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Visit original link: 26. Remove Duplicates from Sorted Array - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions for a better experience!

LeetCode link: 26. Remove Duplicates from Sorted Array, difficulty: Easy.

Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums.

Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things:

• Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums.
• Return k.

Input: nums = [1,1,2]

Output: 2, nums = [1,2,_]

Explanation:

Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.

It does not matter what you leave beyond the returned k (hence they are underscores).

Input: nums = [0,0,1,1,1,2,2,3,3,4]

Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]

Explanation:

Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.

It does not matter what you leave beyond the returned k (hence they are underscores).

• 1 <= nums.length <= 3 * 10^4
• -10^4 <= nums[i] <= 10^4
• nums is sorted in non-decreasing order.

The numbers in the array are already sorted, so any duplicate values must appear consecutively.

To remove duplicates, we need to keep every number that is different from the previous one, and discard the rest.

Since the array needs to be modified in place, two pointers are required: one moves fast and the other moves slow.

the fast pointer increases its position by 1 each time, while the slow pointer marks the current position to be modified.

int slow = 0; // slow pointer
// ...

for (int fast = 1; fast < array_length; fast++) { // This line is the key!
    if (condition_1) {
        // ...
        continue; // 'continue' is better than 'else' because 'else' will introduce more indents
    }

    if (condition_2) {
        // ...
        continue; // 'continue' is better than 'else'
    }

    // condition_3
    // ...
    slow++;
    // ...
}

return something;

• Time complexity: O(N).
• Space complexity: O(1).

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        slow = 1

        # nums = [0,0,1,1,1,2,2,3,3,4]
        for fast in range(1, len(nums)):
            if nums[fast] == nums[slow - 1]:
                continue

            nums[slow] = nums[fast]
            slow += 1

        return slow

# @param {Integer[]} nums
# @return {Integer}
def remove_duplicates(nums)
  slow = 1

  # nums = [0,0,1,1,1,2,2,3,3,4]
  (1...nums.size).each do |fast|
    if nums[fast] == nums[slow - 1]
      next
    end

    nums[slow] = nums[fast]
    slow += 1
  end

  slow
end

class Solution {
    public int removeDuplicates(int[] nums) {
        var slow = 1;

        // nums = [0,0,1,1,1,2,2,3,3,4]
        for (var fast = 1; fast < nums.length; fast++) {
            if (nums[fast] != nums[slow - 1]) {
                nums[slow] = nums[fast];
                slow++;
            }
        }

        return slow;
    }
}

// Welcome to create a PR to complete the code of this language, thanks!

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Visit original link: 26. Remove Duplicates from Sorted Array - LeetCode Python/Java/C++/JS/C#/Go/Ruby Solutions for a better experience!

GitHub repository: leetcode-python-java.