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Sep 30, 2023
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feat: add solutions to lc problem: No.2139 #1725

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128 changes: 123 additions & 5 deletions solution/2100-2199/2139.Minimum Moves to Reach Target Score/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -65,6 +65,22 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一：倒推 + 贪心**

我们不妨从最终的状态开始倒推，假设最终的状态为 $target$，那么我们可以得到 $target$ 的前一个状态为 $target - 1$ 或者 $target / 2$，这取决于 $target$ 的奇偶性以及 $maxDoubles$ 的值。

如果 $target=1$，那么不需要任何操作，直接返回 $0$ 即可。

如果 $maxDoubles=0$，那么我们只能使用递增操作，因此我们需要 $target-1$ 次操作。

如果 $target$ 是偶数且 $maxDoubles>0$，那么我们可以使用加倍操作，因此我们需要 $1$ 次操作，然后递归求解 $target/2$ 和 $maxDoubles-1$。

如果 $target$ 是奇数，那么我们只能使用递增操作，因此我们需要 $1$ 次操作，然后递归求解 $target-1$ 和 $maxDoubles$。

时间复杂度 $O(\min(\log target, maxDoubles))$，空间复杂度 $O(\min(\log target, maxDoubles))$。

我们也可以将上述过程改为迭代的方式，这样可以避免递归的空间开销。

<!-- tabs:start -->

### **Python3**
Expand All @@ -83,6 +99,21 @@ class Solution:
return 1 + self.minMoves(target - 1, maxDoubles)
```

```python
class Solution:
def minMoves(self, target: int, maxDoubles: int) -> int:
ans = 0
while maxDoubles and target > 1:
ans += 1
if target % 2 == 1:
target -= 1
else:
maxDoubles -= 1
target >>= 1
ans += target - 1
return ans
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->
Expand All @@ -104,20 +135,65 @@ class Solution {
}
```

```java
class Solution {
public int minMoves(int target, int maxDoubles) {
int ans = 0;
while (maxDoubles > 0 && target > 1) {
++ans;
if (target % 2 == 1) {
--target;
} else {
--maxDoubles;
target >>= 1;
}
}
ans += target - 1;
return ans;
}
}
```

### **C++**

```cpp
class Solution {
public:
int minMoves(int target, int maxDoubles) {
if (target == 1) return 0;
if (maxDoubles == 0) return target - 1;
if (target % 2 == 0 && maxDoubles) return 1 + minMoves(target >> 1, maxDoubles - 1);
if (target == 1) {
return 0;
}
if (maxDoubles == 0) {
return target - 1;
}
if (target % 2 == 0 && maxDoubles > 0) {
return 1 + minMoves(target >> 1, maxDoubles - 1);
}
return 1 + minMoves(target - 1, maxDoubles);
}
};
```

```cpp
class Solution {
public:
int minMoves(int target, int maxDoubles) {
int ans = 0;
while (maxDoubles > 0 && target > 1) {
++ans;
if (target % 2 == 1) {
--target;
} else {
--maxDoubles;
target >>= 1;
}
}
ans += target - 1;
return ans;
}
};
```

### **Go**

```go
Expand All @@ -135,12 +211,54 @@ func minMoves(target int, maxDoubles int) int {
}
```

### **TypeScript**
```go
func minMoves(target int, maxDoubles int) (ans int) {
for maxDoubles > 0 && target > 1 {
ans++
if target&1 == 1 {
target--
} else {
maxDoubles--
target >>= 1
}
}
ans += target - 1
return
}
```

<!-- 这里可写当前语言的特殊实现逻辑 -->
### **TypeScript**

```ts
function minMoves(target: number, maxDoubles: number): number {
if (target === 1) {
return 0;
}
if (maxDoubles === 0) {
return target - 1;
}
if (target % 2 === 0 && maxDoubles) {
return 1 + minMoves(target >> 1, maxDoubles - 1);
}
return 1 + minMoves(target - 1, maxDoubles);
}
```

