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feat: add solutions to lc problem: No.2139 (#1725)
No.2139.Minimum Moves to Reach Target Score
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‎solution/2100-2199/2139.Minimum Moves to Reach Target Score/README.md

+123-5
Original file line numberDiff line numberDiff line change
@@ -65,6 +65,22 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：倒推 + 贪心**
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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。
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如果 $target=1$，那么不需要任何操作，直接返回 $0$ 即可。
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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。
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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 $maxDoubles-1$。
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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 $maxDoubles$。
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时间复杂度 $O(\min(\log target, maxDoubles))$，空间复杂度 $O(\min(\log target, maxDoubles))$。
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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。
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<!-- tabs:start -->
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### **Python3**
@@ -83,6 +99,21 @@ class Solution:
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return 1 + self.minMoves(target - 1, maxDoubles)
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```
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```python
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class Solution:
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def minMoves(self, target: int, maxDoubles: int) -> int:
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ans = 0
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while maxDoubles and target > 1:
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ans += 1
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if target % 2 == 1:
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target -= 1
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else:
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maxDoubles -= 1
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target >>= 1
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ans += target - 1
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return ans
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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 -->
@@ -104,20 +135,65 @@ 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 minMoves(int target, int maxDoubles) {
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int ans = 0;
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while (maxDoubles > 0 && target > 1) {
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++ans;
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if (target % 2 == 1) {
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--target;
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} else {
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--maxDoubles;
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target >>= 1;
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}
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}
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ans += target - 1;
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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
110160
class Solution {
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public:
112162
int minMoves(int target, int maxDoubles) {
113-
if (target == 1) return 0;
114-
if (maxDoubles == 0) return target - 1;
115-
if (target % 2 == 0 && maxDoubles) return 1 + minMoves(target >> 1, maxDoubles - 1);
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if (target == 1) {
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return 0;
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}
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if (maxDoubles == 0) {
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return target - 1;
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}
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if (target % 2 == 0 && maxDoubles > 0) {
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return 1 + minMoves(target >> 1, maxDoubles - 1);
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}
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return 1 + minMoves(target - 1, maxDoubles);
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}
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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 minMoves(int target, int maxDoubles) {
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int ans = 0;
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while (maxDoubles > 0 && target > 1) {
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++ans;
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if (target % 2 == 1) {
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--target;
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} else {
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--maxDoubles;
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target >>= 1;
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}
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}
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ans += target - 1;
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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
@@ -135,12 +211,54 @@ func minMoves(target int, maxDoubles int) int {
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}
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```
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### **TypeScript**
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```go
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func minMoves(target int, maxDoubles int) (ans int) {
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for maxDoubles > 0 && target > 1 {
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ans++
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if target&1 == 1 {
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target--
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} else {
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maxDoubles--
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target >>= 1
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}
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}
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ans += target - 1
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return
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}
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```
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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### **TypeScript**
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```ts
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function minMoves(target: number, maxDoubles: number): number {
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if (target === 1) {
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return 0;
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}
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if (maxDoubles === 0) {
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return target - 1;
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}
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if (target % 2 === 0 && maxDoubles) {
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return 1 + minMoves(target >> 1, maxDoubles - 1);
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}
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return 1 + minMoves(target - 1, maxDoubles);
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}
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```
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```ts
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function minMoves(target: number, maxDoubles: number): number {
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let ans = 0;
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while (maxDoubles && target > 1) {
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++ans;
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if (target & 1) {
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--target;
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} else {
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--maxDoubles;
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target >>= 1;
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}
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}
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ans += target - 1;
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return ans;
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}
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```
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### **...**

‎solution/2100-2199/2139.Minimum Moves to Reach Target Score/README_EN.md

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@@ -61,6 +61,22 @@ Double again so x = 10
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## Solutions
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**Solution 1: Backtracking + Greedy**
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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 $maxDoubles$.
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If $target=1$, no operation is needed, and we can return $0$ directly.
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If $maxDoubles=0$, we can only use the increment operation, so we need $target-1$ operations.
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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 $maxDoubles-1$.
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If $target$ is odd, we can only use the increment operation, so we need $1$ operation, and then recursively solve $target-1$ and $maxDoubles$.
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The time complexity is $O(\min(\log target, maxDoubles))$, and the space complexity is $O(\min(\log target, maxDoubles))$.
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We can also change the above process to an iterative way to avoid the space overhead of recursion.
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<!-- tabs:start -->
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### **Python3**
@@ -77,6 +93,21 @@ class Solution:
7793
return 1 + self.minMoves(target - 1, maxDoubles)
7894
```
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```python
97+
class Solution:
98+
def minMoves(self, target: int, maxDoubles: int) -> int:
99+
ans = 0
100+
while maxDoubles and target > 1:
101+
ans += 1
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if target % 2 == 1:
103+
target -= 1
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else:
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maxDoubles -= 1
106+
target >>= 1
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ans += target - 1
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return ans
109+
```
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80111
### **Java**
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82113
```java
@@ -96,20 +127,65 @@ class Solution {
96127
}
97128
```
98129

