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.
Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.
Return k after placing the final result in the first k slots of nums.
Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.
Custom Judge:
The judge will test your solution with the following code:
int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length
int k = removeDuplicates(nums); // Calls your implementation
assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
assert nums[i] == expectedNums[i];
}
If all assertions pass, then your solution will be accepted.
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
-100 <= nums[i] <= 100
nums is sorted in non-decreasing order.
There is a faster solution shown last
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
"""
Go through elements one at a time
if the element next to me is the same,
then mark me as underscore. Otherwise, move on
Then have two pointers: one for marking
and one for finding next number.
Iterate to find the first number, and place it
where the marker is. increment both pointers.
"""
length = len(nums)
for i in range(length - 1):
if nums[i] == nums[i + 1]:
nums[i] = "_"
marker = 0
ptr = 0
while ptr < length:
if nums[ptr] != "_":
nums[marker] = nums[ptr]
marker += 1
ptr += 1
return marker
LeetCode Solution You have an insertIndex that keeps track of which index to insert unique values To find a unique value, compare your current number to the previous number. If they are different, update number at insertIndex with current number.
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
size = len(nums)
insertIndex = 1
for i in range(1, size):
# Found unique element
if nums[i - 1] != nums[i]:
# Updating insertIndex in our main array
nums[insertIndex] = nums[i]
# Incrementing insertIndex count by 1
insertIndex = insertIndex + 1
return insertIndex