leetcode-150

3. Remove Duplicates from Sorted Array

Problem Statement

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.

Solution

Approach: Two-Pointer Technique

Since the array is already sorted in non-decreasing order, all duplicates will be adjacent. This allows us to use a two-pointer approach to remove duplicates efficiently in O(n) time complexity with O(1) extra space.

Steps:

  1. Edge Case: If nums is empty, return 0 since there are no elements to process.

  2. Initialize Pointers:
    • k = 1: This pointer represents the index where the next unique element should be placed.
    • Start iterating from index 1 (since the first element is always unique).
  3. Iterate Over the Array:
    • Loop from index 1 to nums.length - 1.
    • If nums[i] is not equal to nums[i - 1] (i.e., a new unique element is found):
      • Assign nums[i] to nums[k] (move the unique element forward).
      • Increment k (increase the count of unique elements).
  4. Return k: The first k elements in nums are now unique, and we return this count.

Example Walkthrough

Example 1:

Input:
nums = [1,1,2]

Steps:

Output: k = 2, nums = [1,2,_]

Example 2:

Input:
nums = [0,0,1,1,1,2,2,3,3,4]

Steps:

Step i nums[i] nums[i] != nums[i - 1] nums[k] = nums[i] k Updated nums
Start         1 [0,0,1,1,1,2,2,3,3,4]
1 1 0 No Skip 1 [0,0,1,1,1,2,2,3,3,4]
2 2 1 Yes nums[1] = 1 2 [0,1,1,1,1,2,2,3,3,4]
3 3 1 No Skip 2 [0,1,1,1,1,2,2,3,3,4]
4 4 1 No Skip 2 [0,1,1,1,1,2,2,3,3,4]
5 5 2 Yes nums[2] = 2 3 [0,1,2,1,1,2,2,3,3,4]
6 6 2 No Skip 3 [0,1,2,1,1,2,2,3,3,4]
7 7 3 Yes nums[3] = 3 4 [0,1,2,3,1,2,2,3,3,4]
8 8 3 No Skip 4 [0,1,2,3,1,2,2,3,3,4]
9 9 4 Yes nums[4] = 4 5 [0,1,2,3,4,2,2,3,3,4]

Output: k = 5, nums = [0,1,2,3,4,_,_,_,_,_]

Implementation:

function removeDuplicates(nums) {
    if (nums.length === 0) return 0;
    
    let k = 1; // Pointer for the unique elements
    
    for (let i = 1; i < nums.length; i++) {
        if (nums[i] !== nums[i - 1]) {
            nums[k] = nums[i];
            k++;
        }
    }
    
    return k;
}

Example usage:

let nums = [0,0,1,1,1,2,2,3,3,4];
let k = removeDuplicates(nums);
console.log(k, nums.slice(0, k)); // Output: 5, [0,1,2,3,4]

Complexity Analysis

Final Thoughts

This two-pointer method is efficient and optimal for modifying the array in-place. 🚀

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