506. Relative Ranks

https://leetcode.com/problems/relative-ranks/description/
Solution 1. Lambda expression.
    vector<string> findRelativeRanks(vector<int>& nums) {
        int N = nums.size();
        if(N==0) return vector<string> {};
        vector<int> rank2idx(N);
        vector<string> res(N);

        for(int i=0; i < N; i++) rank2idx[i] = i;
        sort(rank2idx.begin(), rank2idx.end(), [&](int i, int j){return nums[i]>nums[j];});
        for(int i=3; i < N; i++) {
            res[rank2idx[i]] = to_string(i+1);
        }
        if(N>0) res[rank2idx[0]] = "Gold Medal";
        if(N>1) res[rank2idx[1]] = "Silver Medal";
        if(N>2) res[rank2idx[2]] = "Bronze Medal";
        return res;
    }

Solution 2.  Using a map.
    vector<string> findRelativeRanks(vector<int>& nums) {
        map<int, int, greater<int>> score2idx;
        vector<string> res(nums.size());
        for(int i=0;i<nums.size();i++){
            score2idx[nums[i]] = i;
        }
        int i=1;
        for(auto &x: score2idx) {
            if(i==1) res[x.second] = "Gold Medal";
            else if(i==2) res[x.second] = "Silver Medal";
            else if(i==3) res[x.second] = "Bronze Medal";
            else res[x.second] = to_string(i);
            i++;
        }
        return res;
    }
Solution 3. Use a priority_queue
    vector<string> findRelativeRanks(vector<int>& nums) {
        priority_queue<pair<int,int>> q;
        vector<string> res(nums.size());
        for(int i=0;i<nums.size();i++){
            q.push(make_pair(nums[i],i));
        }
       
        for(int i=0; i<nums.size();i++) {
            auto x = q.top();
            if(i==0) res[x.second] = "Gold Medal";
            else if(i==1) res[x.second] = "Silver Medal";
            else if(i==2) res[x.second] = "Bronze Medal";
            else res[x.second] = to_string(i+1);
            q.pop();
        }
        return res;
    }