/*
LeeCode 80. Remove Duplicates from Sorted Array II
  
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 * 104
  * -10^9 <= nums[i] <= 10^9
  * nums is sorted in non-decreasing order.

Problem Restatement (for the interview)


  Optimal Two-Pointer Solution (Most Interview-Friendly) C++
    * read pointer iterates through all elements
        for (size_t i = 2; i < nums.size(); i++)
    * write pointer k
       will become new length
  
*/

#include <bits/stdc++.h>
#include <gtest/gtest.h>

using namespace std;

void print(const string& label, vector<int>& v);
void print(const string& label, vector<int>& v1, vector<int>& v2);

class Solution // : public testing::Test
{        
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.size() <= 2) return nums.size();
        
        int k = 2;
        for (size_t i = 2; i < nums.size(); i++)
        {
            if (nums[k-2] != nums[i])
            {
                nums[k++] = nums[i];
            }
        }
        return k;

#ifdef xAI
        if (nums.empty()) return 0;
        int k = 2;  // write pointer (new length)        
        for (int num : nums) // read pointer iterates through all elements
        {                
            // Allow the element if:
            //   - We're still filling the first two positions, OR
            //   - It doesn't match the element two positions back (prevents 3+ duplicates)
            if (k < 2 || num != nums[k - 2])
            {
                nums[k++] = num;
            }
        }
        return k;
#endif
    }
};


int main(int argc, char** argv)
{    
    vector<tuple<vector<int>, int, vector<int>>> testCases = {
        //  vector input,            k   vector output
        {   {1},                     1,  {1} },
        {   {-2,-1},                 2,  {-2,-1} },
        {   {1,1,1,1,2,2,2,2},       4,  {1,1,2,2 } },
        {   {1,1,1,2,2,3},           5,  {1,1,2,2,3 } },
        {   {0,0,0,0,0,1,1,1,2,3,3}, 7,  {0,0,1,1,2,3,3} },
        {   {0,0,0,1,1,1,2,2,2},     6,  {0,0,1,1,2,2} },
    };

    [[maybe_unused]] auto useGoogleTest = [](int argc, char** argv)
    {
        testing::InitGoogleTest(&argc, argv);
    
        int ret = RUN_ALL_TESTS();
    
        return ret;
    };

    Solution s;

    int k;
    
    for (auto test: testCases)
    {
        vector<int> initial(get<0>(test));
        int expected_k = get<1>(test);
        
        k = s.removeDuplicates(get<0>(test));

        cout << setw(20) << "k expected: " << get<1>(test) << endl;
        cout << setw(20) << "k result: " << k << endl;
        EXPECT_EQ(k, expected_k);
        
        auto first = get<0>(test).begin();
        auto last = get<0>(test).begin();
        last += k;
        vector<int> result(first, last);
 
        cout << setw(20) << "vector expected: ";
        for (auto i: get<2>(test)) cout << i << " ";
        cout << endl;
        cout << setw(20) << "vector result: ";
        for (auto i: result) cout << i << " ";
        cout << endl;

        cout << setw(20) << "initial vector: ";
        for (auto i: initial) cout << i << " ";
        cout << endl;
        cout << setw(20) << "modified vector: ";
        for (auto i: get<0>(test)) cout << i << " ";
        cout << endl;

        EXPECT_EQ(result, get<2>(test));
        cout << "---------------------------" << endl;
    }
    
    return 0;
}

void print(const string& label, vector<int>& v)
{
    string s(label + "[");
    for (auto& i: v) s += to_string(i) + ",";
    s += "] ";
    cout << s << endl;
}

void print(const string& label, vector<int>& v1, vector<int>& v2)
{
    string s(label + "[");
    for (auto& i: v1) { s += to_string(i) + ","; }
    s += "] [";
    for (auto& i: v2) { s += to_string(i) + ","; }
    s += "] ";
    cout << s << endl;
}