Description (leetcode)

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.

submission

impl Solution {
    pub fn remove_duplicates(nums: &mut Vec<i32>) -> i32 {
        let mut kept = 0;
        for idx in 0..nums.len() {
            // the trick here is to compare the current number with
            // the previous kept number, we keep the current number
            // if they are not the same or we never kept any number
            if kept == 0 || nums[kept - 1] != nums[idx] {
                nums[kept] = nums[idx];
                kept += 1;
            }
        }
        kept as _
    }
}