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feat: add solutions to lc problem: No.2009.Minimum Number of Operations to Make Array Continuous (doocs#571)
 * feat: add cpp solution to lc problem: No.2009 * feat: add java solution to lc problem: No.2009 * feat: add python3 solution to lc problem: No.2009
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‎solution/2000-2099/2009.Minimum Number of Operations to Make Array Continuous/README.md

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@@ -68,15 +68,66 @@
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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 minOperations(self, nums: List[int]) -> int:
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n = len(nums)
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nums = sorted(set(nums))
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ans = n
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for i, start in enumerate(nums):
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end = start + n - 1
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j = bisect_right(nums, end)
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remainLen = j - i
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ans = min(ans, n - remainLen)
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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 minOperations(int[] nums) {
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int N = nums.length;
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if (N == 1) return 0;
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Arrays.sort(nums);
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int M = 1;
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for (int i = 1; i < N; i++) {
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if (nums[i] != nums[i - 1])
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nums[M++] = nums[i];
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}
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int j = 0;
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int ans = N;
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for (int i = 0; i < M; i++) {
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while (j < M && nums[j] <= N + nums[i] - 1) j++;
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ans = Math.min(ans, N - j + i);
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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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### **C++**
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```cpp
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class Solution {
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public:
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int minOperations(vector<int>& nums) {
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sort(nums.begin(), nums.end());
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int End = unique(nums.begin(), nums.end()) - nums.begin();
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int n = nums.size();
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int len = 0;
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for (int i = 0; i < End; ++i) {
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int temp = upper_bound(nums.begin(), nums.begin() + End, n + nums[i] - 1) - nums.begin() - i;
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len = max(len, temp);
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}
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return n - len;
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}
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};
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```
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### **...**

‎solution/2000-2099/2009.Minimum Number of Operations to Make Array Continuous/README_EN.md

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@@ -63,13 +63,64 @@ The resulting array is [1,2,3,4], which is continuous.
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### **Python3**
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```python
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class Solution:
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def minOperations(self, nums: List[int]) -> int:
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n = len(nums)
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nums = sorted(set(nums))
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ans = n
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for i, start in enumerate(nums):
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end = start + n - 1
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j = bisect_right(nums, end)
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remainLen = j - i
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ans = min(ans, n - remainLen)
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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 minOperations(int[] nums) {
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int N = nums.length;
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if (N == 1) return 0;
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Arrays.sort(nums);
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int M = 1;
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for (int i = 1; i < N; i++) {
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if (nums[i] != nums[i - 1])
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nums[M++] = nums[i];
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}
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int j = 0;
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int ans = N;
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for (int i = 0; i < M; i++) {
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while (j < M && nums[j] <= N + nums[i] - 1) j++;
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ans = Math.min(ans, N - j + i);
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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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### **C++**
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```cpp
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class Solution {
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public:
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int minOperations(vector<int>& nums) {
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sort(nums.begin(), nums.end());
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int End = unique(nums.begin(), nums.end()) - nums.begin();
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int n = nums.size();
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int len = 0;
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for (int i = 0; i < End; ++i) {
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int temp = upper_bound(nums.begin(), nums.begin() + End, n + nums[i] - 1) - nums.begin() - i;
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len = max(len, temp);
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}
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return n - len;
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}
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};
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```
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### **...**

‎solution/2000-2099/2009.Minimum Number of Operations to Make Array Continuous/Solution.cpp

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class Solution {
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public:
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int minOperations(vector<int>& nums) {
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sort(nums.begin(), nums.end());
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int End = unique(nums.begin(), nums.end()) - nums.begin();
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int n = nums.size();
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int len = 0;
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for (int i = 0; i < End; ++i) {
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int temp = upper_bound(nums.begin(), nums.begin() + End, n + nums[i] - 1) - nums.begin() - i;
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len = max(len, temp);
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}
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return n - len;
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}
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};

‎solution/2000-2099/2009.Minimum Number of Operations to Make Array Continuous/Solution.java

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class Solution {
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public int minOperations(int[] nums) {
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int N = nums.length;
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if (N == 1) return 0;
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Arrays.sort(nums);
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int M = 1;
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for (int i = 1; i < N; i++) {
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if (nums[i] != nums[i - 1])
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nums[M++] = nums[i];
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}
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int j = 0;
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int ans = N;
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for (int i = 0; i < M; i++) {
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while (j < M && nums[j] <= N + nums[i] - 1)
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j++;
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ans = Math.min(ans, N - j + i);
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}
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return ans;
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}
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}

‎solution/2000-2099/2009.Minimum Number of Operations to Make Array Continuous/Solution.py

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class Solution:
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def minOperations(self, nums: List[int]) -> int:
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n = len(nums)
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nums = sorted(set(nums))
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ans = n
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for i, start in enumerate(nums):
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end = start + n - 1
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j = bisect_right(nums, end)
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remainLen = j - i
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ans = min(ans, n - remainLen)
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return ans

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