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feat: add solutions to lc problem: No.1406
No.1406.Stone Game III
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‎solution/1400-1499/1406.Stone Game III/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：前缀和 + 记忆化搜索**
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我们先预处理出数组 $stoneValue$ 的前缀和数组 $s$，其中 $s[i]$ 表示 $stoneValue$ 数组前 $i$ 个元素之和。
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接下来，我们设计一个函数 $dfs(i)$，表示当前玩家从 $stoneValue$ 数组下标 $i$ 开始拿石子时，能够获得的最大分数。如果当前玩家拿走了 $stoneValue$ 数组下标 $i$ 开始的 $j$ 堆石子，那么下一个玩家能够获得的最大分数为 $dfs(i + j)$，为了使当前玩家获得的分数最大化，我们需要让 $dfs(i + j)$ 最小化，即 $dfs(i) = s[n] - s[i] - min(dfs(i + j))$，其中 $s[n]$ 表示 $stoneValue$ 数组所有元素之和。
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最后，玩家 $Alice$ 的分数为 $dfs(0)$，玩家 $Bob$ 的分数为 $s[n] - dfs(0)$，如果 $Alice$ 的分数大于 $Bob$ 的分数，那么 $Alice$ 获胜；如果 $Alice$ 的分数小于 $Bob$ 的分数，那么 $Bob$ 获胜；如果 $Alice$ 的分数等于 $Bob$ 的分数，那么平局。
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过程中，我们可以使用记忆化搜索，避免重复计算。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $stoneValue$ 的长度。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def stoneGameIII(self, stoneValue: List[int]) -> str:
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@cache
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def dfs(i: int) -> int:
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if i >= len(stoneValue):
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return 0
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t = min(dfs(i + j) for j in range(1, 4))
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return s[-1] - s[i] - t
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s = list(accumulate(stoneValue, initial=0))
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a = dfs(0)
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b = s[-1] - a
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if a == b:
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return 'Tie'
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return 'Alice' if a > b else 'Bob'
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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 -->
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```java
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class Solution {
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private int n;
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private int[] s;
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private Integer[] f;
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public String stoneGameIII(int[] stoneValue) {
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n = stoneValue.length;
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s = new int[n + 1];
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f = new Integer[n];
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for (int i = 1; i <= n; ++i) {
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s[i] = s[i - 1] + stoneValue[i - 1];
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}
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int a = dfs(0);
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int b = s[n] - a;
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return a == b ? "Tie" : a > b ? "Alice" : "Bob";
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}
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private int dfs(int i) {
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if (i >= n) {
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return 0;
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}
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if (f[i] != null) {
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return f[i];
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}
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int t = 1 << 30;
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for (int j = 1; j < 4; ++j) {
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t = Math.min(t, dfs(i + j));
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}
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f[i] = s[n] - s[i] - t;
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return f[i];
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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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string stoneGameIII(vector<int>& stoneValue) {
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int n = stoneValue.size();
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int s[n + 1];
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s[0] = 0;
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for (int i = 1; i <= n; ++i) {
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s[i] = s[i - 1] + stoneValue[i - 1];
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}
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int f[n];
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memset(f, 0x3f, sizeof(f));
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function<int(int)> dfs = [&](int i) -> int {
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if (i >= n) {
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return 0;
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}
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if (f[i] != 0x3f3f3f3f) {
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return f[i];
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}
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int t = 1 << 30;
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for (int j = 1; j < 4; ++j) {
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t = min(t, dfs(i + j));
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}
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return f[i] = s[n] - s[i] - t;
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};
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int a = dfs(0);
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int b = s[n] - a;
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return a == b ? "Tie" : (a > b ? "Alice" : "Bob");
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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 stoneGameIII(stoneValue []int) string {
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n := len(stoneValue)
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s := make([]int, n+1)
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for i, x := range stoneValue {
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s[i+1] = s[i] + x
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}
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const inf = 1 << 30
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f := make([]int, n)
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for i := range f {
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f[i] = inf
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}
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var dfs func(int) int
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dfs = func(i int) int {
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if i >= n {
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return 0
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}
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if f[i] != inf {
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return f[i]
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}
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t := inf
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for j := 1; j <= 3; j++ {
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t = min(t, dfs(i+j))
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}
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f[i] = s[n] - s[i] - t
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return f[i]
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}
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a := dfs(0)
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b := s[n] - a
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if a == b {
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return "Tie"
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}
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if a > b {
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return "Alice"
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}
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return "Bob"
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1400-1499/1406.Stone Game III/README_EN.md

