80. Remove Duplicates from Sorted Array II

Description

Given a sorted array nums, remove the duplicates in-place such that duplicates appeared at most twice and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Given nums = [1, 1, 1, 2, 2, 3],

Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.

It doesn't matter what you leave beyond the returned length.

Example 2

Given nums = [0, 0, 1, 1, 1, 1, 2, 3, 3],

Your function should return length = 7, with the first seven elements of nums being modified to 0, 0, 1, 1, 2, 3 and 3 respectively.

It doesn't matter what values are set beyond the returned length.

Solutions

Solution 1

/**
 * Time complexity : O(n) since we process each element exactly once.
 * Space complexity : O(1)
 */

class Solution {
    public int removeDuplicates(int[] nums) {
        if(nums == null || nums.length == 0) return 0;
        int j = 1, count = 1;
        for(int i = 1; i < nums.length; i++) {
            if(nums[i-1] == nums[i]) {
                count++;
            } else {
                count = 1;
            }
            if(count <= 2) nums[j++] = nums[i];
        }
        return j;
    }
}

Solution 2

/**
 * Time complexity : O(n) since we process each element exactly once.
 * Space complexity : O(1)
 */

class Solution {
    public int removeDuplicates(int[] nums) {
        if(nums==null) return 0;
        if(nums.length <= 2) {
            return nums.length;
        }
        int i = 1, j = 2;
        while (j < nums.length) {
            if(nums[j] == nums[i] && nums[j] == nums[i-1]) {
                j++;
            } else {
                i++;
                nums[i] = nums[j];
                j++;
            }
        }
        return i + 1;
    }
}

Follow up