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LeetCode 27: Remove Element Solution

Master LeetCode problem 27 (Remove Element), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

27. Remove Element

Easy
ArrayTwo Pointers

Problem Explanation

Explanation

To solve this problem, we can use a two-pointer approach where one pointer iterates through the array while the other pointer keeps track of the next position to overwrite when the element needs to be removed. We iterate through the array, and whenever we encounter an element equal to the given value, we skip it. If the element is not equal to the given value, we copy it to the position indicated by the second pointer. This way, we effectively remove the specified value from the array in-place.

Algorithm

  1. Initialize two pointers i and k at the start of the array.
  2. Iterate through the array using pointer i.
  3. If nums[i] is equal to val, skip this element.
  4. If nums[i] is not equal to val, copy nums[i] to nums[k] and increment k.
  5. Continue this process until i reaches the end of the array.
  6. Return the value of k as the length of the modified array.

Time Complexity

The time complexity of this algorithm is O(n), where n is the number of elements in the input array nums.

Space Complexity

The algorithm has a space complexity of O(1) as it modifies the input array in-place without using any extra space.

Solution Code

class Solution {
    public int removeElement(int[] nums, int val) {
        int k = 0;
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] != val) {
                nums[k] = nums[i];
                k++;
            }
        }
        return k;
    }
}

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The time complexity for LeetCode 27 (Remove Element) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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