#   # 80. Remove Duplicates from Sorted Array II

## Problem statement

Given an integer array `nums` sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. The relative order of the elements should be kept the same.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array `nums`. More formally, if there are `k` elements after removing the duplicates, then the first `k` elements of `nums` should hold the final result. It does not matter what you leave beyond the first `k` elements.

Return `k` after placing the final result in the first `k` slots of `nums`.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with `O(1)` extra memory.

### Custom Judge:

The judge will test your solution with the following code:

``````int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
assert nums[i] == expectedNums[i];
}
``````

If all assertions pass, then your solution will be accepted.

### Example 1

``````Input: nums = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).
``````

### Example 2

``````Input: nums = [0,0,1,1,1,1,2,3,3]
Output: 7, nums = [0,0,1,1,2,3,3,_,_]
Explanation: Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).
``````

### Constraints

• `1 <= nums.length <= 3 * 10^4`.

• `-10^4 <= nums[i] <= 10^4`.

• `nums` is sorted in non-decreasing order.

## Solution 1: Erasing the duplicates

In order for each unique element appears at most twice, you have to erase the further appearances if they exist.

Since the array `nums` is sorted, you can determine that existence by checking if `nums[i] == nums[i-2]` for `2 <= i < nums.length`.

### Code

``````#include <vector>
#include <iostream>
using namespace std;
int removeDuplicates(vector<int>& nums) {
int i = 2;
while (i < nums.size()) {
if (nums[i] == nums[i-2]) {
int j = i;
while (j < nums.size() && nums[j] == nums[i]) {
j++;
}
nums.erase(nums.begin() + i, nums.begin() + j);
} else {
i++;
}
}
return nums.size();
}
void printResult(const int k, const vector<int>& nums) {
cout << k << ", [";
for (int i = 0; i < k ; i++) {
cout << nums[i] << ",";
}
cout << "]\n";
}
int main() {
vector<int> nums{1,1,1,2,2,3};
printResult(removeDuplicates(nums), nums);
nums = {0,0,1,1,1,1,2,3,3};
printResult(removeDuplicates(nums), nums);
}
``````
``````Output:
5, [1,1,2,2,3,]
7, [0,0,1,1,2,3,3,]
``````

### Complexity

• Runtime:

• Worst case `O(N*N/3)`, where `N = nums.size()`. The complexity of the `erase()` method is linear in `N`. The worst case is when `erase()` is called maximum `N/3` times.
``````    Example of the worst case:
nums = [1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6].
``````
• On average `O(N)` since the number of `erase()` calls is `O(1)`.
• Extra space `O(1)`.

## Solution 2: Reassigning the satisfying elements

You might need to avoid the `erase()` method in the solution above to reduce the complexity. Moreover, the problem only cares about the first `k` elements of the array `nums` after removing the duplicates.

If you look at the final result after removing duplication, the expected `nums` satisfies

``````nums[i] > nums[i-2] for 2 <= i < nums.length.
``````

You can use this invariant to reassign the array `nums` only the satisified elements.

### Code

``````#include <vector>
#include <iostream>
using namespace std;
int removeDuplicates(vector<int>& nums) {
if (nums.size() <= 2) {
return nums.size();
}
int k = 2;
int i = 2;
while (i < nums.size()) {
if (nums[i] > nums[k - 2]) {
nums[k++] = nums[i];
}
i++;
}
return k;
}
void printResult(const int k, const vector<int>& nums) {
cout << k << ", [";
for (int i = 0; i < k ; i++) {
cout << nums[i] << ",";
}
cout << "]\n";
}
int main() {
vector<int> nums{1,1,1,2,2,3};
printResult(removeDuplicates(nums), nums);
nums = {0,0,1,1,1,1,2,3,3};
printResult(removeDuplicates(nums), nums);
}
``````
``````Output:
Output:
5, [1,1,2,2,3,]
7, [0,0,1,1,2,3,3,]
``````

### Complexity

• Runtime: `O(N)`, where `N = nums.size()`.

• Extra space: `O(1)`.

## References

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