#include #include using namespace std; /* 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 -104 <= nums[i] <= 104 nums is sorted in non-decreasing order. */ class Solution // : public testing::Test { public: int removeDuplicates(vector& nums) { if (nums.size() == 1) return 1; // Use two pointers (indices) for traversing the array size_t idx0 = 0, idx1 = 0; // If there are more than two of the same element shift // duplicate to the end of the arrray auto shiftToEnd = [&nums](size_t idx) { while (idx < nums.size() - 1) { int tmp = nums[idx]; nums[idx] = nums[idx+1]; nums[idx+1] = tmp; idx++; } }; while (idx0 < nums.size() - 1 && nums[idx0] <= nums[idx0 + 1]) { while (nums[idx1] == nums[idx0]) { idx1++; } // Test for more that two of the same element if (idx1 - idx0 > 1 ) { while (idx1 - 1 - idx0 > 1) { shiftToEnd(idx1 - 1); idx1--; } } idx0 += 1; idx1 = idx0; } return idx0 + 1; } }; int main(int argc, char** argv) { vector, int, vector>> 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 } }, { {0,0,1,1,1,1,2,3,3}, 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 initial(get<0>(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, get<1>(test)); auto first = get<0>(test).begin(); auto last = get<0>(test).begin(); last += k; vector 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; }