Priority queue.cpp

/*
Print Binary Tree levels in sorted order 
Medium Accuracy: 55.08% Submissions: 1401 Points: 4
Given an array arr[] which contains data of N nodes of Complete Binary tree in level order fashion. The task is to print the level order traversal in sorted order.

Example 1:

Input:
N = 7
arr[] = {7 6 5 4 3 2 1}
Output:
7
5 6
1 2 3 4
Explanation: The formed Binary Tree is:
 7
 / \
 6 5
 / \ / \
 4 3 2 1
Example 2:

Input:
N = 6
arr[] = {5 6 4 9 2 1}
Output:
5
4 6
1 2 9
Explanation: The formed Binary Tree is:
 5
 / \
 6 4
 / \ / 
 9 2 1 
*/
 vector <vector <int>> binTreeSortedLevels (int arr[], int n)
 {vector<vector<int>> ans;
 int height= log2(n);
 priority_queue<int,vector<int>, greater<int>> pq;
 
 int idx=0;
 for(int i=0;i<=height;i++)
 {
 vector<int> temp;
 for(int j=0;j<pow(2,i) && idx < n;j++)
 {
 pq.push(arr[idx]);
 idx++;
 }
 while(pq.size() > 0)
 {
 temp.push_back(pq.top());
 pq.pop();
 }
 ans.push_back(temp);
 }
 
 return ans;
 // Your code here
 }
};
/*
Geek and Contest || 
Easy Accuracy: 74.26% Submissions: 1128 Points: 2
Geek hosted a contest and N students participated in it. The score of each student is given by an integer array arr[]. The task is to print the number of each student (indexes) in the order they appear on the scoreboard.
A student with a maximum score appears first. If two people have the same score then a higher indexed student appears first.

Example 1:

Input:
N = 5
arr[] = {450, 230, 730, 230, 150}
Output:
3 1 4 2 5
Explanation:
1. The top scorer is at 3rd index, therefore 3 
at the first index.
2. The second-highest scorer is at 1st index therefore
 1 is at the second index.
3. The third-highest scorer is at 4th index therefore
 4 at the thirst index.
*/
#include <bits/stdc++.h>
using namespace std;
int main() {
	int t;
	cin>>t;
	while(t--){
	 int n;
	 cin>>n;
	 priority_queue<pair<int,int>>q;
	 for(int i=1;i<=n;i++){
	 int x;
	 cin>>x;
	 q.push({x,i});
	 }
	 while(!q.empty()){
	 cout<<q.top().second<<" ";
	 q.pop();
	 }
	 
	 cout<<endl;
	 
	}
	
	
	return 0;
}