feat: add solutions to lc problem: No.2171 (doocs#2234)

No.2171.Removing Minimum Number of Magic Beans

Author: yanglbme

solution/2100-2199/2171.Removing Minimum Number of Magic Beans/README.md

@@ -57,28 +57,29 @@
 
 ## 解法
 
-### 方法一：排序求和
+### 方法一：排序 + 枚举
+
+我们可以将所有袋子中的魔法豆按照从小到大的顺序排列，然后枚举每个袋子的魔法豆数目 $beans[i]$ 作为最终袋子中魔法豆数目，那么一共剩余的魔法豆数目为 $beans[i] \times (n - i)$，因此需要拿出的魔法豆数目为 $s - beans[i] \times (n - i)$，其中 $s$ 为所有袋子中魔法豆的总数。我们求出所有方案中需要拿出的魔法豆数目的最小值即可。
+
+时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为袋子的数目。
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def minimumRemoval(self, beans: List[int]) -> int:
         beans.sort()
-        ans = s = sum(beans)
-        n = len(beans)
-        for i, v in enumerate(beans):
-            ans = min(ans, s - v * (n - i))
-        return ans
+        s, n = sum(beans), len(beans)
+        return min(s - x * (n - i) for i, x in enumerate(beans))
 ```
 
 ```java
 class Solution {
     public long minimumRemoval(int[] beans) {
         Arrays.sort(beans);
         long s = 0;
-        for (int v : beans) {
-            s += v;
+        for (int x : beans) {
+            s += x;
         }
         long ans = s;
         int n = beans.length;
@@ -98,7 +99,9 @@ public:
         long long s = accumulate(beans.begin(), beans.end(), 0ll);
         long long ans = s;
         int n = beans.size();
-        for (int i = 0; i < n; ++i) ans = min(ans, s - 1ll * beans[i] * (n - i));
+        for (int i = 0; i < n; ++i) {
+            ans = min(ans, s - 1ll * beans[i] * (n - i));
+        }
         return ans;
     }
 };
@@ -108,27 +111,26 @@ public:
 func minimumRemoval(beans []int) int64 {
 	sort.Ints(beans)
 	s := 0
-	for _, v := range beans {
-		s += v
+	for _, x := range beans {
+		s += x
 	}
 	ans := s
 	n := len(beans)
-	for i, v := range beans {
-		ans = min(ans, s-v*(n-i))
+	for i, x := range beans {
+		ans = min(ans, s-x*(n-i))
 	}
 	return int64(ans)
 }
 ```
 
 ```ts
 function minimumRemoval(beans: number[]): number {
-    const n = beans.length;
-    let sum = beans.reduce((a, c) => a + c, 0);
     beans.sort((a, b) => a - b);
-    let ans = sum;
-    for (let i = 0; i < n; i++) {
-        let num = beans[i];
-        ans = Math.min(sum - num * (n - i), ans);
+    const s = beans.reduce((a, b) => a + b, 0);
+    const n = beans.length;
+    let ans = s;
+    for (let i = 0; i < n; ++i) {
+        ans = Math.min(ans, s - beans[i] * (n - i));
     }
     return ans;
 }

solution/2100-2199/2171.Removing Minimum Number of Magic Beans/README_EN.md

@@ -53,28 +53,29 @@ There are no other solutions that removes 7 beans or fewer.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Sorting + Enumeration
+
+We can sort all the beans in the bags in ascending order, and then enumerate the number of beans $beans[i]$ in each bag as the final number of beans in the bag. The total remaining number of beans is $beans[i] \times (n - i)$, so the number of beans that need to be taken out is $s - beans[i] \times (n - i)$, where $s$ is the total number of beans in all bags. We need to find the minimum number of beans that need to be taken out among all schemes.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the number of bags.
 
