Leetcode problem 72 - Remove Duplicates from Sorted Array

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).

/**
 * @param {number[]} nums
 * @return {number}
 */
var removeDuplicates = function (nums) {
  // Initialize a pointer 'k' to track the position for unique elements.
  let k = 1;

  // Iterate through the array starting from the second element.
  for (let i = 1; i < nums.length; i++) {
    // Check if the current element is different from the previous one.
    if (nums[i] !== nums[i - 1]) {
      // If different, update the array at position 'k' and increment 'k'.
      nums[k] = nums[i];
      k++;
    }
  }

  // 'k' now represents the length of the modified array with duplicates removed.
  return k;
};

The time complexity of the provided removeDuplicates function is O(n), where ‘n’ is the length of the input array nums. The algorithm iterates through the array once.