diff --git a/solution/2100-2199/2151.Maximum Good People Based on Statements/README.md b/solution/2100-2199/2151.Maximum Good People Based on Statements/README.md
index 659b191fe6d76..5340b3481f703 100644
--- a/solution/2100-2199/2151.Maximum Good People Based on Statements/README.md
+++ b/solution/2100-2199/2151.Maximum Good People Based on Statements/README.md
@@ -86,7 +86,7 @@ tags:
- 在认为 0 是坏人但说真话的情况下，这组玩家中没有一个好人。
- 说假话。在这种情况下，1 是好人。
- 在认为 0 是坏人且说假话的情况下，这组玩家中只有一个好人。
-在最佳情况下，至多有一个好人，所以返回 1 。
+在最佳情况下，至多有一个好人，所以返回 1 。
注意，能得到此结论的方法不止一种。
@@ -120,17 +120,17 @@ tags:
```python
class Solution:
def maximumGood(self, statements: List[List[int]]) -> int:
- def check(mask):
+ def check(mask: int) -> int:
cnt = 0
- for i, s in enumerate(statements):
- if (mask >> i) & 1:
- for j, v in enumerate(s):
- if v < 2 and ((mask >> j) & 1) != v:
+ for i, row in enumerate(statements):
+ if mask >> i & 1:
+ for j, x in enumerate(row):
+ if x < 2 and (mask >> j & 1) != x:
return 0
cnt += 1
return cnt
- return max(check(mask) for mask in range(1, 1 << len(statements)))
+ return max(check(i) for i in range(1, 1 << len(statements)))
```
#### Java
diff --git a/solution/2100-2199/2151.Maximum Good People Based on Statements/README_EN.md b/solution/2100-2199/2151.Maximum Good People Based on Statements/README_EN.md
index c6fafdd848c24..99da6be57830d 100644
--- a/solution/2100-2199/2151.Maximum Good People Based on Statements/README_EN.md
+++ b/solution/2100-2199/2151.Maximum Good People Based on Statements/README_EN.md
@@ -116,17 +116,17 @@ Note that there is more than one way to arrive at this conclusion.
```python
class Solution:
def maximumGood(self, statements: List[List[int]]) -> int:
- def check(mask):
+ def check(mask: int) -> int:
cnt = 0
- for i, s in enumerate(statements):
- if (mask >> i) & 1:
- for j, v in enumerate(s):
- if v < 2 and ((mask >> j) & 1) != v:
+ for i, row in enumerate(statements):
+ if mask >> i & 1:
+ for j, x in enumerate(row):
+ if x < 2 and (mask >> j & 1) != x:
return 0
cnt += 1
return cnt
- return max(check(mask) for mask in range(1, 1 << len(statements)))
+ return max(check(i) for i in range(1, 1 << len(statements)))
```
#### Java
diff --git a/solution/2100-2199/2151.Maximum Good People Based on Statements/Solution.py b/solution/2100-2199/2151.Maximum Good People Based on Statements/Solution.py
index c24df6f17b498..b6567e50c9bd5 100644
--- a/solution/2100-2199/2151.Maximum Good People Based on Statements/Solution.py
+++ b/solution/2100-2199/2151.Maximum Good People Based on Statements/Solution.py
@@ -1,13 +1,13 @@
class Solution:
def maximumGood(self, statements: List[List[int]]) -> int:
- def check(mask):
+ def check(mask: int) -> int:
cnt = 0
- for i, s in enumerate(statements):
- if (mask >> i) & 1:
- for j, v in enumerate(s):
- if v < 2 and ((mask >> j) & 1) != v:
+ for i, row in enumerate(statements):
+ if mask >> i & 1:
+ for j, x in enumerate(row):
+ if x < 2 and (mask >> j & 1) != x:
return 0
cnt += 1
return cnt
- return max(check(mask) for mask in range(1, 1 << len(statements)))
