To all viewers who can watch mock interviews without being scared..u all strong hearted people deserve my respect..hats off! ... every time i watch a mock interview it feels so fucking scary i directly jump to leetcode to practice more xD
one way to do that is , whenever you find a number duplicated more than once , you can set the hashmap[number] = 1 , and then consider the rest as part of the answer , because whatever happens we know that if number is duplicated C times , we know for sure that C-1 operations will be needed .
Also thank you so so much for making such mock interviews. I am imagining myself being an interviewee and getting goosebumps while trying to think of an answer.... I would highly recommend this channel and the mock interviews ( including the system design mock interviews ) for anyone looking to become more confident in interviews :)
Good you discussed the edge case too. One solution is to simply binary search at each index which is always n log n and doesn't depend on the values of A[i]. To handle duplicates, we can do this thing on an unique array i.e. remove duplicates before doing the binary search. After that we traverse every index of that unique array and check that how many indexes on the right of it are covered by binary search with the upper bound as A[curr_index] + n - 1. All the rest of the elements needs to be changed. Also, thanks for doing it. It takes lots of efforts.
For the edge case, we can simply maintain 0 or 1 like the element is present or not because if any element is present more than one time then its duplicates must need to be converted to some other number. However, the approach discussed will give a TLE on LeetCode as it will depend on the maximum number in the array that is given as 1e9. This is the code for the approach discussed: class Solution { public int minOperations(int[] nums) { int n = nums.length; int mini = nums[0]; int maxi = nums[0]; for(int i = 0;i < n;i++){ mini = Math.min(mini,nums[i]); //1 maxi = Math.max(maxi,nums[i]); //1000 } HashSetset = new HashSet(); for(int num: nums){ set.add(num); } int ans = n; int curr = 0; int num = mini; int k = 0; for(num = mini,k = 0;k < n;k++,num++){ if(!set.contains(num)){ curr++; } } ans = Math.min(ans,curr); while(num
Hashing can drastically bring the TC from O(N^2) to O(N) but with a cost which is if Arr[i]>10**6 then the code will collapse. So with the constraints available on leetcode(same question obviously) the optimized TC comes down to O(NlogN) and SC to O(N). In my opinion this question doesn't qualify to be in the hard section of leetcode. Here is a piece of code that qualified AC. def minOperations(self, nums: List[int]) -> int: innitial_size = len(nums) nums = sorted(list(set(nums))) unique_array_size = len(nums) maxi = 0 for i in range(unique_array_size): pos = bisect.bisect_left(nums,nums[i]+innitial_size) if pos>=len(nums) or nums[pos]!=(nums[i]+innitial_size-1): pos = pos-1 else: pos = pos elements_within_range = (pos-i)+1 maxi = max(maxi,elements_within_range) elements_off_range = innitial_size-maxi return elements_off_range
Another way to handle duplicates in this problem is To remove all the duplicate elements from the nums array, (keeping their singular instances in the array) Let the number of elements removed be "k" Find the answer "R", for the same question considering the shorter nums array consisting of unique elements. return R + k
O(nlogn) Solution: def minOperations(nums): nums.sort() n = len(nums) max_len = 0 start = 0 for end in range(n): while start < end and nums[end] - nums[start] >= n: start += 1 max_len = max(max_len, end - start + 1) return n - max_len
I guess all this assigning max+1 is not required. Once can used actual hsh array instead of what Luv bhaiya used(Frequency array) , Just firstly making a frq array , then changing it to cumulative , then for all v[i] finding min of hsh[v[i]+n] - hsh[v[i]-1] just like he did. It will look pretty clean. Moreover , This solution is for when limit of all elements in array is 10^5 and with time complexity O(n) . (but the actual question has 10^9 range , Luv bhaiya assumed it at 28:37 ) For that time complexitiy : O(nlogn) , for every v[i] find lower_bound of v[i]+n-1 and that will help. My code if someone requires for 10^5 range: const int MAXN = 1e5 + 1; int hsh[MAXN]; int main(){ int n; cin >> n; vector v(n); for(int i=0;i> v[i];
Sample Edge cases to consider [-1, 5, 99, 100] ===> [97, 98, 99, 100] {2} [1, 1, 1, 2] ==> [1, 2, 3, 4] or [-1, 0, 1, 2] {2} My approach to first find max and min if max - min < num.len -1 in this case answer should be number of duplicates because we can surely fit those duplicates in the range in case if max -min > num.len -1 then we have to find the max number of missing unique element in the range of n to n + num.len - 1 if these numbers are less than duplicates then duplicates is the answer else these numbers - duplicates we can store each count in a hash map we can find the numbers in the range by simply by storing the hash element before that number this algo would be most optimal in case of time complexity and give correct answer Sample Edge cases to consider
Came up with both the approaches O(n) and O(nlogn) by myself ...first thought of the same hashmapping and prefix summing but later realised that N would exceed the space limit and hence came up with sorting and binary searching for lower/uppr bounds and then finding out where it lies and calculating the missing elements in btw :) but i somehow missed considering the case when duplicates are present in the input array...❤
//sort the array (nlogn) // find max and min (n) int i=min, int j=max int count=0; while(//some condition) { arr[j]=abs[length+min]; count++; if(arr[j]=0){ j--; max=arr[j]; } } return count;
Total T.C.= nlogn Though I need to filter out the if my array contains duplicates initially. That will require a separate check outside of this while loop . But my t.c. will remains same.
