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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 + 3 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 + 0 comments

        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;

}
      • 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++; }

    • 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;

      }