+ return max(check(i) for i in range(1, 1 << len(statements)))
diff --git a/solution/2900-2999/2944.Minimum Number of Coins for Fruits/README.md b/solution/2900-2999/2944.Minimum Number of Coins for Fruits/README.md
index 167074aa40056..b8198b0178a64 100644
--- a/solution/2900-2999/2944.Minimum Number of Coins for Fruits/README.md
+++ b/solution/2900-2999/2944.Minimum Number of Coins for Fruits/README.md
@@ -111,16 +111,16 @@ tags:
### 方法一：记忆化搜索
-我们定义一个函数 $dfs(i)$，表示从第 $i$ 个水果开始购买所有水果所需要的最少金币数。那么答案就是 $dfs(1)$。
+我们定义一个函数 $\textit{dfs}(i)$，表示从第 $i$ 个水果开始购买所有水果所需要的最少金币数。那么答案就是 $\textit{dfs}(1)$。
-函数 $dfs(i)$ 的执行逻辑如下：
+函数 $\textit{dfs}(i)$ 的执行逻辑如下：
-- 如果 $i \times 2 \geq n$，说明只要买第 $i - 1$ 个水果即可，剩余的水果都可以免费获得，所以返回 $prices[i - 1]$。
-- 否则，我们可以购买水果 $i$，然后在接下来的 $i + 1$ 到 $2i + 1$ 个水果中选择一个水果 $j$ 开始购买，那么 $dfs(i) = prices[i - 1] + \min_{i + 1 \le j \le 2i + 1} dfs(j)$。
+- 如果 $i \times 2 \geq n$，说明只要买第 $i - 1$ 个水果即可，剩余的水果都可以免费获得，所以返回 $\textit{prices}[i - 1]$。
+- 否则，我们可以购买水果 $i$，然后在接下来的 $i + 1$ 到 $2i + 1$ 个水果中选择一个水果 $j$ 开始购买，那么 $\textit{dfs}(i) = \textit{prices}[i - 1] + \min_{i + 1 \le j \le 2i + 1} \textit{dfs}(j)$。
为了避免重复计算，我们使用记忆化搜索的方法，将已经计算过的结果保存起来，下次遇到相同的情况时，直接返回结果即可。
-时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $prices$ 的长度。
+时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{prices}$ 的长度。
@@ -250,11 +250,11 @@ function minimumCoins(prices: number[]): number {
与方法一类似，我们定义 $f[i]$ 表示从第 $i$ 个水果开始购买所有水果所需要的最少金币数。那么答案就是 $f[1]$。
-状态转移方程为 $f[i] = \min_{i + 1 \le j \le 2i + 1} f[j] + prices[i - 1]$。
+状态转移方程为 $f[i] = \min_{i + 1 \le j \le 2i + 1} f[j] + \textit{prices}[i - 1]$。
-在实现上，我们从后往前计算，并且可以直接在数组 $prices$ 上进行状态转移，这样可以节省空间。
+时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{prices}$ 的长度。
-时间复杂度 $O(n^2)$，其中 $n$ 为数组 $prices$ 的长度。空间复杂度 $O(1)$。
+在代码实现上，我们可以直接使用 $\textit{prices}$ 数组来存储 $f$ 数组，那么空间复杂度可以优化到 $O(1)$。
@@ -334,9 +334,9 @@ function minimumCoins(prices: number[]): number {
我们观察方法二中的状态转移方程，可以发现，对于每个 $i$，我们需要求出 $f[i + 1], f[i + 2], \cdots, f[2i + 1]$ 的最小值，并且随着 $i$ 的减小，这些值的范围也在减小。这实际上是求一个单调收窄的滑动窗口的最小值，我们可以使用单调队列来优化。
-我们从后往前计算，维护一个单调递增的队列 $q$，队列中存储的是下标。如果 $q$ 的队首元素大于 $i \times 2 + 1$，说明 $i$ 之后的元素都不会被用到，所以我们将队首元素出队。如果 $i$ 不大于 $(n - 1) / 2$，那么我们可以将 $prices[q[0] - 1]$ 加到 $prices[i - 1]$ 上，然后将 $i$ 加入队尾。如果 $q$ 的队尾元素对应的水果价格大于等于 $prices[i - 1]$，那么我们将队尾元素出队，直到队尾元素对应的水果价格小于 $prices[i - 1]$ 或者队列为空，然后将 $i$ 加入队尾。
+我们从后往前计算，维护一个单调递增的队列 $q$，队列中存储的是下标。如果 $q$ 的队首元素大于 $i \times 2 + 1$，说明 $i$ 之后的元素都不会被用到，所以我们将队首元素出队。如果 $i$ 不大于 $(n - 1) / 2$，那么我们可以将 $\textit{prices}[q[0] - 1]$ 加到 $\textit{prices}[i - 1]$ 上，然后将 $i$ 加入队尾。如果 $q$ 的队尾元素对应的水果价格大于等于 $\textit{prices}[i - 1]$，那么我们将队尾元素出队，直到队尾元素对应的水果价格小于 $\textit{prices}[i - 1]$ 或者队列为空，然后将 $i$ 加入队尾。
-时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $prices$ 的长度。
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{prices}$ 的长度。
diff --git a/solution/2900-2999/2944.Minimum Number of Coins for Fruits/README_EN.md b/solution/2900-2999/2944.Minimum Number of Coins for Fruits/README_EN.md
index c269d8314e373..0de4665842e66 100644
--- a/solution/2900-2999/2944.Minimum Number of Coins for Fruits/README_EN.md
