RemoveElement.java

package Lc27_RemoveElement;

/**
 * 27. Remove Element
 *
 * Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the
 * elements may be changed. Then return the number of elements in nums which are not equal to val.
 *
 * Consider the number of elements in nums which are not equal to val be k, to get accepted, you need to do the
 * following things:
 *
 * - Change the array nums such that the first k elements of nums contain the elements which are not equal to val.
 * The remaining elements of nums are not important as well as the size of nums.
 * - Return k.
 *
 * Custom Judge:
 *
 * The judge will test your solution with the following code:
 *
 * int[] nums = [...]; // Input array
 * int val = ...; // Value to remove
 * int[] expectedNums = [...]; // The expected answer with correct length.
 * // It is sorted with no values equaling val.
 *
 * int k = removeElement(nums, val); // Calls your implementation
 *
 * assert k == expectedNums.length;
 * sort(nums, 0, k); // Sort the first k elements of nums
 * for (int i = 0; i < actualLength; i++) {
 * assert nums[i] == expectedNums[i];
 * }
 * If all assertions pass, then your solution will be accepted.
 *
 *
 *
 * Example 1:
 * Input: nums = [3,2,2,3], val = 3
 * Output: 2, nums = [2,2,_,_]
 * Explanation: Your function should return k = 2, with the first two elements of nums being 2.
 * It does not matter what you leave beyond the returned k (hence they are underscores).
 *
 * Example 2:
 * Input: nums = [0,1,2,2,3,0,4,2], val = 2
 * Output: 5, nums = [0,1,4,0,3,_,_,_]
 * Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.
 * Note that the five elements can be returned in any order.
 * It does not matter what you leave beyond the returned k (hence they are underscores).
 *
 *
 * Constraints:
 * 0 <= nums.length <= 100
 * 0 <= nums[i] <= 50
 * 0 <= val <= 100
 */
public class RemoveElement {

    /**
     * Soulution 1: Two Pointers (while loop)
     * Time complexity : O(n)
     * Space complexity : O(1)
     */
    public int solution1(int[] nums, int val) {
        int n = nums.length;
        if (n == 0) return 0;
        int fast = 0, slow = 0;
        while (fast < n) {
            if (nums[fast] != val) {
                nums[slow] = nums[fast];
                slow++;
            }
            fast++;
        }
        return slow;
    }

    /**
     * Soulution 1: Two Pointers (for loop)
     * Time complexity : O(n)
     * Space complexity : O(1)
     */
    public int solution1_1(int[] nums, int val) {
        int n = nums.length;
        if (n == 0) return 0;
        int index = 0;
        for (int i = 0; i < n; i++) {
            if (nums[i] != val) {
                nums[index] = nums[i];
                index++;
            }
        }
        return index;
    }

    /**
     * Soulution 2: Two Pointers - when elements to remove are rare
     * Time complexity : O(n)
     * Space complexity : O(1)
     */
    public int solution2(int[] nums, int val) {
        int n = nums.length;
        if (n == 0) return 0;
        int i = 0;
        while (i < n) {
            if (nums[i] == val) {
                nums[i] = nums[n - 1];
                n--;
            } else {
                i++;
            }
        }
        return n;
    }
}