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  1. Practice
  2. Algorithms
  3. Warmup
  4. Birthday Cake Candles
  5. Discussions

Birthday Cake Candles

  • jord_fumar + 3 comments

    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:

    1. Sort the array from min to max (quickSort)
    2. Parse from the very end of the array and iterate count++.
    3. When your values are no longer identical, stop.
    • jagmeetsingh + 5 comments

      I tried this

      static int birthdayCakeCandles(int[] a) {
        Arrays.sort(a);  //double pivot QuicKSort
        int count = 0;
        int i = a.length - 1;
        int tallest = a[i];
        while (a[i--] == tallest) {
            count++;
        }
        return count;
    }

      It is failing the test case

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

      • x_ireesh_x + 0 comments

        Pass all Testcases..

        static int birthdayCakeCandles(int[] ar) {
    Arrays.sort(ar);
    int result = 1;
    int decriment = 2;
    while((decriment <= ar.length) && (ar[ar.length-1] == ar[ar.length-decriment])){
        result++;
        decriment++;
    }
    return result;
}
      • rishavtandon93 + 1 comment

        Try This

        static int birthdayCakeCandles(int[] ar) {

            int count =0;
    Arrays.sort(ar);
    int maxHeight = ar[(ar.length)-1];
    for(int i=0;i<ar.length;i++)
    {
        if(ar[i]==maxHeight)
        {
            count++;
        }
    }

    return count;

}
        • Kanahaiya + 1 comment

          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.

          • wdodman + 0 comments

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

      • Amit_Sharma + 0 comments

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

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

      • dpkcareergreatn1 + 0 comments

        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

      • wdodman + 1 comment

        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; }

        • Kanahaiya + 0 comments

          Hi,

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

          https://youtu.be/1gxFE9EfanE

          Any comments and feedback will be appreciated.

    • bertgayus + 0 comments

      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

    • matthew_zilkie + 0 comments
      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;

      }