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feat: add solutions to lc problem: No.1509
No.1509.Minimum Difference Between Largest and Smallest Value in Three Moves
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‎solution/1500-1599/1509.Minimum Difference Between Largest and Smallest Value in Three Moves/README.md

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@@ -52,22 +52,104 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：排序 + 贪心**
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我们可以先判断数组长度是否小于 $5$，如果小于 $5$，那么直接返回 $0$。
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否则，我们将数组排序，然后贪心地选择数组左边最小的 $l=[0,..3]$ 个数和右边最小的 $r = 3 - l$ 个数，取其中最小的差值 $nums[n - 1 - r] - nums[l]$ 即可。
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时间复杂度 $O(n \times \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组 `nums` 的长度。
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相似题目：
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- [2567. 修改两个元素的最小分数](/solution/2500-2599/2567.Minimum%20Score%20by%20Changing%20Two%20Elements/README.md)
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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 minDifference(self, nums: List[int]) -> int:
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n = len(nums)
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if n < 5:
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return 0
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nums.sort()
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ans = inf
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for l in range(4):
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r = 3 - l
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ans = min(ans, nums[n - 1 - r] - nums[l])
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return ans
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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 minDifference(int[] nums) {
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int n = nums.length;
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if (n < 5) {
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return 0;
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}
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Arrays.sort(nums);
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long ans = 1L << 60;
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for (int l = 0; l <= 3; ++l) {
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int r = 3 - l;
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ans = Math.min(ans, (long) nums[n - 1 - r] - nums[l]);
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}
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return (int) ans;
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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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int minDifference(vector<int>& nums) {
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int n = nums.size();
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if (n < 5) {
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return 0;
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}
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sort(nums.begin(), nums.end());
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long long ans = 1L << 60;
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for (int l = 0; l <= 3; ++l) {
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int r = 3 - l;
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ans = min(ans, 1LL * nums[n - 1 - r] - nums[l]);
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}
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return ans;
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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 minDifference(nums []int) int {
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n := len(nums)
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if n < 5 {
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return 0
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}
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sort.Ints(nums)
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ans := 1 << 60
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for l := 0; l <= 3; l++ {
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r := 3 - l
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ans = min(ans, nums[n-1-r]-nums[l])
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}
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return ans
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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/1500-1599/1509.Minimum Difference Between Largest and Smallest Value in Three Moves/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -62,13 +62,83 @@ After performing 3 moves, the difference between the minimum and maximum is 7 -
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### **Python3**
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```python
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class Solution:
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def minDifference(self, nums: List[int]) -> int:
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n = len(nums)
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if n < 5:
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return 0
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nums.sort()
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ans = inf
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for l in range(4):
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r = 3 - l
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ans = min(ans, nums[n - 1 - r] - nums[l])
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return ans
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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 minDifference(int[] nums) {
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int n = nums.length;
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if (n < 5) {
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return 0;
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}
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Arrays.sort(nums);
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long ans = 1L << 60;
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for (int l = 0; l <= 3; ++l) {
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int r = 3 - l;
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ans = Math.min(ans, (long) nums[n - 1 - r] - nums[l]);
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}
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return (int) ans;
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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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int minDifference(vector<int>& nums) {
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int n = nums.size();
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if (n < 5) {
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return 0;
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}
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sort(nums.begin(), nums.end());
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long long ans = 1L << 60;
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for (int l = 0; l <= 3; ++l) {
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int r = 3 - l;
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ans = min(ans, 1LL * nums[n - 1 - r] - nums[l]);
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}
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return ans;
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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 minDifference(nums []int) int {
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n := len(nums)
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if n < 5 {
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return 0
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}
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sort.Ints(nums)
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ans := 1 << 60
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for l := 0; l <= 3; l++ {
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r := 3 - l
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ans = min(ans, nums[n-1-r]-nums[l])
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}
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return ans
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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/1500-1599/1509.Minimum Difference Between Largest and Smallest Value in Three Moves/Solution.cpp

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class Solution {
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public:
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int minDifference(vector<int>& nums) {
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int n = nums.size();
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if (n < 5) {
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return 0;
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}
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sort(nums.begin(), nums.end());
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long long ans = 1L << 60;
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for (int l = 0; l <= 3; ++l) {
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int r = 3 - l;
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ans = min(ans, 1LL * nums[n - 1 - r] - nums[l]);
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}
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return ans;
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}
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};

‎solution/1500-1599/1509.Minimum Difference Between Largest and Smallest Value in Three Moves/Solution.go

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Original file line numberDiff line numberDiff line change
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func minDifference(nums []int) int {
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n := len(nums)
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if n < 5 {
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return 0
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}
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sort.Ints(nums)
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ans := 1 << 60
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for l := 0; l <= 3; l++ {
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r := 3 - l
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ans = min(ans, nums[n-1-r]-nums[l])
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}
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return ans
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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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}

‎solution/1500-1599/1509.Minimum Difference Between Largest and Smallest Value in Three Moves/Solution.java

Lines changed: 15 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,15 @@
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class Solution {
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public int minDifference(int[] nums) {
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int n = nums.length;
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if (n < 5) {
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return 0;
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}
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Arrays.sort(nums);
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long ans = 1L << 60;
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for (int l = 0; l <= 3; ++l) {
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int r = 3 - l;
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ans = Math.min(ans, (long) nums[n - 1 - r] - nums[l]);
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}
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return (int) ans;
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}
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}

‎solution/1500-1599/1509.Minimum Difference Between Largest and Smallest Value in Three Moves/Solution.py

Lines changed: 11 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,11 @@
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class Solution:
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def minDifference(self, nums: List[int]) -> int:
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n = len(nums)
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if n < 5:
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return 0
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nums.sort()
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ans = inf
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for l in range(4):
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r = 3 - l
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ans = min(ans, nums[n - 1 - r] - nums[l])
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return ans

‎solution/2500-2599/2567.Minimum Score by Changing Two Elements/README.md

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Original file line numberDiff line numberDiff line change
@@ -69,6 +69,10 @@
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时间复杂度 $O(n \log n)$，空间复杂度 $O(\log n)$。其中 $n$ 为数组 `nums` 的长度。
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相似题目：
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- [1509. 三次操作后最大值与最小值的最小差](/solution/1500-1599/1509.Minimum%20Difference%20Between%20Largest%20and%20Smallest%20Value%20in%20Three%20Moves/README.md)
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<!-- tabs:start -->
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### **Python3**

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