Algorithm Day113 - Remove Element

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🚀 Day113 - LeetCode 27: Remove Element

🔍 Problem

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place.
Return the number of remaining elements. Order may change.

🧠 Approach

  • Use two pointers
  • fast scans array
  • slow records valid element positions
  • When nums[fast] != val, copy to nums[slow]

✅ Complexity

  • Time: O(n)
  • Space: O(1)

💻 Java Code

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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;
    }
}

📝 Notes

  • We don’t need an extra array
  • Just overwrite values in-place
  • slow pointer gives the new length

Keep grinding — one step closer every day! 💪🔥