 <!-- tabs:start -->
 
 ```python
 class Solution:
     def minimumRemoval(self, beans: List[int]) -> int:
         beans.sort()
-        ans = s = sum(beans)
-        n = len(beans)
-        for i, v in enumerate(beans):
-            ans = min(ans, s - v * (n - i))
-        return ans
+        s, n = sum(beans), len(beans)
+        return min(s - x * (n - i) for i, x in enumerate(beans))
 ```
 
 ```java
 class Solution {
     public long minimumRemoval(int[] beans) {
         Arrays.sort(beans);
         long s = 0;
-        for (int v : beans) {
-            s += v;
+        for (int x : beans) {
+            s += x;
         }
         long ans = s;
         int n = beans.length;
@@ -94,7 +95,9 @@ public:
         long long s = accumulate(beans.begin(), beans.end(), 0ll);
         long long ans = s;
         int n = beans.size();
-        for (int i = 0; i < n; ++i) ans = min(ans, s - 1ll * beans[i] * (n - i));
+        for (int i = 0; i < n; ++i) {
+            ans = min(ans, s - 1ll * beans[i] * (n - i));
+        }
         return ans;
     }
 };
@@ -104,27 +107,26 @@ public:
 func minimumRemoval(beans []int) int64 {
 	sort.Ints(beans)
 	s := 0
-	for _, v := range beans {
-		s += v
+	for _, x := range beans {
+		s += x
 	}
 	ans := s
 	n := len(beans)
-	for i, v := range beans {
-		ans = min(ans, s-v*(n-i))
+	for i, x := range beans {
+		ans = min(ans, s-x*(n-i))
 	}
 	return int64(ans)
 }
 ```
 
 ```ts
 function minimumRemoval(beans: number[]): number {
-    const n = beans.length;
-    let sum = beans.reduce((a, c) => a + c, 0);
     beans.sort((a, b) => a - b);
-    let ans = sum;
-    for (let i = 0; i < n; i++) {
-        let num = beans[i];
-        ans = Math.min(sum - num * (n - i), ans);
+    const s = beans.reduce((a, b) => a + b, 0);
+    const n = beans.length;
+    let ans = s;
+    for (let i = 0; i < n; ++i) {
+        ans = Math.min(ans, s - beans[i] * (n - i));
     }
     return ans;
 }

solution/2100-2199/2171.Removing Minimum Number of Magic Beans/Solution.cpp

@@ -5,7 +5,9 @@ class Solution {
         long long s = accumulate(beans.begin(), beans.end(), 0ll);
         long long ans = s;
         int n = beans.size();
-        for (int i = 0; i < n; ++i) ans = min(ans, s - 1ll * beans[i] * (n - i));
+        for (int i = 0; i < n; ++i) {
+            ans = min(ans, s - 1ll * beans[i] * (n - i));
+        }
         return ans;
     }
 };

solution/2100-2199/2171.Removing Minimum Number of Magic Beans/Solution.go

@@ -1,13 +1,13 @@
 func minimumRemoval(beans []int) int64 {
 	sort.Ints(beans)
 	s := 0
-	for _, v := range beans {
-		s += v
+	for _, x := range beans {
+		s += x
 	}
 	ans := s
 	n := len(beans)
-	for i, v := range beans {
-		ans = min(ans, s-v*(n-i))
+	for i, x := range beans {
+		ans = min(ans, s-x*(n-i))
 	}
 	return int64(ans)
 }

solution/2100-2199/2171.Removing Minimum Number of Magic Beans/Solution.java

@@ -2,8 +2,8 @@ class Solution {
     public long minimumRemoval(int[] beans) {
         Arrays.sort(beans);
         long s = 0;
-        for (int v : beans) {
-            s += v;
+        for (int x : beans) {
+            s += x;
         }
         long ans = s;
         int n = beans.length;

solution/2100-2199/2171.Removing Minimum Number of Magic Beans/Solution.py

@@ -1,8 +1,5 @@
 class Solution:
     def minimumRemoval(self, beans: List[int]) -> int:
         beans.sort()
-        ans = s = sum(beans)
-        n = len(beans)
-        for i, v in enumerate(beans):
-            ans = min(ans, s - v * (n - i))
-        return ans
+        s, n = sum(beans), len(beans)
+        return min(s - x * (n - i) for i, x in enumerate(beans))

solution/2100-2199/2171.Removing Minimum Number of Magic Beans/Solution.ts

@@ -1,11 +1,10 @@
 function minimumRemoval(beans: number[]): number {
-    const n = beans.length;
-    let sum = beans.reduce((a, c) => a + c, 0);
     beans.sort((a, b) => a - b);
-    let ans = sum;
-    for (let i = 0; i < n; i++) {
-        let num = beans[i];
-        ans = Math.min(sum - num * (n - i), ans);
+    const s = beans.reduce((a, b) => a + b, 0);
+    const n = beans.length;
+    let ans = s;
+    for (let i = 0; i < n; ++i) {
+        ans = Math.min(ans, s - beans[i] * (n - i));
     }
     return ans;
 }