"I look so stupid kya"... That was hilarious :D. Having been on the interviewer seat. I too have heard stuff like these, especially when part of campus hiring events :D
@@KeertiPurswani liked the video only because of how you chose to handle this (rest of the video is superb too) but this set you apart as it felt like a more real experience :)
if the array is [2,7,8,9] then we need only one operation to covert 2--> 6 so it becomes continuous, but you consider the min in array then it will be 2 then it requires 3 operations to convert 7,8,9
Not sure if you have covered this already as a part of the feedback ( since I am commenting as I am watching ), but at 12:02, I felt that asking the interviewer an example again when the interviewer asks you an example is something which could have been avoided. Ideally it would be good if in the beginning of the interview, the candidate enumerates all possible edge cases he/she can think of. Thoughts?
This can also be one solution 1. Sort the array 2. Find min as a[0] 3. Subtract min-1 to all numbers 4. Now we have min as 1 and our max can he n. 5. Make another array as visited. 6. Ittrate in the array and for each value in the array mark visited[value] =1. 7. At last count number of zeros in visited array. Is it correct?? 😊
sort the array and count increasing elements from minimum element that to be every increased element should have a difference of one with previousElement. then (arr.size() - countOfIncreasing elements) do similar approch this taking maximum element and count decreasing element with difference of 1 with previous elements then again get (arr.size() - countOfmaximumelement) finally get minimum of both operation. it will give us answer complexity TC O(nlongn) SC O(1)
For first question first take set and store all uniquely and copy them in new array now we can sort them first and then find differnce between two consecutives now what we want is maximum continuous 1s in differnce and we will count maximum length of continuous 1s now we got our starting point that is minimum element which tells us the ending point as well now using map we can see either can we use the element or need to use operation Hope this get right Pls let me know mam
Can we do like...first we create a map and store frequency of every element of the array and then check for every element from the minimum like let's say the map is mp and if for every (1
Having helpful interviewers who provide hints boosts confidence in high stress interview settings. Ironically, it always turned out bad for me because the more helpful interviewers are, the quicker they provide hints. Even if I solved the complete question, using up that hint has always worked against me. P.S. - The expression of ma'am at 9:23 is enough to predict the result in an actual interview.😂
A simple solution to this would have been to sort the array and the just keep count of elements which are not contiguous to its left neighbour. for eg: [1 10 100 1000]:here ans is 3. this is because 3 elements (i.e 10 100 and 1000)are not contiguous to each other ( considering first element is always contiguous). similarly [1,2,100,1000]: here answer will be 2. this is again because there are only two elements(i.e 100 and 1000) which are not contiguous to their left neighbour. If this is wrong please correct me...thanks
- poorly handles case of elements far away from each other O(MAX- MIN) > O(N) (which he claims his time complexity to be) - uses extra space (doesn't work when MAX > 10^7) - doesn't handle the case of negative elements probably the worst approach one could've thought of. this is a simple sort + two-pointer approach
Instead of de-duplicating can we not just store 1 for the element instead of its frequency? Any duplicate element in the array comes at the cost of missing out on a consecutive element, so I think that'll be taken into account anyway.
My approach,after taking the inputs in array,First find the min element in the array, The maximum element is that array.length,Then use unordered_map and store all the value and it's frequencies,after that Create s foor loop range between min to max element,after that use find function (o(1)),if the element is present or not,if present then no operation else count++, lastly print the count..is it a good approach?
