80. Remove Duplicates from Sorted Array II

Given a sorted array nums , remove the duplicates in- place such that duplicates appeared at most twice and return the new length.

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

Clarification:

Confused why the returned value is an integer, but your answer is an array?

Note that the input array is passed in by reference , which means a modification to the input array will be known to the caller.

Internally you can think of this:

// **nums** is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to **nums** in your function would be known by the caller.
// using the length returned by your function, it prints the first **len** elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

Example 1:

Input: nums = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3]
Explanation: Your function should return length = **5** , with the first five elements of _nums_ being **1, 1, 2, 2** and **3** respectively. It doesn't matter what you leave beyond the returned length.

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 length = **7** , with the first seven elements of _nums_ being modified to  **0** , **0** , **1** , **1** , **2** , **3** and  **3** respectively. It doesn't matter what values are set beyond the returned length.

Constraints:

• 0 <= nums.length <= 3 * 104
• -104 <= nums[i] <= 104
• nums is sorted in ascending order.

# @lc code=start
using LeetCode

function remove_duplicates2!(nums::Vector{Int})::Int
    j = 3
    for i in 3:length(nums)
        if nums[i] != nums[j - 2]
            nums[j] = nums[i]
            j += 1
        end
    end

    return j - 1
end
# @lc code=end

remove_duplicates2! (generic function with 1 method)

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