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@@ -59,13 +59,140 @@ Remember that both play optimally so here Alice will choose the scenario that ma
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### **Python3**
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```python
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class Solution:
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def stoneGameIII(self, stoneValue: List[int]) -> str:
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@cache
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def dfs(i: int) -> int:
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if i >= len(stoneValue):
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return 0
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t = min(dfs(i + j) for j in range(1, 4))
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return s[-1] - s[i] - t
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s = list(accumulate(stoneValue, initial=0))
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a = dfs(0)
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b = s[-1] - a
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if a == b:
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return 'Tie'
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return 'Alice' if a > b else 'Bob'
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```
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### **Java**
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```java
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class Solution {
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private int n;
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private int[] s;
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private Integer[] f;
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public String stoneGameIII(int[] stoneValue) {
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n = stoneValue.length;
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s = new int[n + 1];
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f = new Integer[n];
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for (int i = 1; i <= n; ++i) {
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s[i] = s[i - 1] + stoneValue[i - 1];
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}
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int a = dfs(0);
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int b = s[n] - a;
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return a == b ? "Tie" : a > b ? "Alice" : "Bob";
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}
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private int dfs(int i) {
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if (i >= n) {
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return 0;
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}
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if (f[i] != null) {
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return f[i];
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}
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int t = 1 << 30;
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for (int j = 1; j < 4; ++j) {
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t = Math.min(t, dfs(i + j));
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}
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f[i] = s[n] - s[i] - t;
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return f[i];
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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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string stoneGameIII(vector<int>& stoneValue) {
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int n = stoneValue.size();
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int s[n + 1];
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s[0] = 0;
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for (int i = 1; i <= n; ++i) {
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s[i] = s[i - 1] + stoneValue[i - 1];
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}
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int f[n];
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memset(f, 0x3f, sizeof(f));
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function<int(int)> dfs = [&](int i) -> int {
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if (i >= n) {
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return 0;
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}
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if (f[i] != 0x3f3f3f3f) {
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return f[i];
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}
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int t = 1 << 30;
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for (int j = 1; j < 4; ++j) {
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t = min(t, dfs(i + j));
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}
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return f[i] = s[n] - s[i] - t;
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};
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int a = dfs(0);
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int b = s[n] - a;
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return a == b ? "Tie" : (a > b ? "Alice" : "Bob");
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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 stoneGameIII(stoneValue []int) string {
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n := len(stoneValue)
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s := make([]int, n+1)
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for i, x := range stoneValue {
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s[i+1] = s[i] + x
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}
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const inf = 1 << 30
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f := make([]int, n)
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for i := range f {
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f[i] = inf
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}
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var dfs func(int) int
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dfs = func(i int) int {
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if i >= n {
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return 0
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}
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if f[i] != inf {
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return f[i]
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}
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t := inf
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for j := 1; j <= 3; j++ {
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t = min(t, dfs(i+j))
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}
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f[i] = s[n] - s[i] - t
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return f[i]
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}
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a := dfs(0)
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b := s[n] - a
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if a == b {
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return "Tie"
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}
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if a > b {
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return "Alice"
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}
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return "Bob"
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1400-1499/1406.Stone Game III/Solution.cpp

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class Solution {
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public:
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string stoneGameIII(vector<int>& stoneValue) {
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int n = stoneValue.size();
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int s[n + 1];
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s[0] = 0;
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for (int i = 1; i <= n; ++i) {
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s[i] = s[i - 1] + stoneValue[i - 1];
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}
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int f[n];
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memset(f, 0x3f, sizeof(f));
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function<int(int)> dfs = [&](int i) -> int {
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if (i >= n) {
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return 0;
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}
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if (f[i] != 0x3f3f3f3f) {
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return f[i];
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}
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int t = 1 << 30;
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for (int j = 1; j < 4; ++j) {
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t = min(t, dfs(i + j));
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}
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return f[i] = s[n] - s[i] - t;
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};
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int a = dfs(0);
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int b = s[n] - a;
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return a == b ? "Tie" : (a > b ? "Alice" : "Bob");
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}
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};

‎solution/1400-1499/1406.Stone Game III/Solution.go

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func stoneGameIII(stoneValue []int) string {
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n := len(stoneValue)
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s := make([]int, n+1)
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for i, x := range stoneValue {
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s[i+1] = s[i] + x
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}
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const inf = 1 << 30
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f := make([]int, n)
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for i := range f {
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f[i] = inf
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}
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var dfs func(int) int
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dfs = func(i int) int {
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if i >= n {
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return 0
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}
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if f[i] != inf {
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return f[i]
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}
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t := inf
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for j := 1; j <= 3; j++ {
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t = min(t, dfs(i+j))
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}
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f[i] = s[n] - s[i] - t
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return f[i]
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}
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a := dfs(0)
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b := s[n] - a
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if a == b {
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return "Tie"
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}
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if a > b {
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return "Alice"
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}
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return "Bob"
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}
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func min(a, b int) int {
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if a < b {
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return a
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}
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return b
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}

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