No Suppose after sorting nums becomes (1, 5,6,7,8,13) . So according to you the transformation becomes 1,2,3,4,5,6 in this case ans will be 3 as three elements are changed to 2,3 and 4 .but the actual answer is 2 , as we can change 1 -> 4 and 13 -> 9 so nums becomes 4,5,6,7,8,9.
13:17 Since he is fixing the max or the min, his hash map represents the frequencies of the elements in the range [min, min + n - 1] ( or [max - n + 1, max] if max is fixed ), hence hash map will still be of size n, not max - min, right?
Hey @keerti try to interview one mock with youtuber @william lin , we can learn a lot from him 🙃 i feel you,me and your audience will be surprised to the extreme 🌝
public static int complete(int[] arr) { int count = 0; int n = arr.length; Arrays.sort(arr); int max = arr[0] + n - 1; for(int i = 1 ; i < n ; i++) { if(arr[i] == arr[i-1] || arr[i] > max) count++; } return count; } public static void main(String[] args){ int[] arr = {3,3,4,5}; System.out.println(complete(arr)); } }
Could be a frequency array from highest to lowest value then sliding window of reuqired size and then minimum number of zeroes amongst all sliding window would be the answer
If we make all hash tables second as zero initially and if duplicates are present it will not matters if check only condition that if(mp[first or i] ==0){ count++; } else{ //Nothing to do } Am I right?
I don't want to show off... But Didi I watched your last google interviews and don't know how.. But i successfully thought of efficent solution within 20 mins
My Answer to first question: int makeContiNuous(int arr[],int n){ sort(arr,arr+n); int operations=0; unordered_sets; int maxi=arr[n-1]; int mini=arr[0]; int reqMax=mini+n-1; int toBeChanges=0; for(int i=0;i
I don't know why this question is quoted as "Hard" in Leetcode. Though this mock interview session is awesome. Such sessions are goldmines for many aspirants.
Yeah this question is pretty easily a decent medium , it should not be tagged as hard, the funny part is that he wasnt even able to solve this lol and he teaches cp and also has a course lol.
@@TheArchit1 dont act you can solve all the questions everytime you see them lol, khud karne baithega toh aukaat dikhegi, yaha youtube comments mein sab raja hai
wait is he a great competitive programmer? how can he be so without first defining 'n' in asymptotic time complexity for his algorithm. He says its O(n), Kreeti says n is the length of nums, then how come he reuse it as the (max-min) as n again? Literally they just blow off the very meaning of asymptotic time. Might be good at submitting solutions to codeforces, but that doesn't make you a great competitive programmer until you understand the true meaning of what you say.
can you please make a dedicated video on c++ and oops question but not that are definition based. something that involves pointer and all . also get well soon :)
11:25 if I entered 7 at the place of 4 , then 5 is not the biggest number it will be 7 . Then it will not continuous..... I didn't watch , whole video , you just told that at 11:25 interval is this logic same for all cases...so I commented , Please don't mind if I wrong.
In the whole interview Luv seemed overly aggressive and harsh in the way he speaks. He seems to fill the gap with over-explaining things. And "You know what is hash term? Right" was completely disrespectful towards the interviewer, it shows how he perceives himself with respect to the other person and it is clearly visible in other parts of the video as well. But in reality he is nothing more than a mediocre programmer with average problem solving skills. This question was not even that hard, took me about 2 and a half minutes to come up with a optimized working solution and I don't even consider myself good at cp.
Lol he as cracked directi...He is top ....so he is doing all those so quickly....I don't know which company your r working in.....bring some students who r not working and conduct mock interviews...
"yeah yeah the memory is there, but we are getting O(n) solution", such a competetive coder moment xD
lol😂😂
To all viewers who can watch mock interviews without being scared..u all strong hearted people deserve my respect..hats off! ...
every time i watch a mock interview it feels so fucking scary i directly jump to leetcode to practice more xD
Samee😂💯
one way to do that is , whenever you find a number duplicated more than once , you can set the hashmap[number] = 1 , and then consider the rest as part of the answer , because whatever happens we know that if number is duplicated C times , we know for sure that C-1 operations will be needed .
