To all those who are just starting their coding journey, you might not understand the code sometimes or even the question. But, that is completely fine. Just stick with it since you have chosen this journey with a purpose and have patience, which in my opinion, is the most important that a programmer must have. Be patient, give yourself time and practice whenever you can.
Same thing but in just slight more understanding way var removeDuplicates = function (A) { let pointer = 0; let i = 0; while (i < A.length) { let start = i; while (A[i] == A[i + 1]) { i++; } let end = i; const count = Math.min(2, end - start + 1); for (let x = 0; x < count; x++) { A[pointer] = A[end]; pointer++; } i++; } return pointer; };
I agree. It's pretty similar to his solution from his Remove Duplicates from Sorted Array video (th-cam.com/video/DEJAZBq0FDA/w-d-xo.html). It was my inital apporach when solving this problem!
There is a way simpler solution and more intuitive from my perspective, I can't say that your explanation was really helpful this time, but I really appreciate what you are doing for us. def removeDuplicates(nums: List[int]) -> int: if len(nums)
Way easier solution from ChatGPT, easy to understand and much more intuitive class Solution: def removeDuplicates(self, nums: List[int]) -> int: if len(nums)
An alternative solution building off of the Remove Duplicates from Sorted Array I solution: class Solution: def removeDuplicates(self, nums: List[int]) -> int: left = 1 freq = 1
for i in range(1, len(nums)): if nums[i] > nums[left - 1]: nums[left] = nums[i] left += 1 freq = 1 elif nums[i] == nums[left - 1] and freq < 2: nums[left] = nums[i] left += 1 freq += 1 return left Instead of nested loops I added a var that keeps track of how many times a value is added and an elif that adds up to two of a given value.
class Solution: def removeDuplicates(self, nums: List[int]) -> int: l = 1 count = 1 for r in range(1, len(nums)): if nums[r] != nums[r-1]: count = 0 if count < 2: nums[l] = nums[r] l += 1 count += 1 return l Initialize count to 1 and both pointers to the first index. We may need to reset the count and write in the same iteration when we find a new element. First check if the current element is a duplicate. If it isn't reset the count to 0. Then if the count is < 2 write the current element to the left pointer and increment the left pointer. Let the for loop handle the right pointer increment. return the left pointer.
Love your videos, they are very insightful and helpful. I'm very greatful for you. I would like this time to share my solution to it, as for the first time ever it has a cleaner approach. Same idea with L and R pointers. with each iteration replace L with R, then check if L is valid, as follows: class Solution: def removeDuplicates(self, nums: List[int]) -> int: l = 2 for r in range(2, len(nums)): nums[l] = nums[r] if nums[l] != nums[l - 2]: l += 1 return l
I found it easier to use a set to solve this question and the first version of this question as well. I just check if the element we are currently at is already in both sets. If its in neither or in just one of the sets, then I added it to the other set and I swap the element at r and the element at i. But if the element is already in both sets, then I only increment r and don't do any swapping. Here is the code: sett1= set() sett2= set() i = 0 for r in range(len(nums)): if nums[r] not in sett2: if nums[r] not in sett1: #case where its not in either set sett1.add(nums[r]) nums[i], nums[r] = nums[r], nums[i] i+=1 else: #case where its not in set 2 but its in set1 sett2.add(nums[r]) nums[i], nums[r] = nums[r], nums[i] i+=1 #case where its in both sets return i
My solution is very similar to the 1st version of the problem but with a slight twist. Runtime beats 98% ====================== def removeDuplicates(self, nums: List[int]) -> int: l = 2 flag = 0 for r in range(2, len(nums)): if nums[r] != nums[r-2-flag]: nums[l] = nums[r] l +=1 else: flag +=1 return l
For those, who want a more stricter solution. The relative order is not preserved in this solution, if there are 4 3's [2, 3,3,3,3], we should consider the relative ordering where the first two 3's are considered, but the your solution kinda considers the last occurence of 3.
Thank You for all the videos but what are your thoughts on below solution? I avoid another for loop. I use your r+1 logic. class Solution: def removeDuplicates(self, nums: List[int]) -> int: r=1 while r < len(nums)-1: while r+1
There's a bug in this problem(the problem itself, not the solution): It says it wants to remove duplicates in the title and it says in the description that each element will appear *at most* twice. It's quite normal to reason that this means that any element, even which originally has 2 copies of a given value in it, can be trimmed down to 1 copy. But that is forbidden: you have to leave *both* copies in the output for it to be what is expected. So, the description needs to be changed to "at least and at most" twice: If the original has only one of the given, fine; but if it contains two or more copies of the same value, you have to leave at least two in and remove the rest. Of course, the value of k needs to be set to factor this in.
