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If you dont mind can you please provide a feedback on my video by commenting or liking & disliking. It will help me and others too to find the solution over internet.

If you dont mind can you please provide a feedback on my video by commenting or liking & disliking. It will help me and others too to find the solution over internet.

Count should be initialized to 1 as there is atleast one tallest. In this case, if the frequency of largest element is 0, count would return 0 which is an error

if you use quicksort u already have O(nlogn) which is worse than just iteratring through the array and saving the highest value and then just iterate a second time incrementing a counter everytime the value is similar to the saaved one. then complexity is O(2n) = O(n).
With n = 100 you would have ~ 600 steps(n*logn = 100 * 6 = 600) with your solution and just 200 steps with mine

Did you try using c++ std::sort(ar.begin(), ar.end()); ? I did and it worked flawlessly.
int birthdayCakeCandles(int n, vector<int> ar) {
// Complete this function
sort(ar.begin(), ar.end());
int max = ar[ar.size() -1];
int count = 0;
for(auto num : ar) {
if(num == max) ++count;
}
return count;

## Birthday Cake Candles

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Be aware of time complexity. In my testcase 4, n was 10000. If you use a sorting algorithm with n^2, you will get a runtime error.

The way I structured my code is:

I tried this

It is failing the test case

100000 followed by 100000 times 9999999 (as height of candles(array values)).

Pass all Testcases..

Try This

static int birthdayCakeCandles(int[] ar) {

Hi,

Your solution is good but it will take O(nlogn) time due to sorting which can be optimized to O(n)

Here is the video explanation of my solution in O(n) time -

https://youtu.be/1gxFE9EfanE

Any comments and feedback will be appreciated.

Thanks for that. I worked out a solution where I do it in O(n) time. Posted on your video.

most welcome. :)

If you dont mind can you please provide a feedback on my video by commenting or liking & disliking. It will help me and others too to find the solution over internet.

thanks

most welcome. :)

If you dont mind can you please provide a feedback on my video by commenting or liking & disliking. It will help me and others too to find the solution over internet.

need to add one more check in while condition for all the array elements with same value

while ( i > -1 && a[i--] == tallest ) { count++; }

Count should be initialized to 1 as there is atleast one tallest. In this case, if the frequency of largest element is 0, count would return 0 which is an error

Mine fails on that too. ar.sort(); ar.reverse(); let m = 0; let k = ar[0]; while (parseInt(ar[m]) === parseInt(k)){ m++; } return m; }

Hi,

Here is the video explanation of my solution in O(n) time -

https://youtu.be/1gxFE9EfanE

Any comments and feedback will be appreciated.

while (

i>-1&&a[i--] == tallest) { count++; }You just forgot to check whether the index i is within the bounds.

if you use quicksort u already have O(nlogn) which is worse than just iteratring through the array and saving the highest value and then just iterate a second time incrementing a counter everytime the value is similar to the saaved one. then complexity is O(2n) = O(n). With n = 100 you would have ~ 600 steps(n*logn = 100 * 6 = 600) with your solution and just 200 steps with mine

}

Thanks a lot!!.I was just wondering where did i go wrong..Thanks a lot again