26. Remove Duplicates from Sorted Array

Question

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 .
Consider the number of unique elements in nums to be k . After removing duplicates, return the number of unique elements k .
The first k elements of nums should contain the unique numbers in sorted order . The remaining elements beyond index k - 1 can be ignored.
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 * 10 4
-100 <= nums[i] <= 100
nums is sorted in non-decreasing order.

Solutions

int removeDuplicates(vector<int>& nums) {
    // two pointer approach:
    // slow pointer points to the place need to be replaced.
    // fast pointer iterate through the nums.
    // It ensurs [0, left) have no duplicated numbers.
    // Edge cases:
    // 1. when no elements in nums -> no for loop process, direct return 0;
    // 2. when there's only one element -> no way duplicated. direct return 1;
    // 3. when all elements are duplicated -> reduced to slow = 1 and return.
    // Time Complexity: O(N)
    // Space Complexity: O(1)
    int n = static_cast<int>(nums.size());
    if (n <= 1)
      return n;
    int slow = 1;
    for (int fast = 1; fast < n; ++fast) {
      if (nums[fast] != nums[slow - 1]) {
        nums[slow] = nums[fast];
        ++slow;
      }
    }
    return slow;
}