# [LeetCode] Top Interview 150 Problems Solving # 26 Remove Duplicates from Sorted Array

# **Understanding the Problem**

A given list `nums` has numbers from -100 to 100 in an arranged order. The task is to change `nums` in place by finding duplicate numbers and removing them. `nums` should stay with unique numbers only. It is not important to return any value from the function as this problem only deals with `nums`.

# **Approach**

Should I go with `remove()` method or `replace()`? `remove()` will cause an error in the loop so I decided to replace the duplicate numbers into something that could be marked as *duplicate* and removed later.

# **Solution**

1. I wanted to keep the unique numbers somewhere as `temp`, to use it for checking other numbers whether they are duplicate or not.
    
2. I would go by `for` loop while checking unique numbers. If the number `nums[i]` is not unique and has been duplicate, I changed it with `-1000`.
    
3. I sorted the number to have `-1000` numbers to the starting of the index.
    
4. Then I redeclared `nums` with the sliced `nums`, which sliced all `-1000` numbers.
    

```python
class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        temp = []
        for i in range(len(nums)):
            if nums[i] not in temp:
                temp.append(nums[i])
            else:
                nums[i] = -1000

        nums.sort()
        count = nums.count(-1000)
        nums[:] = nums[count:]
        return
```

# **Reflection**

Changing the number *in place* is a little bit tricky, and the methods may be not working the way I think it should. I guess it is only the matter of practice.

Another thing that I noticed on LeetCode is, that there is a lack of explanation so in some problems I myself need to find a way to break through the barriers towards the solution, interpreting what the actual instruction is and what to return. This problem too. It tells me to return the `k` which is the length of the unique numbers but it was nothing about the `k`. Well, I guess I will figure solution out myself and just go with it.