Super fantastic explanation ,found best algorithm from u ,watched others tutorials but they are just making it complex,it is very easy to solve via your method and also quite easily understandable
My logic was to check if (nums[i] == nums[i-1]) and (nums[i] == nums[i+1]) then i would push nums[i+1] to the back, but i got lost in the logic behind switching. Any comments/guidance regarding this would be appreciated
you make this question hard, you can make it easier by starting left and right from index 2 and if(nums[right ]!==nums[left -2]){ nums[left ]=nums[right ]; we increment left ++ otherwise right ++ , finally return left
# Another Soluion class Solution: def removeDuplicates(self, nums: List[int]) -> int: i = 0 counter = 0 l = len(nums) while i < (l-1): if nums[i] == nums[i+1] and counter == 1: nums.pop(i+1) l -= 1 elif nums[i] == nums[i+1] and counter == 0: counter += 1 i += 1 else: i += 1 counter = 0 return len(nums)
This is my solution -------------------------------------- class Solution: def removeDuplicates(self, nums: List[int]) -> int: if len(nums) < 2: return len(nums) pointer = 2 for num in nums[2:]: if nums[pointer-2] != num: nums[pointer] = num pointer += 1 return pointer -------------------------------------- So basically, if the length of an array is less than 3, obviously everything is already sorted and correctly placed, therefore we just return it. The if condition in the beginning also allows us to start checking from the third element as well. After that we just start checking our current element and the one that is 2 steps behind. If there are equal, it means that the last two elements in the array are already equal to the current one and therefore we cannot place another one (it would be the third). In other cases, it is possible to add another element with the same value
Here's the way i did it. maybe it's faster def removeDuplicates(self, nums) -> int: j = 1 for i in range(2,len(nums)): x = nums[i] if x != nums[i-1] or x != nums[i-2]: j += 1 nums[j] = x if i == j + 1 and i + 1 < len(nums) and x == nums[i+1]: j += 1 return j+1
There is an easier solution where R (right pointer) loops through the array and just checks whether nums[R] == nums[L] == nums[L-1]. If not: L += 1 nums[L] = nums[R] If yes: continue Return: L+1 JS: var removeDuplicates = function(nums) { if (nums.length < 3) return nums.length // Handle edgecase let replaceIndex = 1 for (let i = 2 ; i < nums.length ; i++) { if (nums[i] != nums[replaceIndex] || nums[replaceIndex] != nums[replaceIndex-1]) { replaceIndex++ nums[replaceIndex] = nums[i] } } return replaceIndex+1 };
An even more intuitive solution. O(n) class Solution: def removeDuplicates(self, nums: List[int]) -> int: if len(nums) == 1: return l = 0 r = 0 while r < len(nums): # edge case if l == 0 and nums[l+1] == nums[l]: # skip the first value to continue as usual r+=1 l+=1 if nums[r] != nums[l]: l+=1 nums[l] = nums[r] # Greedily check for one more duplicate as it is two # if we get it we change else no problem. if r+1 < len(nums) and nums[r+1] == nums[r]: l+=1 nums[l] = nums[r] r+=1 r+=1 return l+1
Great work @NeetCodeIO Was thinking you actually stopped uploading daily leetcode video explanations when I didn't get any notification over the past few hours
whoever strugle to understand this sol i think this easier to come up with int removeDuplicates(vector& nums) { ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); if(nums.size()
Good Evening Sir. I'm an Indian and trying to buy Ur Yearly Plan. I'm getting Ur card is declined. Please help me Sir .. I'm interested to crack MAANG interviews
To all those who are just starting their coding journey, you might not understand the code sometimes or even the question. But, that is completely fine. Just stick with it since you have chosen this journey with a purpose and have patience, which in my opinion, is the most important that a programmer must have. Be patient, give yourself time and practice whenever you can.
Well said!
Thank You very much
thank u, because of that exact same problem i quit leetcode twice before. This time im gonna stick to it no matter wat
Greatly said!
thank you
This solution is kind of a tricky one. I wouldn't be able to come up with this on my own, but yes once you explained it, it makes sense.
couldn't agree more.
that's becoming a pattern the more leetcode questions i come across lol
Same thing but in just slight more understanding way
var removeDuplicates = function (A) {
let pointer = 0;
let i = 0;
while (i < A.length) {
let start = i;
while (A[i] == A[i + 1]) {
i++;
}
let end = i;
const count = Math.min(2, end - start + 1);
for (let x = 0; x < count; x++) {
A[pointer] = A[end];
pointer++;
}
i++;
}
return pointer;
};
Even when explained to me as clearly as this, I always find these loops and pointers a complete mind f**k
def removeDuplicates(self, nums):
j, count = 1, 1
for i in range(1, len(nums)):
if nums[i] == nums[i - 1]:
count += 1
else:
count = 1
if count
I feel like ChatGPT's answer is easier to understand on this one.
def removeDuplicates(self, nums: List[int]) -> int:
left = 2
if len(nums)
This solution is amazing... consider it a fixed-length sliding window
Yeah, ChatGPT's solution is much more intuitive and easy to understand. Still thanks for your efforts Neetcode!
