How to Print decreasing order of the PriorityQueue in O(1) Space

Am solving problems on Heaps and i want the output for a problem in decreasing order, using PriorityQueue.
<br>Input:
<br>1
<br>5 2
<br>12 5 787 1 23
<br>Output:
<br>23 787
<br> Wanted Output:
<br> 787 23

class GFG {
public static void main(String args[])throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while(t-->0) {
// take array and kth element iput
String n_k[] = br.readLine().split(" ");
// store array size in n && kth element to find in k
int n = Integer.parseInt(n_k[0]);
int k = Integer.parseInt(n_k[1]);
// Array String input
String s[] = br.readLine().split(" ");
int d[] = new int[n];
for(int i = 0 ; i < n ; i++) {
d[i] = Integer.parseInt(s[i]);
}
PriorityQueue<Integer> minHeap = new PriorityQueue<Integer>();
// delete the minimum element in the element and just keep 
// k greater element in the minHeap
for(int i = 0 ; i < n; i++) {
minHeap.add(d[i]);
// if size of heap increases pop tht last element 
if(minHeap.size()>k) {
minHeap.poll();
}
}
// print remaining element
//HERE IS THE PROBLEM I WANT IT IN " DECREASING ORDER "
// it gives me Increasing order
while(minHeap.size() > 0) {
System.out.print(minHeap.peek()+" ");
minHeap.poll();
}
System.out.println();
}// end of while
}// end of main
}// end of class

Input:
<br>1
<br>5 2
<br>12 5 787 1 23
<br>Output:
<br>23 787

You can use Stream to print the result:

try (Scanner scan = new Scanner(System.in)) {
int totalCases = scan.nextInt();
while (totalCases-- > 0) {
int n = scan.nextInt();
int k = scan.nextInt();
Queue<Integer> minHeap = new PriorityQueue<>(k);
for (int i = 0; i < n; i++) {
if (minHeap.size() == k)
minHeap.remove();
minHeap.add(scan.nextInt());
}
System.out.println(minHeap.stream()
.sorted(Comparator.reverseOrder())
.map(String::valueOf)
.collect(Collectors.joining(" ")));
}
}

Simply replace your code with below.

While declaring priority queue , you can provide comparator with reverse order.

PriorityQueue<Integer> newHeap = new PriorityQueue(minHeap.size(),Comparator.reverseOrder());

Here i had created one newHeap which will store all existing elements in reverseOrder and then doing all existing operation as it is.

It's giving expected answer.

1
5 2
12 5 787 1 23
O/P
787 23 

This will help you.

  class GFG {
public static void main(String args[]) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
String n_k[] = br.readLine().split(" ");
int n = Integer.parseInt(n_k[0]);
int k = Integer.parseInt(n_k[1]);
String s[] = br.readLine().split(" ");
int d[] = new int[n];
for (int i = 0; i < n; i++) {
d[i] = Integer.parseInt(s[i]);
}
PriorityQueue<Integer> minHeap = new PriorityQueue<Integer>();
for (int i = 0; i < n; i++) {
minHeap.add(d[i]);
// if size of heap increases pop tht last element
if (minHeap.size() > k) {
minHeap.poll();
}
}
PriorityQueue<Integer> newHeap = new PriorityQueue(minHeap.size(), Comparator.reverseOrder());
newHeap.addAll(minHeap);
while (newHeap.size() > 0) {
System.out.print(newHeap.peek() + " ");
newHeap.poll();
}
System.out.println();
}
}
}