80. Remove Duplicates from Sorted Array II #

Problem #

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.

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.

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 modification to the input array will be known to the caller as well.

Internally you can think of this:

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

// 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]);
}

Problem Summary #

Given a sorted array nums, remove duplicates from the elements in the array so that each element in the original array appears at most twice. Finally, return the length of the array after deduplication.

Solution Approach #

• The problem indicates a sorted array, so the easiest approach is generally to think of a two-pointer solution. The key point of two pointers: the conditions for moving the two pointers.
• In this problem, the moving condition is: the fast pointer traverses the array from the beginning, and the slow pointer points to the end of the modified array. Only when the second-to-last number pointed to by the slow pointer is not equal to the number pointed to by the fast pointer do we move the slow pointer and assign the value to it at the same time.
• Handle boundary conditions: when the array has fewer than two elements, do nothing.

Code #

package leetcode

func removeDuplicates(nums []int) int {
	slow := 0
	for fast, v := range nums {
		if fast < 2 || nums[slow-2] != v {
			nums[slow] = v
			slow++
		}
	}
	return slow
}