I agree. It's pretty similar to his solution from his Remove Duplicates from Sorted Array video (th-cam.com/video/DEJAZBq0FDA/w-d-xo.html). It was my inital apporach when solving this problem!
well, in similar question #26, you just change every "left - 2" to "left - 1". that work too.
There is a way simpler solution and more intuitive from my perspective, I can't say that your explanation was really helpful this time, but I really appreciate what you are doing for us.
def removeDuplicates(nums: List[int]) -> int:
if len(nums)
In the Constraints:
1
Way easier solution from ChatGPT, easy to understand and much more intuitive
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
if len(nums)
An alternative solution building off of the Remove Duplicates from Sorted Array I solution:
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
left = 1
freq = 1
for i in range(1, len(nums)):
if nums[i] > nums[left - 1]:
nums[left] = nums[i]
left += 1
freq = 1
elif nums[i] == nums[left - 1] and freq < 2:
nums[left] = nums[i]
left += 1
freq += 1
return left
Instead of nested loops I added a var that keeps track of how many times a value is added and an elif that adds up to two of a given value.
Thank you so much sir, but could you please combine videos in a playlist so that will be easy for us to follow
The fact i can write the solution with O(n^2) but still here to optimize the solution...Thank you bro
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
l = 1
count = 1
for r in range(1, len(nums)):
if nums[r] != nums[r-1]:
count = 0
if count < 2:
nums[l] = nums[r]
l += 1
count += 1
return l
Initialize count to 1 and both pointers to the first index. We may need to reset the count and write in the same iteration when we find a new element. First check if the current element is a duplicate. If it isn't reset the count to 0. Then if the count is < 2 write the current element to the left pointer and increment the left pointer. Let the for loop handle the right pointer increment. return the left pointer.
Love your videos, they are very insightful and helpful. I'm very greatful for you.
I would like this time to share my solution to it, as for the first time ever it has a cleaner approach.
Same idea with L and R pointers. with each iteration replace L with R, then check if L is valid, as follows:
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
l = 2
for r in range(2, len(nums)):
nums[l] = nums[r]
if nums[l] != nums[l - 2]:
l += 1
return l
I found it easier to use a set to solve this question and the first version of this question as well. I just check if the element we are currently at is already in both sets. If its in neither or in just one of the sets, then I added it to the other set and I swap the element at r and the element at i. But if the element is already in both sets, then I only increment r and don't do any swapping. Here is the code:
sett1= set()
sett2= set()
i = 0
for r in range(len(nums)):
if nums[r] not in sett2:
if nums[r] not in sett1:
#case where its not in either set
sett1.add(nums[r])
nums[i], nums[r] = nums[r], nums[i]
i+=1
else:
#case where its not in set 2 but its in set1
sett2.add(nums[r])
nums[i], nums[r] = nums[r], nums[i]
i+=1
#case where its in both sets
return i
simpler version:
function removeDuplicates(nums: number[]): number {
// edge case to return early if 2 or fewer elements
if (nums.length
My solution is very similar to the 1st version of the problem but with a slight twist.
Runtime beats 98%
======================
def removeDuplicates(self, nums: List[int]) -> int:
l = 2
flag = 0
for r in range(2, len(nums)):
if nums[r] != nums[r-2-flag]:
nums[l] = nums[r]
l +=1
else:
flag +=1
return l
For those, who want a more stricter solution.
The relative order is not preserved in this solution, if there are 4 3's [2, 3,3,3,3], we should consider the relative ordering where the first two 3's are considered, but the your solution kinda considers the last occurence of 3.
def removeDuplicates(self, nums):
if len(nums)
Thank You for all the videos but what are your thoughts on below solution? I avoid another for loop. I use your r+1 logic.
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
r=1
while r < len(nums)-1:
while r+1
There's a bug in this problem(the problem itself, not the solution):
It says it wants to remove duplicates in the title and it says in the description that each element will appear *at most* twice.
It's quite normal to reason that this means that any element, even which originally has 2 copies of a given value in it, can be trimmed down to 1 copy. But that is forbidden: you have to leave *both* copies in the output for it to be what is expected.
So, the description needs to be changed to "at least and at most" twice: If the original has only one of the given, fine; but if it contains two or more copies of the same value, you have to leave at least two in and remove the rest. Of course, the value of k needs to be set to factor this in.
