0026 Remove Duplicates

0026 Remove Duplicates#

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.

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.

Examples#

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

Analysis#

  • The key constraint here is we need remove duplicates in-place.

  • Use fast and slow pointer:

    • The fast pointer moves forward, and find a non-duplicated element

    • Then assign the non-duplicate to the slow pointer, and move slow pointer one step forward

    • This makes sure that nums[0:slow] are nonduplicated.

Solution#

def removeDuplicates(nums):
    if len(nums) <= 1:
        return len(nums)

    slow, fast = 0, 1
    while fast < len(nums):
        # slow pointer element is less than the fast pointer element 
        # then no duplicates and moves both forward
        if nums[slow] != nums[fast]:
            slow += 1
            # swap the values at slow and fast
            nums[slow], nums[fast] = nums[fast], nums[slow]

        # move fast
        fast += 1
    
    return slow + 1

# test
nums = [1,1,2]
print(removeDuplicates(nums))

nums = [0,0,1,1,1,2,2,3,3,4]
print(removeDuplicates(nums))

2
5
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