His approach was just wow loved the way how he came up with that hashing approach
True❤️😇
Also thank you so so much for making such mock interviews. I am imagining myself being an interviewee and getting goosebumps while trying to think of an answer.... I would highly recommend this channel and the mock interviews ( including the system design mock interviews ) for anyone looking to become more confident in interviews :)
Thank you so much Pranav. Your love and support means a lot to me. Will keep creating better content❤️😇
Good you discussed the edge case too. One solution is to simply binary search at each index which is always n log n and doesn't depend on the values of A[i]. To handle duplicates, we can do this thing on an unique array i.e. remove duplicates before doing the binary search. After that we traverse every index of that unique array and check that how many indexes on the right of it are covered by binary search with the upper bound as A[curr_index] + n - 1. All the rest of the elements needs to be changed.
Also, thanks for doing it. It takes lots of efforts.
For the edge case, we can simply maintain 0 or 1 like the element is present or not because if any element is present more than one time then its duplicates must need to be converted to some other number.
However, the approach discussed will give a TLE on LeetCode as it will depend on the maximum number in the array that is given as 1e9.
This is the code for the approach discussed:
class Solution {
public int minOperations(int[] nums) {
int n = nums.length;
int mini = nums[0];
int maxi = nums[0];
for(int i = 0;i < n;i++){
mini = Math.min(mini,nums[i]); //1
maxi = Math.max(maxi,nums[i]); //1000
}
HashSetset = new HashSet();
for(int num: nums){
set.add(num);
}
int ans = n;
int curr = 0;
int num = mini;
int k = 0;
for(num = mini,k = 0;k < n;k++,num++){
if(!set.contains(num)){
curr++;
}
}
ans = Math.min(ans,curr);
while(num
Hashing can drastically bring the TC from O(N^2) to O(N) but with a cost which is if Arr[i]>10**6 then the code will collapse. So with the constraints available on leetcode(same question obviously) the optimized TC comes down to O(NlogN) and SC to O(N). In my opinion this question doesn't qualify to be in the hard section of leetcode.
Here is a piece of code that qualified AC.
def minOperations(self, nums: List[int]) -> int:
innitial_size = len(nums)
nums = sorted(list(set(nums)))
unique_array_size = len(nums)
maxi = 0
for i in range(unique_array_size):
pos = bisect.bisect_left(nums,nums[i]+innitial_size)
if pos>=len(nums) or nums[pos]!=(nums[i]+innitial_size-1):
pos = pos-1
else:
pos = pos
elements_within_range = (pos-i)+1
maxi = max(maxi,elements_within_range)
elements_off_range = innitial_size-maxi
return elements_off_range
Another way to handle duplicates in this problem is
To remove all the duplicate elements from the nums array, (keeping their singular instances in the array)
Let the number of elements removed be "k"
Find the answer "R", for the same question considering the shorter nums array consisting of unique elements.
return R + k
These sessions and learnings are so helpful.
Thank you so much to both.
Thank you so much, means a lot to us❤️😇😇
O(nlogn) Solution:
def minOperations(nums):
nums.sort()
n = len(nums)
max_len = 0
start = 0
for end in range(n):
while start < end and nums[end] - nums[start] >= n:
start += 1
max_len = max(max_len, end - start + 1)
return n - max_len
I guess all this assigning max+1 is not required. Once can used actual hsh array instead of what Luv bhaiya used(Frequency array) , Just firstly making a frq array , then changing it to cumulative , then for all v[i] finding min of hsh[v[i]+n] - hsh[v[i]-1] just like he did. It will look pretty clean. Moreover , This solution is for when limit of all elements in array is 10^5 and with time complexity O(n) . (but the actual question has 10^9 range , Luv bhaiya assumed it at 28:37 ) For that time complexitiy : O(nlogn) , for every v[i] find lower_bound of v[i]+n-1 and that will help.
