LeetCode Problem 80 - Remove Duplicates from Sorted Array II

Remove Duplicates from Sorted Array II

Problem

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];
}

Solution

class Solution{
public:
    int removeDuplicates(vector<int> &nums)
    {
        int prev = nums[0];
        int count = 1;
        int k = 1;
        for (int i = 1; i < nums.size(); i++)
            if (prev != nums[i])
                count = 1;
                prev = nums[i];
                nums[k++] = nums[i];
            else if (count++ < 2)
                nums[k++] = nums[i];
        return k;
    }
};

Explanation

We need to track two things here.

  1. How many of a value there is
  2. Where the next insert position needs to be
    We set k equal to one since we know that at a minimum we will return a value of 1.
    if previous does not equal the value at index i, we reset count, update previous, and assign nums[k] to nums[i] then increment k
    Otherwise, we just assign nums[k] to nums[i];