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Commit 973ee6a

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feat: add solutions to lc problem: No.2091
No.2091.Removing Minimum and Maximum From Array
1 parent 503b6be commit 973ee6a

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6 files changed

+240
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lines changed

6 files changed

+240
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‎solution/2000-2099/2091.Removing Minimum and Maximum From Array/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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先找出最小值和最大值的下标 mi, mx。如果 mi 下标大于 mx,则将 mx 与 mi 两数进行交换。
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最小删除的次数,共有 3 种情况:
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1. 从左侧往右依次删除 `nums[mi]` 和 `nums[mx]`
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1. 从右侧往左依次删除 `nums[mx]` 和 `nums[mi]`
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1. 从左侧往右删除 `nums[mi]`,从右侧往左删除 `nums[mx]`
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求这 3 种情况的最小值即可。
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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 minimumDeletions(self, nums: List[int]) -> int:
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mi = mx = 0
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for i, num in enumerate(nums):
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if num < nums[mi]:
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mi = i
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if num > nums[mx]:
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mx = i
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if mi > mx:
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mi, mx = mx, mi
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return min(mx + 1, len(nums) - mi, mi + 1 + len(nums) - mx)
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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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public int minimumDeletions(int[] nums) {
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int mi = 0, mx = 0, n = nums.length;
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for (int i = 0; i < n; ++i) {
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if (nums[i] < nums[mi]) {
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mi = i;
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}
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if (nums[i] > nums[mx]) {
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mx = i;
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}
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}
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if (mi > mx) {
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int t = mx;
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mx = mi;
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mi = t;
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}
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return Math.min(Math.min(mx + 1, n - mi), mi + 1 + n - mx);
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}
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}
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```
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### **TypeScript**
@@ -99,6 +137,56 @@ function minimumDeletions(nums: number[]): number {
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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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int minimumDeletions(vector<int>& nums) {
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int mi = 0, mx = 0, n = nums.size();
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for (int i = 0; i < n; ++i)
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{
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if (nums[i] < nums[mi]) mi = i;
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if (nums[i] > nums[mx]) mx = i;
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}
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if (mi > mx)
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{
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int t = mi;
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mi = mx;
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mx = t;
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}
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return min(min(mx + 1, n - mi), mi + 1 + n - mx);
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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 minimumDeletions(nums []int) int {
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mi, mx, n := 0, 0, len(nums)
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for i, num := range nums {
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if num < nums[mi] {
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mi = i
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}
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if num > nums[mx] {
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mx = i
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}
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}
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if mi > mx {
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mi, mx = mx, mi
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}
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return min(min(mx+1, n-mi), mi+1+n-mx)
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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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### **...**
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```

‎solution/2000-2099/2091.Removing Minimum and Maximum From Array/README_EN.md‎

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@@ -63,13 +63,41 @@ We can remove it with 1 deletion.
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### **Python3**
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```python
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class Solution:
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def minimumDeletions(self, nums: List[int]) -> int:
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mi = mx = 0
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for i, num in enumerate(nums):
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if num < nums[mi]:
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mi = i
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if num > nums[mx]:
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mx = i
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if mi > mx:
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mi, mx = mx, mi
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return min(mx + 1, len(nums) - mi, mi + 1 + len(nums) - mx)
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```
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### **Java**
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```java
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class Solution {
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public int minimumDeletions(int[] nums) {
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int mi = 0, mx = 0, n = nums.length;
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for (int i = 0; i < n; ++i) {
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if (nums[i] < nums[mi]) {
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mi = i;
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}
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if (nums[i] > nums[mx]) {
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mx = i;
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}
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}
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if (mi > mx) {
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int t = mx;
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mx = mi;
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mi = t;
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}
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return Math.min(Math.min(mx + 1, n - mi), mi + 1 + n - mx);
99+
}
100+
}
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```
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### **TypeScript**
@@ -82,13 +110,60 @@ function minimumDeletions(nums: number[]): number {
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let j = nums.indexOf(Math.max(...nums));
83111
let left = Math.min(i, j);
84112
let right = Math.max(i, j);
85-
// 左右 left + 1 + n - right
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// 两个都是左边 left + 1 + right - left = right + 1
87-
// 都是右边 n - right + right - left = n - left
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return Math.min(left + 1 + n - right, right + 1, n - left);
89114
};
90115
```
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### **C++**
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```cpp
120+
class Solution {
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public:
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int minimumDeletions(vector<int>& nums) {
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int mi = 0, mx = 0, n = nums.size();
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for (int i = 0; i < n; ++i)
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{
126+
if (nums[i] < nums[mi]) mi = i;
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if (nums[i] > nums[mx]) mx = i;
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}
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if (mi > mx)
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{
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int t = mi;
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mi = mx;
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mx = t;
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}
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return min(min(mx + 1, n - mi), mi + 1 + n - mx);
136+
}
137+
};
138+
```
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### **Go**
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```go
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func minimumDeletions(nums []int) int {
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mi, mx, n := 0, 0, len(nums)
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for i, num := range nums {
146+
if num < nums[mi] {
147+
mi = i
148+
}
149+
if num > nums[mx] {
150+
mx = i
151+
}
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}
153+
if mi > mx {
154+
mi, mx = mx, mi
155+
}
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return min(min(mx+1, n-mi), mi+1+n-mx)
157+
}
158+
159+
func min(a, b int) int {
160+
if a < b {
161+
return a
162+
}
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return b
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}
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```
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92167
### **...**
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94169
```
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@@ -0,0 +1,18 @@
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class Solution {
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public:
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int minimumDeletions(vector<int>& nums) {
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int mi = 0, mx = 0, n = nums.size();
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for (int i = 0; i < n; ++i)
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{
7+
if (nums[i] < nums[mi]) mi = i;
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if (nums[i] > nums[mx]) mx = i;
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}
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if (mi > mx)
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{
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int t = mi;
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mi = mx;
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mx = t;
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}
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return min(min(mx + 1, n - mi), mi + 1 + n - mx);
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}
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};
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func minimumDeletions(nums []int) int {
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mi, mx, n := 0, 0, len(nums)
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for i, num := range nums {
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if num < nums[mi] {
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mi = i
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}
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if num > nums[mx] {
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mx = i
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}
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}
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if mi > mx {
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mi, mx = mx, mi
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}
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return min(min(mx+1, n-mi), mi+1+n-mx)
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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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class Solution {
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public int minimumDeletions(int[] nums) {
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int mi = 0, mx = 0, n = nums.length;
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for (int i = 0; i < n; ++i) {
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if (nums[i] < nums[mi]) {
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mi = i;
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}
8+
if (nums[i] > nums[mx]) {
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mx = i;
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}
11+
}
12+
if (mi > mx) {
13+
int t = mx;
14+
mx = mi;
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mi = t;
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}
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return Math.min(Math.min(mx + 1, n - mi), mi + 1 + n - mx);
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}
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}
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@@ -0,0 +1,11 @@
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class Solution:
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def minimumDeletions(self, nums: List[int]) -> int:
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mi = mx = 0
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for i, num in enumerate(nums):
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if num < nums[mi]:
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mi = i
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if num > nums[mx]:
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mx = i
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if mi > mx:
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mi, mx = mx, mi
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return min(mx + 1, len(nums) - mi, mi + 1 + len(nums) - mx)

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