LeetCode 26

26. Remove Duplicates from Sorted Array

팁

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Problem

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.

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 = [1,1,2]
**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).

Example 2:

**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).

Constraints:

• 1 <= nums.length <= 3 * 104
• -100 <= nums[i] <= 100
• nums is sorted in non-decreasing order.

Solution

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        
        # Pointer to track the position of the unique elements
        i = 0

        # Iterate through the array starting from the second element
        for j in range(1, len(nums)):
            # If current element is different from the last unique element
            if nums[j] != nums[i]:
                # Move the pointer i forward and replace the next position with nums[j]
                i += 1
                nums[i] = nums[j]
                
        # Return the length of the array with unique elements
        return i + 1    

Time and Space Complexity

• Time complexity: O(n), where n is the length of the input array
• Space complexity: O(1), since only a fixed number of pointers (i, j) and no extra data structures are used (in-place modification)