Given scores ofNathletes, find their relative ranks and the people with the top three highest scores, who will be awarded medals: "Gold Medal", "Silver Medal" and "Bronze Medal".

Example 1:

Input:
 [5, 4, 3, 2, 1]

Output:
 ["Gold Medal", "Silver Medal", "Bronze Medal", "4", "5"]

Explanation:
 The first three athletes got the top three highest scores, so they got "Gold Medal", "Silver Medal" and "Bronze Medal". 


For the left two athletes, you just need to output their relative ranks according to their scores.

Note:

1. N is a positive integer and won't exceed 10,000.
2. All the scores of athletes are guaranteed to be unique.

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Basically this question is to find out thescore->rankingmapping. The easiest way is to sort thosescoresinnums. But we will lose their original order. We can create (score,original index) pairs and sort them byscoredecreasingly. Then we will havescore->ranking(new index) mapping and we can useoriginal indexto create the result.

Time complexity: O(NlgN). Space complexity: O(N). N is the number of scores.

Example:

nums[i]    : [10, 3, 8, 9, 4]
pair[i][0] : [10, 3, 8, 9, 4]
pair[i][1] : [ 0, 1, 2, 3, 4]

After sort:
pair[i][0] : [10, 9, 8, 4, 3]
pair[i][1] : [ 0, 3, 2, 4, 1]

public class Solution {
    public String[] findRelativeRanks(int[] nums) {
        int[][] pair = new int[nums.length][2];

        for (int i = 0; i 
<
 nums.length; i++) {
            pair[i][0] = nums[i];
            pair[i][1] = i;
        }

        Arrays.sort(pair, (a, b) -
>
 (b[0] - a[0]));

        String[] result = new String[nums.length];

        for (int i = 0; i 
<
 nums.length; i++) {
            if (i == 0) {
                result[pair[i][1]] = "Gold Medal";
            }
            else if (i == 1) {
                result[pair[i][1]] = "Silver Medal";
            }
            else if (i == 2) {
                result[pair[i][1]] = "Bronze Medal";
            }
            else {
                result[pair[i][1]] = (i + 1) + "";
            }
        }

        return result;
    }
}

Also we can use an one dimension array. This will save a little bit space but space complexity is still O(n).

public class Solution {
    public String[] findRelativeRanks(int[] nums) {
        Integer[] index = new Integer[nums.length];

        for (int i = 0; i 
<
 nums.length; i++) {
            index[i] = i;
        }

        Arrays.sort(index, (a, b) -
>
 (nums[b] - nums[a]));

        String[] result = new String[nums.length];

        for (int i = 0; i 
<
 nums.length; i++) {
            if (i == 0) {
                result[index[i]] = "Gold Medal";
            }
            else if (i == 1) {
                result[index[i]] = "Silver Medal";
            }
            else if (i == 2) {
                result[index[i]] = "Bronze Medal";
            }
            else {
                result[index[i]] = (i + 1) + "";
            }
        }

        return result;
    }
}