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Debesh P.
Debesh P.

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27. Remove Element | LeetCode | Top Interview 150 | Coding Questions

#leetcode #programming #productivity #beginners

Problem Link

https://leetcode.com/problems/remove-element/

Detailed Step-by-Step Explanation

https://leetcode.com/problems/remove-element/solutions/7439481/most-optimal-solution-beats-500-best-eas-hegz

leetcode 27

Problem in Simple Words

You are given an integer array nums and an integer val.

  • Remove all occurrences of val from the array
  • Do the operation in place
  • Order of elements does not matter
  • Return the number of elements that are not equal to val

What the Problem Is Really Asking

You are not asked to resize the array.

You are only asked to:

  • Rearrange the array so that all elements not equal to val come first
  • Return how many such elements exist

Everything beyond that index does not matter.

Key Idea

We use two pointers:

  • i → scans every element
  • j → tracks where the next valid element should go

Whenever we see a value that is not equal to val, we keep it.

Why This Works

  • We overwrite only values that are meant to be removed
  • All valid elements are packed at the front
  • No extra memory is used

Variables Explained 

int j = 0;
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  • j stores the position where the next valid element should be placed
  • It also ends up being the final count of valid elements

Step-by-Step Logic

We loop through the array:

for (int i = 0; i < n; i++) {
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For every element:

  • If it is not equal to val

    • Copy it to index j
    • Increment j

  • If it equals val

    • Skip it

Full Code 

class Solution {
    public int removeElement(int[] nums, int val) {
        int n = nums.length;
        int j = 0;

        for (int i = 0; i < n; i++) {
            if (nums[i] != val) {
                nums[j] = nums[i];
                j++;
            }
        }

        return j;
    }
}
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Example Walkthrough

Input:

nums = [3,2,2,3]
val = 3
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Step by step:

  • i = 0 → value is 3 → skip
  • i = 1 → value is 2 → keep → nums[0] = 2
  • i = 2 → value is 2 → keep → nums[1] = 2
  • i = 3 → value is 3 → skip

Result:

  • First 2 elements are valid
  • Returned value = 2

Why We Return j

  • j counts how many valid elements were placed
  • It represents the new length of the array
  • Everything after index j - 1 can be ignored

Time and Space Complexity

  • Time: O(n)
  • Space: O(1)

Final Takeaway

This problem is about filtering, not deleting.

Once you realize that the array does not need to shrink,
just be rearranged, the solution becomes straightforward and efficient.

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