My code if someone requires for 10^5 range:
const int MAXN = 1e5 + 1;
int hsh[MAXN];
int main(){
int n;
cin >> n;
vector v(n);
for(int i=0;i> v[i];
int ans = 1e5+5;
for(int i=1;i
Sample Edge cases to consider
[-1, 5, 99, 100] ===> [97, 98, 99, 100] {2}
[1, 1, 1, 2] ==> [1, 2, 3, 4] or [-1, 0, 1, 2] {2}
My approach
to first find max and min
if max - min < num.len -1
in this case answer should be number of duplicates because we can surely fit those duplicates in the range
in case if max -min > num.len -1
then we have to find the max number of missing unique element in the range of n to n + num.len - 1
if these numbers are less than duplicates then duplicates is the answer
else these numbers - duplicates
we can store each count in a hash map
we can find the numbers in the range by simply by storing the hash element before that number
this algo would be most optimal in case of time complexity and give correct answer
Sample Edge cases to consider
Came up with both the approaches O(n) and O(nlogn) by myself ...first thought of the same hashmapping and prefix summing but later realised that N would exceed the space limit and hence came up with sorting and binary searching for lower/uppr bounds and then finding out where it lies and calculating the missing elements in btw :) but i somehow missed considering the case when duplicates are present in the input array...❤
//sort the array (nlogn)
// find max and min (n)
int i=min, int j=max
int count=0;
while(//some condition)
{
arr[j]=abs[length+min];
count++;
if(arr[j]=0){
j--;
max=arr[j];
}
}
return count;
Total T.C.= nlogn
Though I need to filter out the if my array contains duplicates initially. That will require a separate check outside of this while loop . But my t.c. will remains same.
Great content di... really loving the mock interview series. Watched all the videos. Plz never discontinue this mock interview series. ❤️🙏
"I look so stupid kya"... That was hilarious :D. Having been on the interviewer seat. I too have heard stuff like these, especially when part of campus hiring events :D
Hahaha, yeah. Hoping these mock interviews help with such small things😇😇
@@KeertiPurswani liked the video only because of how you chose to handle this (rest of the video is superb too) but this set you apart as it felt like a more real experience :)
Keerti mam.. please help me learn data structure to a professional level where I would be able to attend interviews ...as thorough professional..
@Keerti Purswani
Mam Mock interview of
CODE WITH HARRY
Please
LUV and NISHANT sir are Pro coders
But u are missing another pro coder
Code with Harry
Di your content is so amazing. Every new video just introduces new way of solution.
we can use sliding window to find maximum elements that is inside (maxelement - minelement)
Can you post the link to question if it is present in any platform? that would be really helpful.
It would be really helpful if you also provided the problem link(if exists) in the video description 🙏🙏🙏
if the array is [2,7,8,9] then we need only one operation to covert 2--> 6 so it becomes continuous, but you consider the min in array then it will be 2 then it requires 3 operations to convert 7,8,9
Looking at the level of question, it feels like i should better change my career trajectory, this was so difficult
Not sure if you have covered this already as a part of the feedback ( since I am commenting as I am watching ), but at 12:02, I felt that asking the interviewer an example again when the interviewer asks you an example is something which could have been avoided. Ideally it would be good if in the beginning of the interview, the candidate enumerates all possible edge cases he/she can think of. Thoughts?
This can also be one solution
1. Sort the array
2. Find min as a[0]
3. Subtract min-1 to all numbers
4. Now we have min as 1 and our max can he n.
5. Make another array as visited.
6. Ittrate in the array and for each value in the array mark visited[value] =1.
7. At last count number of zeros in visited array.
Is it correct?? 😊
sort the array and count increasing elements from minimum element that to be every increased element should have a difference of one with previousElement.
then (arr.size() - countOfIncreasing elements)
do similar approch this taking maximum element and count decreasing element with difference of 1 with previous elements then again get (arr.size() - countOfmaximumelement)
finally get minimum of both operation. it will give us answer
complexity
TC O(nlongn)
SC O(1)
Sort the array, remove the duplicats and use the lower_bound
For first question first take set and store all uniquely and copy them in new array now we can sort them first and then find differnce between two consecutives now what we want is maximum continuous 1s in differnce and we will count maximum length of continuous 1s now we got our starting point that is minimum element which tells us the ending point as well now using map we can see either can we use the element or need to use operation
Hope this get right
Pls let me know mam
Can we do like...first we create a map and store frequency of every element of the array and then check for every element from the minimum like let's say the map is mp and if for every (1
I am huge fan of @LUV he is legend in his studies always.
These interviews boosted my confidence a lot
Having helpful interviewers who provide hints boosts confidence in high stress interview settings. Ironically, it always turned out bad for me because the more helpful interviewers are, the quicker they provide hints. Even if I solved the complete question, using up that hint has always worked against me.
P.S. - The expression of ma'am at 9:23 is enough to predict the result in an actual interview.😂
So basically if interviewer gives hint then it is bad?
