Remove Duplicates from Sorted Array (LeetCode)

Problem Description

Problem Statement	Solution Link	LeetCode Profile
Remove Duplicates from Sorted Array	Remove Duplicates from Sorted Array Solution on LeetCode	VijayShankerSharma

Problem 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:

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.

Examples

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.

Approach

To solve the problem, we can use the two-pointer technique. Here is the step-by-step approach:

1. Initialize Pointers:

   • Use uniqueIndex to track the position to place the next unique element.

2. Iterate Through the Array:

   • Traverse the array starting from the second element.
   • If the current element is different from the element at uniqueIndex, move uniqueIndex forward and update it with the current element.

3. Return Result:

   • The number of unique elements is uniqueIndex + 1.

Solution Code

Python

class Solution(object):
    def removeDuplicates(self, nums):
        if len(nums) == 0:
            return 0
        
        unique_index = 0  # Pointer for placing unique elements
        
        for i in range(1, len(nums)):
            if nums[i] != nums[unique_index]:
                # Found a unique element, place it in the next position
                unique_index += 1
                nums[unique_index] = nums[i]
        
        # The number of unique elements is one more than the index of the last unique element
        return unique_index + 1

Java

class Solution {
    public int removeDuplicates(int[] nums) {
        if (nums.length == 0) return 0;
        
        int uniqueIndex = 0; // Pointer for placing unique elements
        
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] != nums[uniqueIndex]) {
                // Found a unique element, place it in the next position
                uniqueIndex++;
                nums[uniqueIndex] = nums[i];
            }
        }
        
        // The number of unique elements is one more than the index of the last unique element
        return uniqueIndex + 1;
    }
}

C++

class Solution {
public:
    int removeDuplicates(vector<int>& nums) {
        if (nums.size() == 0) return 0;
        
        int uniqueIndex = 0; // Pointer for placing unique elements
        
        for (int i = 1; i < nums.size(); i++) {
            if (nums[i] != nums[uniqueIndex]) {
                // Found a unique element, place it in the next position
                uniqueIndex++;
                nums[uniqueIndex] = nums[i];
            }
        }
        
        // The number of unique elements is one more than the index of the last unique element
        return uniqueIndex + 1;
    }
};

Conclusion

The above solution efficiently removes duplicates from a sorted array in-place. It employs a two-pointer technique to achieve a time complexity of $O(N)$ and a space complexity of $O(1)$. This ensures that the algorithm can handle input sizes up to the upper limit specified in the constraints efficiently, providing a simple yet effective approach to solving the problem of removing duplicates from a sorted array.