LeetCode link: 26. Remove Duplicates from Sorted Array, difficulty: Easy.
Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. The relative order of the elements should be kept the same. Then return the number of unique elements in nums.
Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things:
- Change the array
numssuch that the firstkelements ofnumscontain the unique elements in the order they were present innumsinitially. The remaining elements ofnumsare not important as well as the size ofnums. - Return
k.
Example 1:
Input: nums = [1,1,2]
Output: 2, nums = [1,2,_]
Explanation:
Your function should return k = 2, with the first two elements of nums being 1 and 2 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).
Example 2:
Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]
Explanation:
Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).
Constraints:
1 <= nums.length <= 3 * 10^4-10^4 <= nums[i] <= 10^4numsis sorted in non-decreasing order.
Hint 1
In this problem, the key point to focus on is the input array being sorted. As far as duplicate elements are concerned, what is their positioning in the array when the given array is sorted? Look at the image below for the answer. If we know the position of one of the elements, do we also know the positioning of all the duplicate elements?
Hint 2
We need to modify the array in-place and the size of the final array would potentially be smaller than the size of the input array. So, we ought to use a two-pointer approach here. One, that would keep track of the current element in the original array and another one for just the unique elements.
Hint 3
Essentially, once an element is encountered, you simply need to bypass its duplicates and move on to the next unique element.
Intuition
The numbers in the array are already sorted, so any duplicate values must appear consecutively.
To remove duplicates, we need to keep every number that is different from the previous one, and discard the rest.
Since the array needs to be modified in place, two pointers are required: one moves fast and the other moves slow.
the fast pointer increases its position by 1 each time, while the slow pointer marks the current position to be modified.
Pattern of "Fast & Slow Pointers Technique"
int slow = 0; // slow pointer
// ...
for (int fast = 1; fast < array_length; fast++) { // This line is the key!
if (condition_1) {
// ...
continue; // 'continue' is better than 'else' because 'else' will introduce more indents
}
if (condition_2) {
// ...
continue; // 'continue' is better than 'else'
}
// condition_3
// ...
slow++;
// ...
}
return something;
Complexity
Time complexity
N
Space complexity
1
Python #
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
slow = 1
# nums = [0,0,1,1,1,2,2,3,3,4]
for fast in range(1, len(nums)):
if nums[fast] == nums[slow - 1]:
continue
nums[slow] = nums[fast]
slow += 1
return slow
Ruby #
# @param {Integer[]} nums
# @return {Integer}
def remove_duplicates(nums)
slow = 1
# nums = [0,0,1,1,1,2,2,3,3,4]
(1...nums.size).each do |fast|
if nums[fast] == nums[slow - 1]
next
end
nums[slow] = nums[fast]
slow += 1
end
slow
end
Java #
class Solution {
public int removeDuplicates(int[] nums) {
var slow = 1;
// nums = [0,0,1,1,1,2,2,3,3,4]
for (var fast = 1; fast < nums.length; fast++) {
if (nums[fast] != nums[slow - 1]) {
nums[slow] = nums[fast];
slow++;
}
}
return slow;
}
}