost {
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return money
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}
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return money - cost
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}
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```
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### **TypeScript**
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```ts
@@ -106,35 +186,56 @@ function buyChoco(prices: number[], money: number): number {
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}
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```
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```ts
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function buyChoco(prices: number[], money: number): number {
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let [a, b] = [1000, 1000];
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for (const x of prices) {
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if (x < a) {
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b = a;
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a = x;
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} else if (x < b) {
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b = x;
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}
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}
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const cost = a + b;
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return money < cost ? money : money - cost;
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}
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```
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### **Rust**
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```rust
112208
impl Solution {
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pub fn buy_choco(mut prices: Vec<i32>, money: i32) -> i32 {
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prices.sort();
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let sum = prices[0] + prices[1];
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if sum > money {
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let cost = prices[0] + prices[1];
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if cost > money {
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return money;
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}
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money - sum
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money - cost
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}
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}
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```
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```rust
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impl Solution {
128-
pub fn buy_choco(mut prices: Vec<i32>, money: i32) -> i32 {
129-
prices.sort();
130-
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let sum = prices.iter().take(2).sum::<i32>();
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if sum > money {
134-
return money;
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pub fn buy_choco(prices: Vec<i32>, money: i32) -> i32 {
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let mut a = 1000;
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let mut b = 1000;
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for &x in prices.iter() {
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if x < a {
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b = a;
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a = x;
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} else if x < b {
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b = x;
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}
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}
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let cost = a + b;
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if money < cost {
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money
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} else {
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money - cost
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}
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money - sum
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}
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}
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```

‎solution/2700-2799/2706.Buy Two Chocolates/README_EN.md

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@@ -38,6 +38,18 @@
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## Solutions
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**Solution 1: Sorting**
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We can sort the prices of the chocolates in ascending order, and then add the first two prices to get the minimum cost $cost$ of buying two chocolates. If this cost is greater than the money we have, then we return `money`. Otherwise, we return `money - cost`.
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The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Where $n$ is the length of the array `prices`.
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**Solution 2: One-pass Traversal**
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We can find the two smallest prices in one pass, and then calculate the cost.
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The time complexity is $O(n)$, where $n$ is the length of the array `prices`. The space complexity is $O(1)$.
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<!-- tabs:start -->
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### **Python3**
@@ -50,6 +62,19 @@ class Solution:
5062
return money if money < cost else money - cost
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```
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```python
66+
class Solution:
67+
def buyChoco(self, prices: List[int], money: int) -> int:
68+
a = b = inf
69+
for x in prices:
70+
if x < a:
71+
a, b = x, a
72+
elif x < b:
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b = x
74+
cost = a + b
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return money if money < cost else money - cost
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```
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### **Java**
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5580
```java
@@ -62,6 +87,24 @@ class Solution {
6287
}
6388
```
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90+
```java
91+
class Solution {
92+
public int buyChoco(int[] prices, int money) {
93+
int a = 1000, b = 1000;
94+
for (int x : prices) {
95+
if (x < a) {
96+
b = a;
97+
a = x;
98+
} else if (x < b) {
99+
b = x;
100+
}
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}
102+
int cost = a + b;
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return money < cost ? money : money - cost;
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}
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}
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```
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65108
### **C++**
66109

67110
```cpp
@@ -75,6 +118,25 @@ public:
75118
};
76119
```
77120
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```cpp
122+
class Solution {
123+
public:
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int buyChoco(vector<int>& prices, int money) {
125+
int a = 1000, b = 1000;
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for (int x : prices) {
127+
if (x < a) {
128+
b = a;
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a = x;
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} else if (x < b) {
131+
b = x;
132+
}
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}
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int cost = a + b;
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return money < cost ? money : money - cost;
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}
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};
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```
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### **Go**
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80142
```go
@@ -88,6 +150,24 @@ func buyChoco(prices []int, money int) int {
88150
}
89151
```
90152

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```go
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func buyChoco(prices []int, money int) int {
155+
a, b := 1001, 1001
156+
for _, x := range prices {
157+
if x < a {
158+
a, b = x, a
159+
} else if x < b {
160+
b = x
161+
}
162+
}
163+
cost := a + b
164+
if money < cost {
165+
return money
166+
}
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return money - cost
168+
}
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```
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### **TypeScript**
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93173
```ts
@@ -98,35 +178,56 @@ function buyChoco(prices: number[], money: number): number {
98178
}
99179
```
100180

181+
```ts
182+
function buyChoco(prices: number[], money: number): number {
183+
let [a, b] = [1000, 1000];
184+
for (const x of prices) {
185+
if (x < a) {
186+
b = a;
187+
a = x;
188+
} else if (x < b) {
189+
b = x;
190+
}
191+
}
192+
const cost = a + b;
193+
return money < cost ? money : money - cost;
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}
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```
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101197
### **Rust**
102198

103199
```rust
104200
impl Solution {
105201
pub fn buy_choco(mut prices: Vec<i32>, money: i32) -> i32 {
106202
prices.sort();
107-
108-
let sum = prices[0] + prices[1];
109-
if sum > money {
203+
let cost = prices[0] + prices[1];
204+
if cost > money {
110205
return money;
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}
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money - sum
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money - cost
114208
}
115209
}
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```
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118212
```rust
119213
impl Solution {
120-
pub fn buy_choco(mut prices: Vec<i32>, money: i32) -> i32 {
121-
prices.sort();
122-
123-
let sum = prices.iter().take(2).sum::<i32>();
124-
125-
if sum > money {
126-
return money;
214+
pub fn buy_choco(prices: Vec<i32>, money: i32) -> i32 {
215+
let mut a = 1000;
216+
let mut b = 1000;
217+
for &x in prices.iter() {
218+
if x < a {
219+
b = a;
220+
a = x;
221+
} else if x < b {
222+
b = x;
223+
}
224+
}
225+
let cost = a + b;
226+
if money < cost {
227+
money
228+
} else {
229+
money - cost
127230
}
128-
129-
money - sum
130231
}
131232
}
132233
```

‎solution/2700-2799/2706.Buy Two Chocolates/Solution.cpp

+15-7
Original file line numberDiff line numberDiff line change
@@ -1,8 +1,16 @@
1-
class Solution {
2-
public:
3-
int buyChoco(vector<int>& prices, int money) {
4-
sort(prices.begin(), prices.end());
5-
int cost = prices[0] + prices[1];
6-
return money < cost ? money : money - cost;
7-
}
1+
class Solution {
2+
public:
3+
int buyChoco(vector<int>& prices, int money) {
4+
int a = 1000, b = 1000;
5+
for (int x : prices) {
6+
if (x < a) {
7+
b = a;
8+
a = x;
9+
} else if (x < b) {
10+
b = x;
11+
}
12+
}
13+
int cost = a + b;
14+
return money < cost ? money : money - cost;
15+
}
816
};

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