80. Remove Duplicates from Sorted Array II

LeetCode 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 arrayin-place with O(1) extra memory.

Clarification:

Confused why the returned value is an integer, but your answer is an array?

Note that the input array is passed in by reference , which means a modification to the input array will be known to the caller.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy)
int len = removeDuplicates(nums);

// any modification to nums in your function would be known by the caller.
// using the length returned by your function, it prints the first len elements.
for (int i = 0; i < len; i++) {
    print(nums[i]);
}

Example 1:

Input: nums = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3]
Explanation: 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:

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 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.

Constraints:

• 1 <= nums.length <= 3 * 104

• -104 <= nums[i] <= 104

• nums is sorted in ascending order.

Tags

Array, Two Pointers

Solution

This is a general solution for removing duplicates from the given sorted array such that duplicates appeared at most k times and return the new length. We assign a writer pointer w to the first position of the array, and use another pointer to traverse the array.

• Keep the first k elements;

• if the current visited value is not equal to the value at w-k , the position w pointing to will be overwritten by the current value.

Complexity

• Time complexity: $O(n)$

• Space complexity: $O(1)$

Code

func removeDuplicates(nums []int) int {
	process := func (k int) int {
		var w int // writer pointer
		for _, v := range nums {
			if w<k || nums[w-k] != v {
				nums[w] = v
				w++
			}
		}
		return w
	}
	return process(2)
}

Reference

1. 【宫水三叶】关于「删除有序数组重复项」的通解