+++ b/solution/2900-2999/2944.Minimum Number of Coins for Fruits/README_EN.md
@@ -107,7 +107,18 @@ tags:
-### Solution 1
+### Solution 1: Memoization Search
+
+We define a function $\textit{dfs}(i)$ to represent the minimum number of coins needed to buy all the fruits starting from the $i$-th fruit. The answer is $\textit{dfs}(1)$.
+
+The execution logic of the function $\textit{dfs}(i)$ is as follows:
+
+- If $i \times 2 \geq n$, it means that buying the $(i - 1)$-th fruit is sufficient, and the remaining fruits can be obtained for free, so return $\textit{prices}[i - 1]$.
+- Otherwise, we can buy fruit $i$, and then choose a fruit $j$ to start buying from the next $i + 1$ to $2i + 1$ fruits. Thus, $\textit{dfs}(i) = \textit{prices}[i - 1] + \min_{i + 1 \le j \le 2i + 1} \textit{dfs}(j)$.
+
+To avoid redundant calculations, we use memoization to store the results that have already been computed. When encountering the same situation again, we directly return the result.
+
+The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{prices}$.
@@ -231,7 +242,17 @@ function minimumCoins(prices: number[]): number {
-### Solution 2
+### Solution 2: Dynamic Programming
+
+We can rewrite the memoization search in Solution 1 into a dynamic programming form.
+
+Similar to Solution 1, we define $f[i]$ to represent the minimum number of coins needed to buy all the fruits starting from the $i$-th fruit. The answer is $f[1]$.
+
+The state transition equation is $f[i] = \min_{i + 1 \le j \le 2i + 1} f[j] + \textit{prices}[i - 1]$.
+
+The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{prices}$.
+
+In the code implementation, we can directly use the $\textit{prices}$ array to store the $f$ array, thus optimizing the space complexity to $O(1)$.
@@ -307,7 +328,13 @@ function minimumCoins(prices: number[]): number {
-### Solution 3
+### Solution 3: Dynamic Programming + Monotonic Queue Optimization
+
+Observing the state transition equation in Solution 2, we can see that for each $i$, we need to find the minimum value of $f[i + 1], f[i + 2], \cdots, f[2i + 1]$. As $i$ decreases, the range of these values also decreases. This is essentially finding the minimum value in a sliding window with a narrowing range, which can be optimized using a monotonic queue.
+
+We calculate from back to front, maintaining a monotonically increasing queue $q$, where the queue stores indices. If the front element of $q$ is greater than $i \times 2 + 1$, it means that the elements after $i$ will not be used, so we dequeue the front element. If $i$ is not greater than $(n - 1) / 2$, we can add $\textit{prices}[q[0] - 1]$ to $\textit{prices}[i - 1]$, and then add $i$ to the back of the queue. If the fruit price corresponding to the back element of $q$ is greater than or equal to $\textit{prices}[i - 1]$, we dequeue the back element until the fruit price corresponding to the back element is less than $\textit{prices}[i - 1]$ or the queue is empty, then add $i$ to the back of the queue.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{prices}$.