26. Remove Duplicates from Sorted Array #

Problem #

Given a sorted array nums, remove the duplicates in-place such that each element appear only once 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 = [1,1,2],

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

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

Example 2:

Given nums = [0,0,1,1,1,2,2,3,3,4],

Your function should return length = 5, with the first five elements of nums being modified to 0, 1, 2, 3, and 4 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 array so that each element in the original array appears only once. Finally, return the length of the array after deduplication.

Solution Approach #

This problem is very similar to Problem 27. This problem is basically the same as Problem 283 and Problem 27: Problem 283 removes 0s, Problem 27 removes a specified element, and this problem removes duplicate elements. In essence, they are the same.

Here, deletion from the array is not an actual deletion; it just moves the elements to be deleted to the space at the end of the array, and then returns the actual number of elements remaining in the array. When the OJ finally judges the problem, it will read and output the remaining number of elements in the array.

Code #

package leetcode

// Solution 1
func removeDuplicates(nums []int) int {
	if len(nums) == 0 {
		return 0
	}
	last, finder := 0, 0
	for last < len(nums)-1 {
		for nums[finder] == nums[last] {
			finder++
			if finder == len(nums) {
				return last + 1
			}
		}
		nums[last+1] = nums[finder]
		last++
	}
	return last + 1
}

// Solution 2
func removeDuplicates1(nums []int) int {
	if len(nums) == 0 {
		return 0
	}
	length := len(nums)
	lastNum := nums[length-1]
	i := 0
	for i = 0; i < length-1; i++ {
		if nums[i] == lastNum {
			break
		}
		if nums[i+1] == nums[i] {
			removeElement1(nums, i+1, nums[i])
			// fmt.Printf("At this point num = %v length = %v\n", nums, length)
		}
	}
	return i + 1
}

func removeElement1(nums []int, start, val int) int {
	if len(nums) == 0 {
		return 0
	}
	j := start
	for i := start; i < len(nums); i++ {
		if nums[i] != val {
			if i != j {
				nums[i], nums[j] = nums[j], nums[i]
				j++
			} else {
				j++
			}
		}
	}
	return j
}