Remove Element (LeetCode)

Problem Description

Problem Statement	Solution Link	LeetCode Profile
Merge Two Sorted Lists	Merge Two Sorted Lists Solution on LeetCode	VijayShankerSharma

Problem 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 to 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.

Examples

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

Approach

To solve the problem, we can use the following approach:

1. Initialize Two Pointers:

   • Use two pointers, slow and fast, where slow points to the beginning of the array and fast moves forward to explore the array.

2. Iterate Through the Array:

   • While fast is within the bounds of the array:
     • If nums[fast] is not equal to val, copy nums[fast] to nums[slow] and increment both pointers.
     • If nums[fast] is equal to val, just increment fast.

3. Return Result:

   • The new length of the array without the elements equal to val is given by slow.

Solution Code

Python

class Solution:
    def removeElement(self, nums, val):
        # Pointer to place the next non-val element
        k = 0
        
        # Iterate through all elements in the array
        for i in range(len(nums)):
            if nums[i] != val:
                # Place the non-val element at the next position in the array
                nums[k] = nums[i]
                k += 1
        
        return k

def _driver():
    # Example usage
    nums = [3, 2, 2, 3]
    val = 3
    sol = Solution()
    k = sol.removeElement(nums, val)
    print("The number of elements not equal to val:", k)
    print("The modified array:", nums[:k])

if __name__ == "__main__":
    _driver()

Java

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

C++

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

Conclusion

The above solution efficiently removes all occurrences of a specified value from an integer array in-place. It employs a two-pointer technique to achieve a time complexity of $O(N)$ and a space complexity of $O(1)$. This ensures that the algorithm can handle input sizes up to the upper limit specified in the constraints efficiently, providing a simple yet effective approach to solving the problem of removing an element from an array.