Remove Element Problem Leetcode

 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 :-


class Solution {

    public int removeElement(int[] nums, int val) {

        int k = 0;  // Pointer for placing non-val elements


        for (int i = 0; i < nums.length; i++) {

            if (nums[i] != val) {

                nums[k] = nums[i];  // Move valid elements to the front

                k++;  // Increase count of non-val elements

            }

        }


        return k;  // Return the new length of the array

    }

}

Time & Space Complexity

✅ Time Complexity: O(n)O(n) (one pass through the array)
✅ Space Complexity: O(1)O(1) (modifies nums in place, no extra space)

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