80. Remove Duplicates from Sorted Array II

Method 1

Given an integer array nums sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. The relative order of the elements should be kept the same.
Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.
Return k after placing the final result in the first k slots of nums.
Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.
Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
    assert nums[i] == expectedNums[i];
}

If all assertions pass, then your solution will be accepted.

Example 1

Input: nums = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
It does not matter what you leave beyond  the returned k (hence they are underscores).

Example 2

Input: nums = [0,0,1,1,1,1,2,3,3]
Output: 7, nums = [0,0,1,1,2,3,3,_,_]
Explanation: Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.
It does not matter what you leave beyond  the returned k (hence they are underscores).

Method 1

【O(n) time | O(1) space】

package Leetcode.Array;

/**
 * @author zhengxingxing
 * @date 2025/02/09
 */
public class RemoveDuplicatesFromSortedArrayII {

    public static int removeDuplicates(int[] nums) {
        // If the array has less than or equal to 2 elements, no need to remove any duplicates.
        // All elements can be kept, as they satisfy the condition of appearing at most twice.
        if (nums.length <= 2) {
            return nums.length;
        }

        // Two-pointer technique:
        // i acts as the pointer for the new, filtered array where elements are written.
        // Initialize i to 2 because the first two elements are always kept (no extra duplicates to remove).
        int i = 2;

        // Iterate through the array starting from the 3rd element (index 2).
        for (int j = 2; j < nums.length; j++) {
            // nums[j] is the current element being checked.
            // nums[i - 2] is the element at the position 2 steps behind the last written position i.
            // If nums[j] != nums[i - 2], it means nums[j] has not yet appeared more than twice.
            // This allows us to keep nums[j] in the final array.
            if (nums[j] != nums[i - 2]) {
                // Copy nums[j] to the current position i, effectively "keeping" this element.
                nums[i] = nums[j];
                // Increment i, moving to the next position to potentially write to in the filtered array.
                i++;
            }
            // If nums[j] == nums[i - 2], it means nums[j] has already been included twice,
            // so we skip it and do not increment i.
        }

        // Return the new length of the array. The elements from nums[0] to nums[i - 1]
        // are the elements that satisfy the condition of appearing at most twice.
        return i;
    }

    public static void main(String[] args) {
        // Example 1: Input array has multiple duplicates (more than twice).
        int[] nums1 = {1, 1, 1, 2, 2, 3};
        int k1 = removeDuplicates(nums1); // Call the method to remove duplicates.
        System.out.print("New length: " + k1 + ", Array: ");
        printArray(nums1, k1); // Print the modified array up to the new length.

        // Example 2: Input array has a mixture of duplicates and unique elements.
        int[] nums2 = {0, 0, 1, 1, 1, 1, 2, 3, 3};
        int k2 = removeDuplicates(nums2); // Call the method to remove duplicates.
        System.out.print("New length: " + k2 + ", Array: ");
        printArray(nums2, k2); // Print the modified array up to the new length.
    }


    private static void printArray(int[] nums, int length) {
        // Print each of the first `length` elements of the array.
        for (int i = 0; i < length; i++) {
            System.out.print(nums[i] + " ");
        }
        // Print a new line after printing all required elements.
        System.out.println();
    }
}

Method 1

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