Remove Element

Description

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

Solution

Classical Solution

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int nums_size = nums.size();
        for (int i = 0; i < nums_size; i++) {
            // When the element that needs to be removed is found, 
            // the array is collectively moved forward by one position.
            if (nums[i] == val) {
                for (int j = i + 1; j < nums_size; j++) {
                    nums[j - 1] = nums[j];
                }
                // Because the values after subscript i are moved forward by one position, 
                // i is also moved forward by one position.
                i--;
                // size of array -1
                nums_size--;
            }
        }
        return nums_size;
    }
};

• Time complexity: \(O(n^2)\);
• Space complexity: \(O(1)\).

Double Pointer Solution

Define fast and slow pointers:

• Fast pointer: looks for an element of the new array, which is an array that does not contain the target element;
• Slow pointer: points to the location where the subscript of the new array is updated.

class Solution {
public:
    int removeElement(vector<int>& nums, int val) {
        int slowIndex = 0;
        for (int fastIndex = 0; fastIndex < nums.size(); fastIndex++) {
            if (val != nums[fastIndex])
                nums[slowIndex++] = nums[fastIndex];
        }
        return slowIndex;
    }
};

• Time complexity: \(O(n)\);
• Space complexity: \(O(1)\).