@@rocklee3254 anecdotal evidence
A simple solution to this would have been to sort the array and the just keep count of elements which are not contiguous to its left neighbour.
for eg:
[1 10 100 1000]:here ans is 3.
this is because 3 elements (i.e 10 100 and 1000)are not contiguous to each other ( considering first element is always contiguous).
similarly
[1,2,100,1000]: here answer will be 2.
this is again because there are only two elements(i.e 100 and 1000) which are not contiguous to their left neighbour.
If this is wrong please correct me...thanks
What about
[1,2,10,11]
According to you ans would be 1
But ans should be 2
that reaction of Keerti when he said you understand the Hashmap
right
- poorly handles case of elements far away from each other O(MAX- MIN) > O(N) (which he claims his time complexity to be)
- uses extra space (doesn't work when MAX > 10^7)
- doesn't handle the case of negative elements
probably the worst approach one could've thought of.
this is a simple sort + two-pointer approach
Could we have kept the count as 1 if an element is encountered regardless of its frequency? In that case cumulative sum would have been as expected.
right.
Instead of de-duplicating can we not just store 1 for the element instead of its frequency? Any duplicate element in the array comes at the cost of missing out on a consecutive element, so I think that'll be taken into account anyway.
I was thinking of the same.
You have to romove it. Which is one Operation
@@prateekkatiyar9532 removing that duplicate is same as adding a number which is not present. Eg 1,1,3. Removing 1 is same as adding the missing 2.
My approach,after taking the inputs in array,First find the min element in the array, The maximum element is that array.length,Then use unordered_map and store all the value and it's frequencies,after that Create s foor loop range between min to max element,after that use find function (o(1)),if the element is present or not,if present then no operation else count++, lastly print the count..is it a good approach?
I think sortign array and then using lower_bound would have simply solved our problem in time complexity=nlogn
No
Suppose after sorting nums becomes (1, 5,6,7,8,13) . So according to you the transformation becomes 1,2,3,4,5,6 in this case ans will be 3 as three elements are changed to 2,3 and 4 .but the actual answer is 2 , as we can change 1 -> 4 and 13 -> 9 so nums becomes 4,5,6,7,8,9.
I’m not even a coder, but still I’m watching this 😂 idk why!
Luv bro has got good coding skills
I am really happy that I was able to solve this question in 1 try
Does this problem exist on leetcode? If yes, please post the question number. THanks
In the first question can we use this approach like if (min-arr[i])>arr.size-1 increment count?
Thanks mam for being so active on utube !
I just wanted to ask whether the array is sorted or not. If the array is sorted then it could be done in O(n).
Always helpful ☺️✌️many problems I learnt from your channel get good confidence for big product based
Companies.
13:17 Since he is fixing the max or the min, his hash map represents the frequencies of the elements in the range [min, min + n - 1] ( or [max - n + 1, max] if max is fixed ), hence hash map will still be of size n, not max - min, right?
chup reh tu
yup
I thought the same approach with both reverse and forward.
Idk these questions seems to hard for me right now.
Hey @keerti try to interview one mock with youtuber @william lin , we can learn a lot from him 🙃 i feel you,me and your audience will be surprised to the extreme 🌝
At 11:24 approach will not work if 5 is repeated twice. So for the cases when there is repetition of number it will not work.
public static int complete(int[] arr) {
int count = 0;
int n = arr.length;
Arrays.sort(arr);
int max = arr[0] + n - 1;
for(int i = 1 ; i < n ; i++) {
if(arr[i] == arr[i-1] || arr[i] > max)
count++;
}
return count;
}
public static void main(String[] args){
int[] arr = {3,3,4,5};
System.out.println(complete(arr));
}
}
Could be a frequency array from highest to lowest value then sliding window of reuqired size and then minimum number of zeroes amongst all sliding window would be the answer
Mam Your work is awesome 😊😊
but the code area is slightly dark. plz use light theme or increase brightness.
Quite good didi keep bringing didi .I have solved this question with different approach but today I learn new approach which is quite good 🔥
If size of array is 15 and there is an element whose count is 11. Can we convert it into continues according to this question.
Min can also be changed like [2,10,11,12,13] here just 2 is to be changed
Please make a dedicated video on how to make linkedin profile so strong that recruiters of tech giant's approach freshers for job openings
That hashset question was out of the box. One should try in the real interview.
Do provide question link too
Can you please also attach the link to the question in the description box ..it would be helpful?
