this one and yesterday's problem where we assumed our array to be a linked list and tried to find the node where the loop starts (Leetcode 287: find the duplicate number) were kinda hard to think when solving for the first time. One thing I can surely note from this that if the value of integers is always less than the length of the array (or one less), we can use it as an index to the array. Thanks for the explanation!!
you are genus you are amazing you just smoothly put that logic in hour mind that in my dreams i can scream it lout all i am saying is whaattttttta content , you have such a great skills keeep doing , i am greatful for this channel
Example taken in method 2, what if second occurrence of 3 is before 2(i.e, 2 and 3 change their positions of second occurrence, 2 at index 6 and 3 at index 5), at that situation , answer will not be in ascending order. Then we have to sort the returning array again ?
Can't we first sort the array and then iterate over the array and look for duplicates by comparing adjacent elements. Sorting will take O(nlogn) time and iteration will take O(n) time and no extra space is required. So, overall TC is O(n). Do correct me if I am wrong.
That is because the index ranges from 0 to (n-1) But the array elements are from 1 to n So if you do (index + 1) you are referring to the actual elements present in the array.
You are expected to return a result set, no need to count that. When we talk about space complexity, usually it is the extra space you would be needing.
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Sure...do help me out by sharing this video as much as you can :)
@@nikoo28 Okay
this one and yesterday's problem where we assumed our array to be a linked list and tried to find the node where the loop starts (Leetcode 287: find the duplicate number) were kinda hard to think when solving for the first time.
One thing I can surely note from this that if the value of integers is always less than the length of the array (or one less), we can use it as an index to the array.
Thanks for the explanation!!
What was the programming contest you are referring to?
you are genus you are amazing you just smoothly put that logic in hour mind that in my dreams i can scream it lout all i am saying is whaattttttta content , you have such a great skills keeep doing , i am greatful for this channel
thank you so much for these kind and awesome words
Superb Work. Explanation is too good. Very smooth :)
Very well explained bro !!
class Solution {
public:
vector findDuplicates(vector& nums) {
vector result ;
for(int i=0; i
Really helpful. Thanks
why answer is repeating when they occur in the array more than 4 times
great work keep it doing
Thank you tremendously
Heap sort can be use with space comp O(n).
If 0(1)
Arr[arr[i]%n ] = +n
New arr[I]
Again we will bring back original array by
Arr[I]
don't you think you are using o(n) space by modifying the array itself
space remains constant since , seperate or extra space is not utilised
OG Explanation SIr ❤❤
great explanation
Example taken in method 2, what if second occurrence of 3 is before 2(i.e, 2 and 3 change their positions of second occurrence, 2 at index 6 and 3 at index 5), at that situation , answer will not be in ascending order. Then we have to sort the returning array again ?
superb 😎😎😎
bhaisahab mst explaination
Can't we first sort the array and then iterate over the array and look for duplicates by comparing adjacent elements. Sorting will take O(nlogn) time and iteration will take O(n) time and no extra space is required. So, overall TC is O(n).
Do correct me if I am wrong.
Overall TC will be O(nlogn) here.
why we have (index+1) , can anyone explain
That is because the index ranges from 0 to (n-1)
But the array elements are from 1 to n
So if you do (index + 1) you are referring to the actual elements present in the array.
how to implement code for returning -1 when repeating element does not exist?
That will be a very small change in the code. If the resultSet is empty at the end, you can return a “-1”
Doesn't creating a result set consider as extra space?
You are expected to return a result set, no need to count that.
When we talk about space complexity, usually it is the extra space you would be needing.
@@nikoo28 understood 🙌🏻
Please some times name the Alogrithm name also
otherwise everything is good
not every algorithm has a name, and your interviewer will never ask the algorithm name.
If the element is 0 ,index becomes -1. It is showing array index out of bound error. How to solve this issue?
Element will never be 0 as we have range from [1,n]
Good 💥
Bhai, wanna know about desktop space analysis problem
I can’t find the link to the problem. Reach out to me on my email available in channel information.
If the element encounted 0 the. What to do
Then you will have to handle the 0 differently. For this particular problem check the constraints: 1
What is name of the approach
I don't know if there is a specific name to this approach
👍
it will not give sorted array for this input n = 8,
4 3 2 4 8 2 3 1, how to fix it, pls correct me if input is not valid.
What do you mean by not giving a sorted array? You are looking for duplicates..right??