26. Remove Duplicates from Sorted Array

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.

https://leetcode.cn/problems/remove-duplicates-from-sorted-array/

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.

Test Cases

1
2
3

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

solution_test.py

import pytest

from solution import Solution


@pytest.mark.parametrize('nums, expected, expectedNums', [
    ([1,1,2], 2, [1,2]),
    ([0,0,1,1,1,2,2,3,3,4], 5, [0,1,2,3,4]),
])
@pytest.mark.parametrize('sol', [Solution()])
def test_solution(sol, nums, expected, expectedNums):
    nums = nums.copy()
    assert sol.removeDuplicates(nums) == expected
    assert nums[:expected] == expectedNums

Thoughts

用两个指针（数组下标），r 用来读取，l 用来 in-place 写入去重后的数字。

第一个元素不需要去重，所以 l 和 r 可以都从 1 开始。对于当前的 l 和 r，如果 nums[r] 与 nums[r-1] 相等，则是重复的数字，需要丢弃，否则在位置 l 写入 nums[l]。

时间复杂度 O(n)，空间复杂度 O(1)。

Code

solution.py

class Solution:
    def removeDuplicates(self, nums: list[int]) -> int:
        n = len(nums)
        l = r = 1
        while r < n:
            if nums[r] != nums[r - 1]:
                nums[l] = nums[r]
                l += 1
            r += 1

        return l