This can be done in nlogn after sorting and doing operation (in On) and counting em?
If we make all hash tables second as zero initially and if duplicates are present it will not matters if check only condition that
if(mp[first or i] ==0){
count++;
}
else{
//Nothing to do
}
Am I right?
I was just watching his stl series ans i got this video , what a timing 😄
Another nice interview, looking forward for more informative videos from you Keerti 😊😊😍😘
okay , what i can see is the question is somewhat like longest consecutive sequence present in an array
Can someone provide the leetcode question link for this question ??
Can I get a leetcode link to this problem?
By continuous, I think he assumed the numbers would be in a sequential manner.
I don't want to show off... But Didi I watched your last google interviews and don't know how.. But i successfully thought of efficent solution within 20 mins
I have no idea about whats going on.. but I abolutely enjoyed it 😊
You look sick tc. Great video tho✌🏼😅 luv is too good.
Can you also mentioned question link where it asked , so that we could solve.
Can I have the problem link to check whether my solution got accepted or not?
Maam can you share the problem link so that we can practice and see if our approach passes all test cases or not?
leetcode 2009
My Answer to first question:
int makeContiNuous(int arr[],int n){
sort(arr,arr+n);
int operations=0;
unordered_sets;
int maxi=arr[n-1];
int mini=arr[0];
int reqMax=mini+n-1;
int toBeChanges=0;
for(int i=0;i
I don't know why this question is quoted as "Hard" in Leetcode. Though this mock interview session is awesome. Such sessions are goldmines for many aspirants.
Yeah this question is pretty easily a decent medium , it should not be tagged as hard, the funny part is that he wasnt even able to solve this lol and he teaches cp and also has a course lol.
@@TheArchit1 dont act you can solve all the questions everytime you see them lol, khud karne baithega toh aukaat dikhegi, yaha youtube comments mein sab raja hai
dii,
what if i sort the array and check for two elements
such that
if a[i+1]-a[i]!=1
count++;
count will be the answer do it work
@LUV please check this out sir
Luv OP
❤️❤️
I think this will not work for the case
[1,20,21,22,24]
11:10 no it will not work if number repeats 2 or more time 😂😂
18:57 if(count>1)
replaceNeeded += count -1
20:29 what if random number is in array 😂😂
Mam where is the link of this problem ? I want to solve this problem .
we can do this in O(n) space and time by taking two pointer and hash map. can you share me question link so that I can check if it's possible
wait is he a great competitive programmer? how can he be so without first defining 'n' in asymptotic time complexity for his algorithm. He says its O(n), Kreeti says n is the length of nums, then how come he reuse it as the (max-min) as n again? Literally they just blow off the very meaning of asymptotic time. Might be good at submitting solutions to codeforces, but that doesn't make you a great competitive programmer until you understand the true meaning of what you say.
can you please make a dedicated video on c++ and oops question but not that are definition based. something that involves pointer and all .
also get well soon :)
question link? plz
please paste the leetcode link of the question in the description
11:25 if I entered 7 at the place of 4 , then 5 is not the biggest number it will be 7 . Then it will not continuous.....
I didn't watch , whole video , you just told that at 11:25 interval is this logic same for all cases...so I commented , Please don't mind if I wrong.
Fixing minimum doesn't work
my relative told me about code to care challenge 2021 philips can you explain more about it in your next video
In the whole interview Luv seemed overly aggressive and harsh in the way he speaks. He seems to fill the gap with over-explaining things. And "You know what is hash term? Right" was completely disrespectful towards the interviewer, it shows how he perceives himself with respect to the other person and it is clearly visible in other parts of the video as well. But in reality he is nothing more than a mediocre programmer with average problem solving skills. This question was not even that hard, took me about 2 and a half minutes to come up with a optimized working solution and I don't even consider myself good at cp.
Lol he as cracked directi...He is top ....so he is doing all those so quickly....I don't know which company your r working in.....bring some students who r not working and conduct mock interviews...
How about initialising the duplicates as -1....
Leetcode 2009 number problem is the question guys
Can anyone give link to this leetcode problem?
Leet code 2009
Can I get the leetcode link of this question
I heard someone telling him from behind.
Striver bhaiya kb aa rhe 😂
Is this question in any websites like leetcode etc? How can i run my code. I have a different logic
yup, its in leetcode..
Please comment down the problem number on leetcode so that we could try on our own.