Question

Formatted question description: https://leetcode.ca/all/27.html

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).

Constraints:

0 <= nums.length <= 100
0 <= nums[i] <= 50
0 <= val <= 100

Algorithm

A variable is needed for counting, and then the original array is traversed. If the current value is different from the given value, the current value will cover the position of the counting variable, and the counting variable will be incremented by 1.

Code

public class Remove_Element {

    // time: O(N)
    // space: O(N)
	public class Solution {
	    public int removeElement(int[] nums, int val) {
	        if(nums == null || nums.length == 0) {
	            return 0;
	        }

	        int i = 0; // slow pointer
	        int j = 0; // fast pointer

	        while(j < nums.length) {
	            while(j < nums.length && nums[j] == val) {
	                j++;
	            }

	            // protect index j+1
	            if(j < nums.length) {
	                nums[i] = nums[j];

	                i++;
	                j++;
	            }
	        }

	        return i; // i is incremented, but is index. So same as length

	    }
	}

}

############

class Solution {
    public int removeElement(int[] nums, int val) {
        int cnt = 0, n = nums.length;
        for (int i = 0; i < n; ++i) {
            if (nums[i] == val)
                ++cnt;
            else
                nums[i - cnt] = nums[i];
        }
        return n - cnt;
    }
}

// OJ: https://leetcode.com/problems/remove-element/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int removeElement(vector<int>& A, int val) {
        int j = 0, N = A.size();
        for (int i = 0; i < N; ++i) {
            if (A[i] != val) A[j++] = A[i];
        }
        return j;
    }
};

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        cnt, n = 0, len(nums)
        for i in range(n):
            if nums[i] == val:
                cnt += 1
            else:
                nums[i - cnt] = nums[i]
        return n - cnt

############

class Solution(object):
  def removeElement(self, nums, val):
    """
    :type nums: List[int]
    :type val: int
    :rtype: int
    """
    slow = -1
    for i in range(0, len(nums)):
      if nums[i] != val:
        slow += 1
        nums[slow] = nums[i]
    return slow + 1

func removeElement(nums []int, val int) int {
    cnt, n := 0, len(nums)
    for i := 0; i < n; i++ {
        if (nums[i] == val) {
            cnt++
        } else {
            nums[i - cnt] = nums[i]
        }
    }
    return n - cnt
}

/**
 * @param {number[]} nums
 * @param {number} val
 * @return {number}
 */
var removeElement = function (nums, val) {
    let cnt = 0;
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        if (nums[i] == val) ++cnt;
        else nums[i - cnt] = nums[i];
    }
    return n - cnt;
};

class Solution {

    /**
     * @param Integer[] $nums
     * @param Integer $val
     * @return Integer
     */
    function removeElement(&$nums, $val) {
        for ($i = count($nums) - 1; $i >= 0; $i--) {
            if ($nums[$i] == $val) {
                array_splice($nums, $i, 1);
            }
        }
    }
}

impl Solution {
    pub fn remove_element(nums: &mut Vec<i32>, val: i32) -> i32 {
        let mut len = 0;
        for i in 0..nums.len() {
            if nums[i] != val {
                nums[len] = nums[i];
                len += 1;
            }
        }
        len as i32
    }
}

function removeElement(nums: number[], val: number): number {
    let k: number = 0;
    for (const x of nums) {
        if (x !== val) {
            nums[k++] = x;
        }
    }
    return k;
}

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