Relative Ranks
EasyPrompt
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
1stplace athlete's rank is"Gold Medal". - The
2ndplace athlete's rank is"Silver Medal". - The
3rdplace athlete's rank is"Bronze Medal". - For the
4thplace to thenthplace athlete, their rank is their placement number (i.e., thexthplace 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.length1 <= n <= 1040 <= score[i] <= 106- All the values in
scoreare unique.
Approaches
3 approaches with complexity analysis and trade-offs.
This is a straightforward brute-force approach. For each athlete, we iterate through all other athletes to count how many have a higher score. This count directly gives us the rank of the current athlete.
Algorithm
- Create a result array
answerof sizen. - Loop through each score
score[i]in the input array. - For each
score[i], initialize itsrankto 1. - Start a nested loop to compare
score[i]with every other scorescore[j]. - If
score[j]is greater thanscore[i], increment therank. - After the inner loop, the calculated
rankis the placement of the athlete. - Convert the rank number into the corresponding string ("Gold Medal", "Silver Medal", "Bronze Medal", or the number itself as a string).
- Store this rank string in
answer[i]. - Return the
answerarray after the outer loop finishes.
Walkthrough
The algorithm works by taking each score one by one and comparing it against every other score in the array. By counting how many scores are greater than the current score, we can determine its rank. For example, if zero scores are greater, the rank is 1. If one score is greater, the rank is 2, and so on. Once the numeric rank is found, it's converted to the required string format and placed in the result array at the same index as the original score.
class Solution { public String[] findRelativeRanks(int[] score) { int n = score.length; String[] answer = new String[n]; for (int i = 0; i < n; i++) { int rank = 1; for (int j = 0; j < n; j++) { if (score[j] > score[i]) { rank++; } } if (rank == 1) { answer[i] = "Gold Medal"; } else if (rank == 2) { answer[i] = "Silver Medal"; } else if (rank == 3) { answer[i] = "Bronze Medal"; } else { answer[i] = String.valueOf(rank); } } return answer; }}Complexity
Time
O(n^2), where n is the number of athletes. For each athlete, we perform a linear scan of the entire score array.
Space
O(n) to store the result array. If the output array is not considered, the space complexity is O(1).
Trade-offs
Pros
Simple to understand and implement.
Requires no extra space besides the output array.
Cons
Highly inefficient for larger inputs due to its quadratic time complexity.
Likely to cause a 'Time Limit Exceeded' error on most online judges for the given constraints.
A more efficient approach is to sort the scores. However, sorting the original array would lose the initial order of athletes. To solve this, we can store each score along with its original index in a separate data structure (like a 2D array or a list of custom objects), sort this structure based on scores, and then use the stored indices to populate the result array correctly.
Algorithm
- Create a data structure to hold pairs of (score, original_index). A 2D array
int[n][2]is a good choice. - Populate this structure by iterating through the input
scorearray:pairs[i] = {score[i], i}. - Sort the
pairsarray in descending order based on the scores. - Create a result array
answerof sizen. - Iterate through the sorted
pairsarray. The element at indexiin the sorted array corresponds to the athlete with ranki+1. - For each
pairat indexi, get its original index,originalIndex = pair[1]. - Assign the rank based on
i: "Gold Medal" fori=0, "Silver Medal" fori=1, "Bronze Medal" fori=2, andString.valueOf(i+1)for others. - Place the rank string into
answer[originalIndex]. - Return the
answerarray.
Walkthrough
We first create a 2D array, scoreWithIndex, to store pairs of {score, original_index}. After populating this array, we sort it in descending order based on the scores. The sorted array now has the highest score at index 0, the second highest at index 1, and so on. We can then iterate through this sorted array. For each element at index i, we know its rank is i+1. We retrieve its original index and place the corresponding rank string ('Gold Medal', 'Silver Medal', etc.) into our final answer array at that original index.
import java.util.Arrays; class Solution { public String[] findRelativeRanks(int[] score) { int n = score.length; int[][] scoreWithIndex = new int[n][2]; for (int i = 0; i < n; i++) { scoreWithIndex[i][0] = score[i]; scoreWithIndex[i][1] = i; } // Sort in descending order based on score Arrays.sort(scoreWithIndex, (a, b) -> b[0] - a[0]); String[] answer = new String[n]; for (int i = 0; i < n; i++) { int originalIndex = scoreWithIndex[i][1]; if (i == 0) { answer[originalIndex] = "Gold Medal"; } else if (i == 1) { answer[originalIndex] = "Silver Medal"; } else if (i == 2) { answer[originalIndex] = "Bronze Medal"; } else { answer[originalIndex] = String.valueOf(i + 1); } } return answer; }}An alternative with the same complexity is to use a Max-Heap (PriorityQueue) to store the {score, index} pairs, which naturally processes them in descending order of score.
