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

Birthday Cake Candles

  • jord_fumar + 2 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 + 1 comment

      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;
}
    • 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