Remove Element

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 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

Code Template

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        # Your code here
        pass

Solutions

• Solution 1: This approach uses a two-pointer technique to overwrite elements equal to val with elements that are not equal to val. It runs in O(n) time and uses O(1) additional space.
• Solution 2: This approach minimizes the number of writes by swapping unwanted elements with the last element. It also runs in O(n) time and uses O(1) additional space.

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