Remove Duplicates from Sorted Array
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.
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.
My Solution
You can solve this with two pointers. One pointer advances after processing each input. The other pointer advances only if the element at the other pointer differs.
Time complexity would be O(n). Space complexity would be O(1) since the array is modified in place.
class Solution {
public int removeDuplicates(int[] nums) {
int i = 1;
int j = 1;
int n = nums.length;
while (j < n) {
if (nums[j] != nums[i-1]) {
nums[i] = nums[j];
i++;
j++;
}
else {
j++;
}
}
return i;
}
}