You are given an integer array score of size n, where score[i] is the score of the ith athlete in a competition. All the scores are guaranteed to be unique.
The athletes are placed based on their scores, where the 1st place athlete has the highest score, the 2nd place athlete has the 2nd highest score, and so on. The placement of each athlete determines their rank:
The 1st place athlete's rank is "Gold Medal".
The 2nd place athlete's rank is "Silver Medal".
The 3rd place athlete's rank is "Bronze Medal".
For the 4th place to the nth place athlete, their rank is their placement number (i.e., the xth place athlete's rank is "x").
Return an array answer of size n where answer[i] is the rank of the ith athlete.
Example 1:
Input: score = [5,4,3,2,1]
Output: ["Gold Medal","Silver Medal","Bronze Medal","4","5"]
Explanation: The placements are [1st, 2nd, 3rd, 4th, 5th].Example 2:
Input: score = [10,3,8,9,4]
Output: ["Gold Medal","5","Bronze Medal","Silver Medal","4"]
Explanation: The placements are [1st, 5th, 3rd, 2nd, 4th].
Constraints:
- n == score.length
- 1 <= n <= 104
- 0 <= score[i] <= 106
- All the values in score are unique.
My Solution
We can initialize a PriorityQueue with pairs of scores and index of each player. We then add all the pairs to the PriorityQueue. We poll the first three pairs from the PriorityQueue to find the indices at which we should place the “Gold Medal”, “Silver Medal”, and “Bronze Medal” ranks. We also check whether there are at least 2 or 3 elements in scores before doing so due to the constraints. Finally we update all the rest of the positions by polling the PriorityQueue and placing the rank in our output.
Time complexity is O(n * log n) because of the time complexity of the PriorityQueue. Space Complexity is O(n) because of the space required for the PriorityQueue.
class Solution {
public String[] findRelativeRanks(int[] score) {
int n = score.length;
String[] out = new String[n];
PriorityQueue<List<Integer>> pq = new PriorityQueue<>((a,b) -> b.get(1) - a.get(1));
int rem = 4;
for (int i = 0; i < n; i++) {
List<Integer> elem = new ArrayList<>();
elem.add(i);
elem.add(score[i]);
pq.add(elem);
}
List<Integer> first = pq.poll();
out[first.get(0)] = "Gold Medal";
if (n >= 2) {
List<Integer> second = pq.poll();
out[second.get(0)] = "Silver Medal";
}
if (n >= 3) {
List<Integer> third = pq.poll();
out[third.get(0)] = "Bronze Medal";
}
while (!pq.isEmpty()) {
List<Integer> place = pq.poll();
out[place.get(0)] = Integer.toString(rem);
rem++;
}
return out;
}
}