```ts
function minMoves(target: number, maxDoubles: number): number {
let ans = 0;
while (maxDoubles && target > 1) {
++ans;
if (target & 1) {
--target;
} else {
--maxDoubles;
target >>= 1;
}
}
ans += target - 1;
return ans;
}
```

### **...**
Expand Down
126 changes: 123 additions & 3 deletions solution/2100-2199/2139.Minimum Moves to Reach Target Score/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -61,6 +61,22 @@ Double again so x = 10

## Solutions

**Solution 1: Backtracking + Greedy**

Let's start by backtracking from the final state. Assuming the final state is $target$, we can get the previous state of $target$ as $target - 1$ or $target / 2$, depending on the parity of $target$ and the value of $maxDoubles$.

If $target=1$, no operation is needed, and we can return $0$ directly.

If $maxDoubles=0$, we can only use the increment operation, so we need $target-1$ operations.

If $target$ is even and $maxDoubles>0$, we can use the doubling operation, so we need $1$ operation, and then recursively solve $target/2$ and $maxDoubles-1$.

If $target$ is odd, we can only use the increment operation, so we need $1$ operation, and then recursively solve $target-1$ and $maxDoubles$.

The time complexity is $O(\min(\log target, maxDoubles))$, and the space complexity is $O(\min(\log target, maxDoubles))$.

We can also change the above process to an iterative way to avoid the space overhead of recursion.

<!-- tabs:start -->

### **Python3**
Expand All @@ -77,6 +93,21 @@ class Solution:
return 1 + self.minMoves(target - 1, maxDoubles)
```

```python
class Solution:
def minMoves(self, target: int, maxDoubles: int) -> int:
ans = 0
while maxDoubles and target > 1:
ans += 1
if target % 2 == 1:
target -= 1
else:
maxDoubles -= 1
target >>= 1
ans += target - 1
return ans
```

### **Java**

```java
Expand All @@ -96,20 +127,65 @@ class Solution {
}
```

```java
class Solution {
public int minMoves(int target, int maxDoubles) {
int ans = 0;
while (maxDoubles > 0 && target > 1) {
++ans;
if (target % 2 == 1) {
--target;
} else {
--maxDoubles;
target >>= 1;
}
}
ans += target - 1;
return ans;
}
}
```

### **C++**

```cpp
class Solution {
public:
int minMoves(int target, int maxDoubles) {
if (target == 1) return 0;
if (maxDoubles == 0) return target - 1;
if (target % 2 == 0 && maxDoubles) return 1 + minMoves(target >> 1, maxDoubles - 1);
if (target == 1) {
return 0;
}
if (maxDoubles == 0) {
return target - 1;
}
if (target % 2 == 0 && maxDoubles > 0) {
return 1 + minMoves(target >> 1, maxDoubles - 1);
}
return 1 + minMoves(target - 1, maxDoubles);
}
};
```

```cpp
class Solution {
public:
int minMoves(int target, int maxDoubles) {
int ans = 0;
while (maxDoubles > 0 && target > 1) {
++ans;
if (target % 2 == 1) {
--target;
} else {
--maxDoubles;
target >>= 1;
}
}
ans += target - 1;
return ans;
}
};
```

### **Go**

```go
Expand All @@ -127,10 +203,54 @@ func minMoves(target int, maxDoubles int) int {
}
```

```go
func minMoves(target int, maxDoubles int) (ans int) {
for maxDoubles > 0 && target > 1 {
ans++
if target&1 == 1 {
target--
} else {
maxDoubles--
target >>= 1
}
}
ans += target - 1
return
}
```

### **TypeScript**

```ts
function minMoves(target: number, maxDoubles: number): number {
if (target === 1) {
return 0;
}
if (maxDoubles === 0) {
return target - 1;
}
if (target % 2 === 0 && maxDoubles) {
return 1 + minMoves(target >> 1, maxDoubles - 1);
}
return 1 + minMoves(target - 1, maxDoubles);
}
```

```ts
function minMoves(target: number, maxDoubles: number): number {
let ans = 0;
while (maxDoubles && target > 1) {
++ans;
if (target & 1) {
--target;
} else {
--maxDoubles;
target >>= 1;
}
}
ans += target - 1;
return ans;
}
```

### **...**
Expand Down
16 changes: 12 additions & 4 deletions solution/2100-2199/2139.Minimum Moves to Reach Target Score/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -1,9 +1,17 @@
class Solution {
public:
int minMoves(int target, int maxDoubles) {
if (target == 1) return 0;
if (maxDoubles == 0) return target - 1;
if (target % 2 == 0 && maxDoubles) return 1 + minMoves(target >> 1, maxDoubles - 1);
return 1 + minMoves(target - 1, maxDoubles);
int ans = 0;
while (maxDoubles > 0 && target > 1) {
++ans;
if (target % 2 == 1) {
--target;
} else {
--maxDoubles;
target >>= 1;
}
}
ans += target - 1;
return ans;
}
};
21 changes: 11 additions & 10 deletions solution/2100-2199/2139.Minimum Moves to Reach Target Score/Solution.go
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Original file line number Diff line number Diff line change
@@ -1,12 +1,13 @@
func minMoves(target int, maxDoubles int) int {
if target == 1 {
return 0
func minMoves(target int, maxDoubles int) (ans int) {
for maxDoubles > 0 && target > 1 {
ans++
if target&1 == 1 {
target--
} else {
maxDoubles--
target >>= 1
}
}
if maxDoubles == 0 {
return target - 1
}
if target%2 == 0 && maxDoubles > 0 {
return 1 + minMoves(target>>1, maxDoubles-1)
}
return 1 + minMoves(target-1, maxDoubles)
ans += target - 1
return
}
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