Leetcode problem 73 - Remove Duplicates from Sorted Array II

Given an integer array nums sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. 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 = [1,1,1,2,2,3] Output: 5, nums = [1,1,2,2,3,_] Explanation: Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively. It does not matter what you leave beyond the returned k (hence they are underscores). Example 2:

Input: nums = [0,0,1,1,1,1,2,3,3] Output: 7, nums = [0,0,1,1,2,3,3,,] Explanation: Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 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 pointer index to keep track of first two elements
  let k = 2;

  // The index in for loop starts from 2, which is third element in nums array
  for (let i = 2; i < nums.length; i++) {
    // Check if first element is not same with third element
    if (nums[k - 2] !== nums[i]) {
      // if it is true, third element is assigned in nums array at index k plus 1, which means next index of nums array.
      nums[k++] = nums[i];
    }
    // if first two elements are already in the nums array, next element of two elements is ignored.
  }
  // return number of elements in nums array
  return k;
};

The time complexity is O(n), where n is the length of the input array nums. This is because the function uses a single loop that iterates through the array once.