Remove Element
(#27), Easy
Category: Array, Two Pointer
Problem Statement
You are given an array nums and an integer val. Your task is to remove all occurrences of val from the array in-place, and return the number of elements that are not equal to val. The modified array should have all the remaining (non-val) elements at the beginning. The rest of the elements don’t matter. Return the count k, which is the number of elements not equal to val.
Examples
Input: nums = [3,2,2,3], val = 3
Output: 2
Explanation: Modified array: [2,2,_,_]. First 2 elements are not equal to 3.
Input: nums = [0,1,2,2,3,0,4,2], val = 2
Output: 5
Explanation: Modified array: [0,1,3,0,4,_,_,_]. First 5 elements are not equal to 2.
Approach
Approach 1: Two Pointer (Swap Based)
- Initialize two pointers `first = 0` and `sec = 0`.
- Loop while `sec < nums.length`.
- If `nums[sec] !== val`, swap `nums[first]` and `nums[sec]`, then increment `first`.
- Always increment `sec` regardless of condition.
// Approach 1
var removeElement = function(nums, val) {
let first = 0;
let sec = 0;
while (sec < nums.length) {
if (nums[sec] !== val) {
[nums[first], nums[sec]] = [nums[sec], nums[first]];
first++;
}
sec++;
}
return first;
};Approach 2: Cleaner Overwrite
- Initialize a pointer `k = 0`.
- Loop through the array using index `i`.
- If `nums[i] !== val`, set `nums[k] = nums[i]` and increment `k`.
- `k` will be the count of elements not equal to val.
// Approach 2
var removeElement = function(nums, val) {
let k = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] !== val) {
nums[k] = nums[i];
k++;
}
}
return k;
};Complexity
Time: O(n) – We iterate through the array once.
Space: O(1) – Done in-place without extra space.