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Question
Solution

Given a sorted array, remove the duplicates in-place such that each element appear only once and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Example:

Given nums = [1,1,2],

Your function should return length = 2, with the first two elements of nums being 1 and 2 respectively.
It doesn't matter what you leave beyond the new length.

Solution
- Approach #1 (Two Pointers) [Accepted]

Solution

Approach #1 (Two Pointers) [Accepted]

Algorithm

Since the array is already sorted, we can keep two pointers $i$ and $j$ , where $i$ is the slow-runner while $j$ is the fast-runner. As long as $nums[i] = nums[j]$ , we increment $j$ to skip the duplicate.

When we encounter $nums[j] \neq nums[i]$ , the duplicate run has ended so we must copy its value to $nums[i + 1]$ . $i$ is then incremented and we repeat the same process again until $j$ reaches the end of array.

public int removeDuplicates(int[] nums) {
    if (nums.length == 0) return 0;
    int i = 0;
    for (int j = 1; j < nums.length; j++) {
        if (nums[j] != nums[i]) {
            i++;
            nums[i] = nums[j];
        }
    }
    return i + 1;
}

Complexity analysis

Time complextiy : $O(n)$ . Assume that $n$ is the length of array. Each of $i$ and $j$ traverses at most $n$ steps.
Space complexity : $O(1)$ .

26. Remove Duplicates from Sorted Array

Solution

Approach #1 (Two Pointers) [Accepted]