130+
```java
131+
class Solution {
132+
public int minMoves(int target, int maxDoubles) {
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int ans = 0;
134+
while (maxDoubles > 0 && target > 1) {
135+
++ans;
136+
if (target % 2 == 1) {
137+
--target;
138+
} else {
139+
--maxDoubles;
140+
target >>= 1;
141+
}
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}
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ans += target - 1;
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return ans;
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}
146+
}
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```
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99149
### **C++**
100150

101151
```cpp
102152
class Solution {
103153
public:
104154
int minMoves(int target, int maxDoubles) {
105-
if (target == 1) return 0;
106-
if (maxDoubles == 0) return target - 1;
107-
if (target % 2 == 0 && maxDoubles) return 1 + minMoves(target >> 1, maxDoubles - 1);
155+
if (target == 1) {
156+
return 0;
157+
}
158+
if (maxDoubles == 0) {
159+
return target - 1;
160+
}
161+
if (target % 2 == 0 && maxDoubles > 0) {
162+
return 1 + minMoves(target >> 1, maxDoubles - 1);
163+
}
108164
return 1 + minMoves(target - 1, maxDoubles);
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}
110166
};
111167
```
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```cpp
170+
class Solution {
171+
public:
172+
int minMoves(int target, int maxDoubles) {
173+
int ans = 0;
174+
while (maxDoubles > 0 && target > 1) {
175+
++ans;
176+
if (target % 2 == 1) {
177+
--target;
178+
} else {
179+
--maxDoubles;
180+
target >>= 1;
181+
}
182+
}
183+
ans += target - 1;
184+
return ans;
185+
}
186+
};
187+
```
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113189
### **Go**
114190

115191
```go
@@ -127,10 +203,54 @@ func minMoves(target int, maxDoubles int) int {
127203
}
128204
```
129205

206+
```go
207+
func minMoves(target int, maxDoubles int) (ans int) {
208+
for maxDoubles > 0 && target > 1 {
209+
ans++
210+
if target&1 == 1 {
211+
target--
212+
} else {
213+
maxDoubles--
214+
target >>= 1
215+
}
216+
}
217+
ans += target - 1
218+
return
219+
}
220+
```
221+
130222
### **TypeScript**
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132224
```ts
225+
function minMoves(target: number, maxDoubles: number): number {
226+
if (target === 1) {
227+
return 0;
228+
}
229+
if (maxDoubles === 0) {
230+
return target - 1;
231+
}
232+
if (target % 2 === 0 && maxDoubles) {
233+
return 1 + minMoves(target >> 1, maxDoubles - 1);
234+
}
235+
return 1 + minMoves(target - 1, maxDoubles);
236+
}
237+
```
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239+
```ts
240+
function minMoves(target: number, maxDoubles: number): number {
241+
let ans = 0;
242+
while (maxDoubles && target > 1) {
243+
++ans;
244+
if (target & 1) {
245+
--target;
246+
} else {
247+
--maxDoubles;
248+
target >>= 1;
249+
}
250+
}
251+
ans += target - 1;
252+
return ans;
253+
}
134254
```
135255

136256
### **...**

‎solution/2100-2199/2139.Minimum Moves to Reach Target Score/Solution.cpp

+12-4
Original file line numberDiff line numberDiff line change
@@ -1,9 +1,17 @@
11
class Solution {
22
public:
33
int minMoves(int target, int maxDoubles) {
4-
if (target == 1) return 0;
5-
if (maxDoubles == 0) return target - 1;
6-
if (target % 2 == 0 && maxDoubles) return 1 + minMoves(target >> 1, maxDoubles - 1);
7-
return 1 + minMoves(target - 1, maxDoubles);
4+
int ans = 0;
5+
while (maxDoubles > 0 && target > 1) {
6+
++ans;
7+
if (target % 2 == 1) {
8+
--target;
9+
} else {
10+
--maxDoubles;
11+
target >>= 1;
12+
}
13+
}
14+
ans += target - 1;
15+
return ans;
816
}
917
};

‎solution/2100-2199/2139.Minimum Moves to Reach Target Score/Solution.go

+11-10
Original file line numberDiff line numberDiff line change
@@ -1,12 +1,13 @@
1-
func minMoves(target int, maxDoubles int) int {
2-
if target == 1 {
3-
return 0
1+
func minMoves(target int, maxDoubles int) (ans int) {
2+
for maxDoubles > 0 && target > 1 {
3+
ans++
4+
if target&1 == 1 {
5+
target--
6+
} else {
7+
maxDoubles--
8+
target >>= 1
9+
}
410
}
5-
if maxDoubles == 0 {
6-
return target - 1
7-
}
8-
if target%2 == 0 && maxDoubles > 0 {
9-
return 1 + minMoves(target>>1, maxDoubles-1)
10-
}
11-
return 1 + minMoves(target-1, maxDoubles)
11+
ans += target - 1
12+
return
1213
}

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