Leetcode problem 8 - Remove Element

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the elements may be changed. Then return the number of elements in nums which are not equal to val.

Consider the number of elements in nums which are not equal to val 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 elements which are not equal to val. 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 val = …; // Value to remove int[] expectedNums = […]; // The expected answer with correct length. // It is sorted with no values equaling val.

int k = removeElement(nums, val); // Calls your implementation

assert k == expectedNums.length; sort(nums, 0, k); // Sort the first k elements of nums for (int i = 0; i < actualLength; i++) { assert nums[i] == expectedNums[i]; } If all assertions pass, then your solution will be accepted.

Example 1:

Input: nums = [3,2,2,3], val = 3 Output: 2, nums = [2,2,,] Explanation: Your function should return k = 2, with the first two elements of nums being 2. It does not matter what you leave beyond the returned k (hence they are underscores). Example 2:

Input: nums = [0,1,2,2,3,0,4,2], val = 2 Output: 5, nums = [0,1,4,0,3,,,_] Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4. Note that the five elements can be returned in any order. It does not matter what you leave beyond the returned k (hence they are underscores).

/**
 * @param {number[]} nums
 * @param {number} val
 * @return {number}
 */
var removeElement = function (nums, val) {
  // Initialize a pointer 'k' to keep track of the current position
  let k = 0;

  // Iterate through each element in the 'nums' array.
  for (let i = 0; i < nums.length; i++) {
    // Check if the current element is not equal to the target value 'val'
    if (nums[i] !== val) {
      // Move it to the position indicated by the 'k' pointer and increment 'k'.
      nums[k] = nums[i];
      k++;
    }
  }

  // The value of 'k' now represents the length of the modified array where 'val' is removed.
  return k;
};

Inside the function, I initialize a counter k to keep track of elements that are not equal to val. I iterate through the nums array using a pointer i.

For each element nums[i], if it is not equal to val, we move it to the left side of the array by assigning it to nums[k]. Then, we increment k by 1.

After iterating through the entire array, the first k elements will contain the elements not equal to val. The remaining elements of nums can be ignored, as well as the size of nums.

Finally, I return the value of k, which represents the count of elements in nums that are not equal to val.

The time complexity of this algorithm is O(n), where n is the length of the ‘nums’ array. The algorithm iterates through each element of the array once.