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Remove Duplicates from Sorted Array⚓︎

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Description⚓︎

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:

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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.

Solution⚓︎

Double Pointer⚓︎

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class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int slowIndex = 1, fastIndex;
        for (fastIndex = 1; fastIndex < nums.size(); fastIndex++) {
            if (nums[fastIndex] != nums[fastIndex - 1])
                nums[slowIndex++] = nums[fastIndex];
        }
        return slowIndex;
    }
};

Time complexity: \(O(n)\)

Another way to handle index:

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class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        int slowIndex = 0;
        for (int fastIndex = 0; fastIndex < nums.size(); fastIndex++) {
            if (!fastIndex || nums[fastIndex] != nums[fastIndex - 1])
                nums[slowIndex++] = nums[fastIndex];
        }
        return slowIndex;
    }
};

Library Functions⚓︎

  • The unique function rearranges the elements in nums so that all unique elements appear at the start of the vector, and then returns an iterator to the new logical end of the range.
  • The distance function then calculates how many elements are between the start of the vector and this new logical end, which is the count of unique elements.
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class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        return distance(nums.begin(), unique(nums.begin(), nums.end()));
    }
};