27. Remove Element

Question

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).

Solution

Using double pointer method, if left element equal the integer val , swap left element and right element, and move forward left pointer, at the same time move backward right pointer.

My first version, there are some issue here

1. Naming issue
2. Shouldn’t using swap function
3. Need to simplify this piece of code a bit.

First Version - Swift

class Solution {
    func swap(_ nums: inout[Int], _ indexA: Int, _ indexB: Int) {
        var tmp = nums[indexA]
        nums[indexA] = nums[indexB]
        nums[indexB] = tmp
    }

    func removeElement(_ nums: inout [Int], _ val: Int) -> Int {
        var swapLastIndex = nums.count - 1

        if (nums.count <= 0) {
            return 0
        }

        for i in 0..<nums.count {
            while(swapLastIndex >= 0 && nums[swapLastIndex] == val) {
                swapLastIndex -= 1
            }

            if (swapLastIndex < 0 || i >= swapLastIndex) {
                return swapLastIndex + 1
            }

            if (nums[i] == val) {
                swap(&nums, i, swapLastIndex)
                swapLastIndex -= 1
            }
        }

        return 0
    }
}

After optimizing code

Swift

class Solution {
    func removeElement(_ nums: inout [Int], _ val: Int) -> Int {
        var right = nums.count - 1
        var i = 0

        while i <= right {
            if nums[i] == val {
                nums[i] = nums[right]
                right -= 1
            } else {
                i += 1
            }
        }

        return right + 1
    }
}

Javascript

/**
 * @param {number[]} nums
 * @param {number} val
 * @return {number}
 */
var removeElement = function (nums, val) {
    let right = nums.length - 1
    let left = 0
    while (left <= right) {
        if (nums[left] == val) {
            nums[left] = nums[right]
            right--
        } else {
            left++
        }
    }

    return right + 1
};

Inspiration

Javascript is different grammar with C language:

In C language we can use nums[left++] = nums[right—]. it means:

nums[left] = nums[right]
left++
right—

But in Javascript, nums[left++] = nums[++left]. So it’s different between C & Javascript grammar.