Interview question: javascript implement array remove element with T(n)=O(n); S(n)=O(1)

removeElement.js

/**
 * From: https://leetcode.com/problems/remove-element/
 * Question:
 * Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The relative order of the elements may be changed.
 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 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).

 Constraints:
 0 <= nums.length <= 100
 0 <= nums[i] <= 50
 0 <= val <= 100
 * @param {number[]} nums
 * @param {number} val
 * @return {number}
 * my solution
 */
var removeElement = function (nums, val) {
    /**
     * I use optimized way:
     * for ex [1,2,3], var is 1, if remove 1 and become [2,3], it need mv 2 itmes, the better way is: mv 3 replace 1 and become [3,2] it only mv 1 item
     if near the array start element == val(for ex first element == val) , use regular remove need move all other elements to prev pos, so I use better solution, when remove element, mv end element to instead removed element pos
     */
    var leftIdx = 0
    var rightIdx = nums.length - 1
    var shouldBreak = () => {
        return leftIdx > rightIdx
    }
    while (true) {
        // find last retained element from right side
        while (true) {
            if (shouldBreak()) {
                break
            }
            if (nums[rightIdx] != val) {
                break
            }
            rightIdx--
        }
        // find target from left side
        while (true) {
            if (shouldBreak()) {
                break
            }
            if (nums[leftIdx] == val) {
                nums[leftIdx] = nums[rightIdx]
                leftIdx++
                rightIdx--
                break
            }
            leftIdx++
        }
        if (shouldBreak()) {
            break
        }
    }
    return leftIdx
}
console.log(removeElement([3, 2, 2, 3], 3))
/**
 * official solution
 */

removeElement2.js

var removeElement = function (nums, val) {
    var leftIdx = 0
    var rightIdx = nums.length - 1
    while (true) {
        if (nums[leftIdx] == val) {
            // if nums[rightIdx] == val, next time nums[leftIdx] = nums[rightIdx - 1], find until nums[leftidx] != val, to run leftIdx++
            nums[leftIdx] = nums[rightIdx]
            rightIdx--
        } else {
            leftIdx++
        }
        if (leftIdx > rightIdx) {
            break
        }
    }
    return leftIdx
}
console.log(removeElement([0, 1, 2, 2, 3, 0, 4, 2], 2))

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