feat: add solutions to lc problem: No.1553

solution/1500-1599/1553.Minimum Number of Days to Eat N Oranges/README.md

@@ -69,13 +69,30 @@
 
 ### 方法一：记忆化搜索
 
+根据题目描述，对于每个 $n$，我们可以选择三种方式之一：
+
+1. 将 $n$ 减少 $1$；
+2. 如果 $n$ 能被 $2$ 整除，将 $n$ 的值除以 $2$；
+3. 如果 $n$ 能被 $3$ 整除，将 $n$ 的值除以 $3$。
+
+因此，问题等价于求解通过上述三种方式，将 $n$ 减少到 $0$ 的最少天数。
+
+我们设计一个函数 $dfs(n)$，表示将 $n$ 减少到 $0$ 的最少天数。函数 $dfs(n)$ 的执行过程如下：
+
+1. 如果 $n < 2$，返回 $n$；
+2. 否则，我们可以先通过 $n \bmod 2$ 次操作 $1$，将 $n$ 减少到 $2$ 的倍数，然后执行操作 $2$，将 $n$ 减少到 $n/2$；我们也可以先通过 $n \bmod 3$ 次操作 $1$，将 $n$ 减少到 $3$ 的倍数，然后执行操作 $3$，将 $n$ 减少到 $n/3$。我们选择上述两种方式中最少的一种，即 $1 + \min(n \bmod 2 + dfs(n/2), n \bmod 3 + dfs(n/3))$。
+
+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 $dfs(n)$ 的值存储在哈希表中。
+
+时间复杂度 $O(\log^2 n)$，空间复杂度 $O(\log^2 n)$。
+
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def minDays(self, n: int) -> int:
         @cache
-        def dfs(n):
+        def dfs(n: int) -> int:
             if n < 2:
                 return n
             return 1 + min(n % 2 + dfs(n // 2), n % 3 + dfs(n // 3))
@@ -115,8 +132,12 @@ public:
     }
 
     int dfs(int n) {
-        if (n < 2) return n;
-        if (f.count(n)) return f[n];
+        if (n < 2) {
+            return n;
+        }
+        if (f.count(n)) {
+            return f[n];
+        }
         int res = 1 + min(n % 2 + dfs(n / 2), n % 3 + dfs(n / 3));
         f[n] = res;
         return res;
@@ -140,6 +161,23 @@ func minDays(n int) int {
 }
 ```
 
+```ts
+function minDays(n: number): number {
+    const f: Record<number, number> = {};
+    const dfs = (n: number): number => {
+        if (n < 2) {
+            return n;
+        }
+        if (f[n] !== undefined) {
+            return f[n];
+        }
+        f[n] = 1 + Math.min((n % 2) + dfs((n / 2) | 0), (n % 3) + dfs((n / 3) | 0));
+        return f[n];
+    };
+    return dfs(n);
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/1500-1599/1553.Minimum Number of Days to Eat N Oranges/README_EN.md

@@ -53,15 +53,32 @@ You need at least 3 days to eat the 6 oranges.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Memoization Search
+
+According to the problem description, for each $n$, we can choose one of three ways:
+
+1. Decrease $n$ by $1$;
+2. If $n$ can be divided by $2$, divide the value of $n$ by $2$;
+3. If $n$ can be divided by $3$, divide the value of $n$ by $3$.
+
+Therefore, the problem is equivalent to finding the minimum number of days to reduce $n$ to $0$ through the above three ways.
+
+We design a function $dfs(n)$, which represents the minimum number of days to reduce $n$ to $0$. The execution process of the function $dfs(n)$ is as follows:
+
+1. If $n < 2$, return $n$;
+2. Otherwise, we can first reduce $n$ to a multiple of $2$ by $n \bmod 2$ operations of $1$, and then perform operation $2$ to reduce $n$ to $n/2$; we can also first reduce $n$ to a multiple of $3$ by $n \bmod 3$ operations of $1$, and then perform operation $3$ to reduce $n$ to $n/3$. We choose the minimum of the above two ways, that is, $1 + \min(n \bmod 2 + dfs(n/2), n \bmod 3 + dfs(n/3))$.
+
+To avoid repeated calculations, we use the method of memoization search and store the calculated values of $dfs(n)$ in a hash table.
+
+The time complexity is $O(\log^2 n)$, and the space complexity is $O(\log^2 n)$.
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def minDays(self, n: int) -> int:
         @cache
-        def dfs(n):
+        def dfs(n: int) -> int:
             if n < 2:
                 return n
             return 1 + min(n % 2 + dfs(n // 2), n % 3 + dfs(n // 3))
@@ -101,8 +118,12 @@ public:
     }
 
     int dfs(int n) {
-        if (n < 2) return n;
-        if (f.count(n)) return f[n];
+        if (n < 2) {
+            return n;
+        }
+        if (f.count(n)) {
+            return f[n];
+        }
         int res = 1 + min(n % 2 + dfs(n / 2), n % 3 + dfs(n / 3));
         f[n] = res;
         return res;
@@ -126,6 +147,23 @@ func minDays(n int) int {
 }
 ```
 
+```ts
+function minDays(n: number): number {
+    const f: Record<number, number> = {};
+    const dfs = (n: number): number => {
+        if (n < 2) {
+            return n;
+        }
+        if (f[n] !== undefined) {
+            return f[n];
+        }
+        f[n] = 1 + Math.min((n % 2) + dfs((n / 2) | 0), (n % 3) + dfs((n / 3) | 0));
+        return f[n];
+    };
+    return dfs(n);
+}
+```
+
 <!-- tabs:end -->
 
 <!-- end -->

solution/1500-1599/1553.Minimum Number of Days to Eat N Oranges/Solution.cpp

@@ -7,8 +7,12 @@ class Solution {
     }
 
     int dfs(int n) {
-        if (n < 2) return n;
-        if (f.count(n)) return f[n];
+        if (n < 2) {
+            return n;
+        }
+        if (f.count(n)) {
+            return f[n];
+        }
         int res = 1 + min(n % 2 + dfs(n / 2), n % 3 + dfs(n / 3));
         f[n] = res;
         return res;

solution/1500-1599/1553.Minimum Number of Days to Eat N Oranges/Solution.py

@@ -1,7 +1,7 @@
 class Solution:
     def minDays(self, n: int) -> int:
         @cache
-        def dfs(n):
+        def dfs(n: int) -> int:
             if n < 2:
                 return n
             return 1 + min(n % 2 + dfs(n // 2), n % 3 + dfs(n // 3))

solution/1500-1599/1553.Minimum Number of Days to Eat N Oranges/Solution.ts

@@ -0,0 +1,14 @@
+function minDays(n: number): number {
+    const f: Record<number, number> = {};
+    const dfs = (n: number): number => {
+        if (n < 2) {
+            return n;
+        }
+        if (f[n] !== undefined) {
+            return f[n];
+        }
+        f[n] = 1 + Math.min((n % 2) + dfs((n / 2) | 0), (n % 3) + dfs((n / 3) | 0));
+        return f[n];
+    };
+    return dfs(n);
+}