5. Remove Element
easyAsked at CourseraRemove all instances of a value in-place — Coursera uses this to test two-pointer fluency for filtering enrollment lists.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given an array nums and integer val, remove all occurrences of val in-place. Return the new length; order of remaining elements may change.
Constraints
0 <= nums.length <= 1000 <= nums[i], val <= 50
Examples
Example 1
nums = [3,2,2,3], val = 32, nums = [2,2,_,_]Example 2
nums = [0,1,2,2,3,0,4,2], val = 25, nums = [0,1,3,0,4,_,_,_]Approaches
1. Filter and copy back
Filter to a new array, copy back.
- Time
- O(n)
- Space
- O(n)
const f = nums.filter(x => x !== val);
for (let i = 0; i < f.length; i++) nums[i] = f[i];
return f.length;Tradeoff:
2. Two pointers
Write pointer advances only on keep; iterate read pointer once.
- Time
- O(n)
- Space
- O(1)
function removeElement(nums, val) {
let k = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] !== val) {
nums[k] = nums[i];
k++;
}
}
return k;
}Tradeoff:
Coursera-specific tips
Coursera reviewers map this to filtering unenrolled learners from a roster array in-place — say it out loud so they see you connect the abstract pattern to a real data ops chore.
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