26. Remove Duplicates from Sorted Array

Remove Duplicates from Sorted Array

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.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

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) Two Pointer

The given array is a sorted array, so duplicate numbers will be group together. First, we can handle the edge case of an empty array. Then we can start iterating elements from the head.

The second pointer will go through all elements in the list, and the first pointer only move when meeting the unique number.

Code:

var removeDuplicates = function(nums) {
   // Handling Edge Case
   if(nums.length === 0 ) return 0
     let p1 = 0
     for(let p2 = 1; p2< nums.length; p2++){
        if(nums[p1] !== nums[p2]){
            p1++;
            nums[p1] = nums[p2]
        }       
     }
    return p1 +1        
}

Besides, we can also start iterator from the end of array. If the value of two pointer is the same, then remove the element.

var removeDuplicates = function(nums) {
    // Handling Edge Case
    if(nums.length === 0 ) return 0
    for(let i = nums.length-1; i > 0;i--){
        if(nums[i]===nums[i-1]){
            nums.splice(i,1)
        }
    }
    return nums.length
};

Complexity 

Time Complexity
Time complexity will be O(n).


Space Complexity
We are not using any extra space, therefore, the space complexity should also be O(1).

Conclusion

That’s all folks! In this post, we solved LeetCode problem #26. Remove Duplicates from Sorted Array

I hope you have enjoyed this post. Feel free to share your thoughts on this.

You can find the complete source code on my GitHub repository. If you like what you learn. feel free to fork 🔪 and star ⭐ it.

In this blog, I have tried to collect & present the most important points to consider when improving Data structure and logic, feel free to add, edit, comment, or ask. For more information please reach me here
Happy coding!