Complexity
Time
O(n log n), dominated by the sorting step. Populating the initial and final arrays takes O(n) time.
Space
O(n) to store the `scoreWithIndex` array or a similar data structure.
Trade-offs
Pros
Significantly faster than the brute-force approach for larger inputs.
A standard and reliable method for ranking problems where original order matters.
Cons
Requires extra space to store the score-index pairs.
This is the most efficient approach, which leverages the constraints on the score values (0 <= score[i] <= 10^6). Instead of a comparison-based sort, we can use an array as a direct map (or a hash map) to link scores to their original indices. This allows us to find the ranks in linear time.
Algorithm
- Find the maximum score (
maxScore) in the input array. - Create an integer array
scoreToIndexof sizemaxScore + 1. This array will map a score to its original index. Initialize it with a sentinel value like -1. - Iterate through the input
scorearray and populate the map:scoreToIndex[score[i]] = i. - Create the result array
answerof sizen. - Initialize a rank counter,
rank = 1. - Iterate downwards from
maxScoreto0. - If
scoreToIndex[s]is not the sentinel value, it means a scoresexisted in the input. - Retrieve its original index:
originalIndex = scoreToIndex[s]. - Assign the appropriate rank string based on the
rankcounter toanswer[originalIndex]. - Increment the
rankcounter. - Return the
answerarray.
Walkthrough
The core idea is similar to Counting Sort. We first find the maximum score to determine the size of a helper array, scoreToIndex. This array will use the score value as an index and store the original athlete's index as the value. After populating this map in O(n) time, we can iterate from the maxScore down to 0. When we find a valid entry in our scoreToIndex array, we know we've found the next highest score. We assign it the current rank, place the rank string in the final answer array at the correct original index, and then increment our rank counter for the next score we find.
import java.util.Arrays; class Solution { public String[] findRelativeRanks(int[] score) { int n = score.length; int maxScore = 0; for (int s : score) { maxScore = Math.max(maxScore, s); } // scoreToIndex[s] stores the original index of score s int[] scoreToIndex = new int[maxScore + 1]; Arrays.fill(scoreToIndex, -1); // Initialize with a sentinel value for (int i = 0; i < n; i++) { scoreToIndex[score[i]] = i; } String[] answer = new String[n]; int rank = 1; for (int s = maxScore; s >= 0; s--) { if (scoreToIndex[s] != -1) { // If score s exists int originalIndex = scoreToIndex[s]; if (rank == 1) { answer[originalIndex] = "Gold Medal"; } else if (rank == 2) { answer[originalIndex] = "Silver Medal"; } else if (rank == 3) { answer[originalIndex] = "Bronze Medal"; } else { answer[originalIndex] = String.valueOf(rank); } rank++; } } return answer; }}Complexity
Time
O(n + M), where `n` is the number of scores and `M` is the maximum score. Finding max score is O(n), populating the map is O(n), and iterating through the map is O(M).
Space
O(n + M), where `n` is the number of scores and `M` is the maximum score value. This is for the `scoreToIndex` mapping array and the result array.
Trade-offs
Pros
Fastest possible approach with linear time complexity.
Avoids the O(n log n) overhead of comparison-based sorting.
Cons
Space complexity is dependent on the maximum value of the score, which could be very large in other problems.
Can be memory-intensive if the range of scores is much larger than the number of scores.
Solutions
Solution
class Solution {public String[] findRelativeRanks(int[] score) { int n = score.length; Integer[] idx = new Integer[n]; for (int i = 0; i < n; ++i) { idx[i] = i; } Arrays.sort(idx, (i1, i2)->score[i2] - score[i1]); String[] ans = new String[n]; String[] top3 = new String[]{"Gold Medal", "Silver Medal", "Bronze Medal"}; for (int i = 0; i < n; ++i) { ans[idx[i]] = i < 3 ? top3[i] : String.valueOf(i + 1); } return ans; }}class Solution: def findRelativeRanks(self, score: List[int]) -> List[str]: n = len(score) idx = list(range(n)) idx . sort(key=lambda x: - score[x]) top3 = ['Gold Medal', 'Silver Medal', 'Bronze Medal'] ans = [None] * n for i in range(n): ans[idx[i]] = top3[i] if i < 3 else str(i + 1) return ansclass Solution {public: vector<string> findRelativeRanks(vector<int> &score) { int n = score.size(); vector<pair<int, int>> idx; for (int i = 0; i < n; ++i) idx.push_back(make_pair(score[i], i)); sort(idx.begin(), idx.end(), [&](const pair<int, int> &x, const pair<int, int> &y) { return x.first > y.first; }); vector<string> ans(n); vector<string> top3 = {"Gold Medal", "Silver Medal", "Bronze Medal"}; for (int i = 0; i < n; ++i) ans[idx[i].second] = i < 3 ? top3[i] : to_string(i + 1); return ans; }};Video walkthrough
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