Problem

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.

Examples

Example 1:

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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:

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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].

Solutions

Video explanation

Here is the video explaining below methods in detail. Please check it out:

Method 1 - Sorting value and index

Code

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public String[] findRelativeRanks(int[] score) {
	int n = score.length;
	
	int[][] pair = new int[n][2];
	for (int i = 0; i < n; i++) {
		pair[i][0] = score[i];
		pair[i][1] = i;
	}

	Arrays.sort(pair, (a, b) -> b[0] - a[0]);
	
	String[] ans = new String[n];
	
	for (int i = 0; i < n; i++) {
		int idx = pair[i][1];
		if (i == 0) {
			ans[idx] = "Gold Medal"; 
		}  else if (i == 1) {
			ans[idx] = "Silver Medal"; 
		} else if (i == 2) {
			ans[idx] = "Bronze Medal"; 
		} else {
			ans[idx] = (i+1) + "";
		}
	}
	
	return ans;
}

Complexity

  • ⏰ Time complexity: O(nlogn) - sorting the array
  • 🧺 Space complexity: O(n)

Method 2 - Using Priority Queue OR Max Heap

Create the max heap, add the array index to priority queue, and use the comparator on array[idx]. Now, we start polling value from the heap, getting the max value, till we empty the queue.

Code

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public String[] findRelativeRanks(int[] score) {
        int n = score.length;
        
        PriorityQueue<Integer> pq = new PriorityQueue<Integer>((a,b) -> score[b] - score[a]);
        for (int i = 0; i < n; i++) {
            pq.add(i);
        }

        int i = 1;
        String[] ans = new String[n];
        
        while(!pq.isEmpty()) {
            int idx = pq.poll();
            if (i == 1) {
                ans[idx] = "Gold Medal";
            } else if (i == 2) {
                ans[idx] = "Silver Medal";
            } else if (i == 3) {
                ans[idx] = "Bronze Medal";
            } else {
                ans[idx] = "" + i;
            }
            i++;
        }
        
        return ans;      
}

Complexity

  • ⏰ Time complexity: O(n log n) - n elements are polled from queue, and each polling takes log n time.
  • 🧺 Space complexity: O(n) for storing values in heap.