# 506.relative-ranks

## Statement

• Difficulty: Easy
• Tag: `数组` `排序` `堆（优先队列）`

• 名次第 `1` 的运动员获金牌 `"Gold Medal"`
• 名次第 `2` 的运动员获银牌 `"Silver Medal"`
• 名次第 `3` 的运动员获铜牌 `"Bronze Medal"`
• 从名次第 `4` 到第 `n` 的运动员，只能获得他们的名次编号（即，名次第 `x` 的运动员获得编号 `"x"`）。

``````输入：score = [5,4,3,2,1]

``````输入：score = [10,3,8,9,4]

``````

• `n == score.length`
• `1 <= n <= 104`
• `0 <= score[i] <= 106`
• `score` 中的所有值 互不相同

• Difficulty: Easy
• Tag: `Array` `Sorting` `Heap (Priority Queue)`

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.

## Solution

``````#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif

class Solution {
public:
vector<string> findRelativeRanks(vector<int> &score) {
map<int, int> m;
for (int i = 0; i < score.size(); i++) {
m[score[i]] = i;
}

auto f = [](int ix) -> string {
if (ix == 1) {
return "Gold Medal";
}

if (ix == 2) {
return "Silver Medal";
}

if (ix == 3) {
return "Bronze Medal";
}

};

auto res = vector<string>(score.size(), "");

int ix = score.size();
for (const auto &[k, v] : m) {
res[v] = f(ix);
--ix;
}

return res;
}
};

#ifdef LOCAL

int main() {
return 0;
}

#endif
``````