[Java] Leetcode 26/80. Remove Duplicates from Sorted Array I/II [Two Pointers #3]

[@EricProgramming]

class SolutionNo1 {
public int removeDuplicates(int[] nums) {
int L = 0, R = 1;
int N = nums.length;
//L

[@ketanmuttha1066]

Thanks, Eric. I was trying to understand the reason to have this condition in your solution1
nums[++L] = nums[R]; ? why did you modify the array?

[@ingluissantana]

The 26 was easy but the 80 is still killing me. :P

[@vishalupadhayay6391]

I was doing something very similar but then some test cases! uff!

[@degea9]

what tool did you use to draw on the web browser's screen ?

[@EricProgramming]

It used to call web paint, but I think their interface has changed.

[@pro_pranay]

Hash set?

[@luz8011]

wow, thank you so much!

[@wonghuang1830]

Thank you!

[@Rob-J-BJJ]

class Solution {
public int removeDuplicates(int[] nums) {
int k = 1;
int L = 0;
int R = 1;
int N = nums.length;
while(R < N) {
if (nums[L] == nums[R]) {
R++;


}
else if (nums[L] != nums[R]) {
L++;
int temp = nums[L];
nums[L] = nums[R];
// nums[R] = temp;
k++;


}
}

return k;
}
}

[@Rob-J-BJJ]

how come i didnt do a base case and still passed all test cases?

[@imram4u]

Below is the python solution I tried to write as per the explanation. But it is throwing a wrong answer. can someone correct it pls
class Solution:
def removeDuplicates(self, nums: List[int]) -> int:
n = len(nums)
left =0
right = 1
count = 0
while(right < n):
if (nums[left] != nums[right]):
left +=1
nums[left] = nums[right]
count = 0
elif (nums[left] == nums[right] and count < 1):
count +=1
nums[left + 1] = nums[right]
right +=1
return left + 1

[@chaitanyasharma6270]

public int removeDuplicates(int[] nums) {
if(nums.length

[@sulkhan5302]

Honestly, I love watching your videos but this video really wasn't as clearly explained as the others (specifically the second problem).

[@vishalupadhayay6391]

It could have been better! Agree.

[@vishalupadhayay6391]

Agree! Specially problem no 80