26. Remove Duplicates from Sorted Array

• Link: https://leetcode.com/problems/remove-duplicates-from-sorted-array/description/

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. 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:

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 * 104
• -100 <= nums[i] <= 100
• nums is sorted in non-decreasing order.

Approach 1: Brute Force

Intuition

• We have to think of a data structure that does not store duplicate elements.
• So we can use Hash set.
• Hash set only stores unique elements

Algorithm

1. Declare a HashSet
2. Run a for loop starting to the end
3. Put every element of the array in the set.
4. Store size of the set in a variable K.
5. Now put all elements of the set in the array from the starting of the array.
6. Return K.

Code

class Solution {
public:
	int removeDuplicates(vector<int>& nums) {
		set<int> s;
		for (int i = 0 ; i < nums.size() ; i++) {
			s.insert(nums[i]);
		}
		
		int k = s.size();
		int i = 0;
		
		for (int x: s) {
			nums[i++] = x;
		}

		return k;
	}
}

Complexity Analysis

• Time Complexity: $O(n\log n)$ + $O(n)$
• Space Complexity: $O(n)$

Approach 2: Optimal

Intuition

• We can think of using two pointer i & j, we move j till we don't get a number arr[j] which is different from arr[i].
• As we got a unique number we will increase the i pointer & update its value by arr[j].

Algorithm

1. Take a variable i=0
2. Use a for loop by using a variable j from 1 to length of the array.
3. If arr[j] != arr[i], increase i and update arr[i] = arr[j]
4. After completion of the loop return i+1, i.e, size of the array of unique elements

Code

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

Complexity Analysis

• Time Complexity: $O(n)$
• Space Complexity: $O(1)$