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

Intuition

The intuition behind this solution is to iterate through the array and keep track of two pointers: index and i. The index pointer represents the position where the next non-target element should be placed, while the i pointer iterates through the array elements. By overwriting the target elements with non-target elements, the solution effectively removes all occurrences of the target value from the array.

Approach

1. Initialize index to 0, which represents the current position for the next non-target element.
2. Iterate through each element of the input array using the i pointer.
3. For each element nums[i], check if it is equal to the target value.
   • If nums[i] is not equal to val, it means it is a non-target element.
   • Set nums[index] to nums[i] to store the non-target element at the current index position.
   • Increment index by 1 to move to the next position for the next non-target element.
4. Continue this process until all elements in the array have been processed.
5. Finally, return the value of index, which represents the length of the modified array.

Complexity

• Time complexity:

O(n)

• Space complexity:

O(1)

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

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

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

Happy Learning – If you require any further information, feel free to contact me.