Super fantastic explanation ,found best algorithm from u ,watched others tutorials but they are just making it complex,it is very easy to solve via your method and also quite easily understandable
simple solution, but the intutition behind it is tough
class Solution {
public:
int removeDuplicates(vector& nums) {
int len = nums.size();
if (len
My logic was to check
if (nums[i] == nums[i-1]) and (nums[i] == nums[i+1])
then i would push nums[i+1] to the back, but i got lost in the logic behind switching. Any comments/guidance regarding this would be appreciated
Is the time complexity O(n)^2 ? Since we are using 2 loops for the worst case?
you make this question hard, you can make it easier by starting left and right from index 2 and if(nums[right ]!==nums[left -2]){
nums[left ]=nums[right ]; we increment left ++ otherwise right ++ , finally return left
Thanks for the solution. Very easy to follow and reason.
i came up with an NlogN solution but broke my head trying to move the duplicate spots to the end,,,,,this one is very smart :0
Thank you for your efforts for the fellow community !🤟
# Another Soluion
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
i = 0
counter = 0
l = len(nums)
while i < (l-1):
if nums[i] == nums[i+1] and counter == 1:
nums.pop(i+1)
l -= 1
elif nums[i] == nums[i+1] and counter == 0:
counter += 1
i += 1
else:
i += 1
counter = 0
return len(nums)
at 0.55 why is it better to shift the value to the end instead of popping because both have time complexity of O(n)?
This is my solution
--------------------------------------
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
if len(nums) < 2:
return len(nums)
pointer = 2
for num in nums[2:]:
if nums[pointer-2] != num:
nums[pointer] = num
pointer += 1
return pointer
--------------------------------------
So basically, if the length of an array is less than 3, obviously everything is already sorted and correctly placed, therefore we just return it. The if condition in the beginning also allows us to start checking from the third element as well. After that we just start checking our current element and the one that is 2 steps behind. If there are equal, it means that the last two elements in the array are already equal to the current one and therefore we cannot place another one (it would be the third). In other cases, it is possible to add another element with the same value
Here's the way i did it. maybe it's faster
def removeDuplicates(self, nums) -> int:
j = 1
for i in range(2,len(nums)):
x = nums[i]
if x != nums[i-1] or x != nums[i-2]:
j += 1
nums[j] = x
if i == j + 1 and i + 1 < len(nums) and x == nums[i+1]:
j += 1
return j+1
Beautiful 🙌
well explained helped a lot!
My solution took 50 minutes and 38 lines of code.
There is an easier solution where R (right pointer) loops through the array and just checks whether nums[R] == nums[L] == nums[L-1].
If not:
L += 1
nums[L] = nums[R]
If yes: continue
Return: L+1
JS:
var removeDuplicates = function(nums) {
if (nums.length < 3) return nums.length // Handle edgecase
let replaceIndex = 1
for (let i = 2 ; i < nums.length ; i++) {
if (nums[i] != nums[replaceIndex] || nums[replaceIndex] != nums[replaceIndex-1]) {
replaceIndex++
nums[replaceIndex] = nums[i]
}
}
return replaceIndex+1
};
An even more intuitive solution. O(n)
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
if len(nums) == 1:
return
l = 0
r = 0
while r < len(nums):
# edge case
if l == 0 and nums[l+1] == nums[l]:
# skip the first value to continue as usual
r+=1
l+=1
if nums[r] != nums[l]:
l+=1
nums[l] = nums[r]
# Greedily check for one more duplicate as it is two
# if we get it we change else no problem.
if r+1 < len(nums) and nums[r+1] == nums[r]:
l+=1
nums[l] = nums[r]
r+=1
r+=1
return l+1
Great work @NeetCodeIO
Was thinking you actually stopped uploading daily leetcode video explanations when I didn't get any notification over the past few hours
is there any reason for not uploading daily challenges ?
whoever strugle to understand this sol
i think this easier to come up with
int removeDuplicates(vector& nums) {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
if(nums.size()
Good question
l,c = 1,1
for r in range(1,len(nums)):
if(nums[r]!=nums[r-1]):
c = 1
nums[l]=nums[r]
l += 1
else:
if c
geez before 5m you make this so redundant tbh that removes clarity
my solution : runtime 57 ms. mem 16.64 mb
code
for n in nums:
if nums.count(n)>2:
while nums.count(n)>2:
nums.remove(n)
return len(nums)
I would rather prefer to use a dictionary & if value for key > 2, pop the value & finally return the len(nums).
are these questions prasent in NeetCodeAll ?
Will not prasent 😀
Doesn't this have a crazy time complexity like O(n^2) or even O(n^3) since he has a nested while loop and a for loop in within it?
Please make a video on sliding window median
You rock!
Good Evening Sir.
I'm an Indian and trying to buy Ur Yearly Plan. I'm getting Ur card is declined. Please help me Sir .. I'm interested to crack MAANG interviews
Hi, please email me (neetcodebusiness@gmail.com)
Can be done in O(n)...
Todays Daily Challenge?
🙆♂️
This is nice solution