We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Implementation
  4. Climbing the Leaderboard
  5. Discussions

Climbing the Leaderboard

Sort 1812 Discussions, By:

  • Tomheza
     + 110 comments

    I have one tip for you guys: go from the lowest to highest and start at index that you checked last, last time that way you wont need to go through entire array every time!

  • Kanahaiya
     + 14 comments

    Hello friends,

    climbing the leaderboard can be optimized by applying binary search

    If interested to know more about the generic algorithm in details-

    click here for the video explanation of generic algorithm with complexity analysis.

    or you can click on the image too to follow youtube tutorial.

    Here is the working solution:-

    source code :

    static int[] climbingLeaderboard(int[] scores, int[] alice) {
	int n = scores.length;
	int m = alice.length;

	int res[] = new int[m];
	int[] rank = new int[n];

	rank[0] = 1;

	for (int i = 1; i < n; i++) {
		if (scores[i] == scores[i - 1]) {
			rank[i] = rank[i - 1];
		} else {
			rank[i] = rank[i - 1] + 1;
		}
	}

	for (int i = 0; i < m; i++) {
		int aliceScore = alice[i];
		if (aliceScore > scores[0]) {
			res[i] = 1;
		} else if (aliceScore < scores[n - 1]) {
			res[i] = rank[n - 1] + 1;
		} else {
			int index = binarySearch(scores, aliceScore);
			res[i] = rank[index];

		}
	}
	return res;

}

private static int binarySearch(int[] a, int key) {

	int lo = 0;
	int hi = a.length - 1;

	while (lo <= hi) {
		int mid = lo + (hi - lo) / 2;
		if (a[mid] == key) {
			return mid;
		} else if (a[mid] < key && key < a[mid - 1]) {
			return mid;
		} else if (a[mid] > key && key >= a[mid + 1]) {
			return mid + 1;
		} else if (a[mid] < key) {
			hi = mid - 1;
		} else if (a[mid] > key) {
			lo = mid + 1;
		}
	}
	return -1;
}

    Would really appreciate your feedback like, dislike , comment etc. on my video.

    Do not forget to upvote, if you find it useful.

  • quachtridat
     + 26 comments

    My solution was to use stack and to store only distinct numbers.

    #include <stack>
#include <iostream>
using namespace std;

int main(){
  unsigned long n, m, i, tmp;
  cin >> n;
  stack<unsigned long> scores;
  for (i = 0; i < n; ++i) {
    cin >> tmp;
    if (scores.empty() || scores.top() != tmp) scores.push(tmp);
  }
  cin >> m;
  for (i = 0; i < m; ++i) {
    cin >> tmp;
    while (!scores.empty() && tmp >= scores.top()) scores.pop();
    cout << (scores.size() + 1) << endl;
  }
  return 0;
}

    One more good solution without using stack is to store only distinct numbers like above, then we can use binary search to find out the rank, as nelson_matos brought in at https://www.hackerrank.com/challenges/climbing-the-leaderboard/forum/comments/271253.

  • albiewalbie
     + 22 comments

    Python 2

    n = int(raw_input().strip())
scores = map(int,raw_input().strip().split(' '))
m = int(raw_input().strip())
alice = map(int,raw_input().strip().split(' '))

leaderboard = sorted(set(scores), reverse = True)
l = len(leaderboard)

for a in alice:
    while (l > 0) and (a >= leaderboard[l-1]):
        l -= 1
    print l+1

  • nelson_matos
     + 6 comments

    A good way to solve the problem is to store distinct values of the scores with O(N) complexity. For example, Test Case scores are: 100 100 50 40 40 20 10

    You only need to store: 100 50 40 20 10

    Then use a simple binary search catered to a descending list which returns the corresponding index.

Load more conversations