Priority Queue using Doubly Linked List

Last Updated : 20 Mar, 2023

Given Nodes with their priority, implement a priority queue using doubly linked list.

push(): This function is used to insert a new data into the queue.
pop(): This function removes the element with the lowest priority value from the queue.
peek() / top(): This function is used to get the lowest priority element in the queue without removing it from the queue.

Approach :

1. Create a doubly linked list having fields info(hold the information of the Node), priority(hold the priority of the Node), prev(point to previous Node), next(point to next Node).
2. Insert the element and priority in the Node.
3. Arrange the Nodes in the increasing order of priority.

Below is the implementation of above steps :

C++

// C++ code to implement priority
// queue using doubly linked list
#include <bits/stdc++.h>
using namespace std;
 
// Linked List Node
struct Node {
    int info;
    int priority;
    struct Node *prev, *next;
};
 
// Function to insert a new Node
void push(Node** fr, Node** rr, int n, int p)
{
    Node* news = (Node*)malloc(sizeof(Node));
    news->info = n;
    news->priority = p;
 
    // If linked list is empty
    if (*fr == NULL) {
        *fr = news;
        *rr = news;
        news->next = NULL;
    }
    else {
        // If p is less than or equal front
        // node's priority, then insert at
        // the front.
        if (p <= (*fr)->priority) {
            news->next = *fr;
            (*fr)->prev = news->next;
            *fr = news;
        }
 
        // If p is more rear node's priority,
        // then insert after the rear.
        else if (p > (*rr)->priority) {
            news->next = NULL;
            (*rr)->next = news;
            news->prev = (*rr)->next;
            *rr = news;
        }
 
        // Handle other cases
        else {
 
            // Find position where we need to
            // insert.
            Node* start = (*fr)->next;
            while (start->priority > p)
                start = start->next;
            (start->prev)->next = news;
            news->next = start->prev;
            news->prev = (start->prev)->next;
            start->prev = news->next;
        }
    }
}
 
// Return the value at rear
int peek(Node* fr) { return fr->info; }
 
bool isEmpty(Node* fr) { return (fr == NULL); }
 
// Removes the element with the
// least priority value from the list
int pop(Node** fr, Node** rr)
{
    Node* temp = *fr;
    int res = temp->info;
    (*fr) = (*fr)->next;
    free(temp);
    if (*fr == NULL)
        *rr = NULL;
    return res;
}
 
// Driver code
int main()
{
    Node *front = NULL, *rear = NULL;
    push(&front, &rear, 2, 3);
    push(&front, &rear, 3, 4);
    push(&front, &rear, 4, 5);
    push(&front, &rear, 5, 6);
    push(&front, &rear, 6, 7);
    push(&front, &rear, 1, 2);
 
    cout << pop(&front, &rear) << endl;
    cout << peek(front);
 
    return 0;
}

C

// C code to implement priority
// queue using doubly linked list
#include <stdio.h>
#include <stdlib.h>
 
// Linked List Node
struct Node {
    int info;
    int priority;
    struct Node *prev, *next;
};
 
// Function to insert a new Node
void push(struct Node** fr, struct Node** rr, int n, int p)
{
    struct Node* news
        = (struct Node*)malloc(sizeof(struct Node*));
    news->info = n;
    news->priority = p;
 
    // If linked list is empty
    if (*fr == NULL) {
        *fr = news;
        *rr = news;
        news->next = NULL;
    }
    else {
        // If p is less than or equal front
        // node's priority, then insert at
        // the front.
        if (p <= (*fr)->priority) {
            news->next = *fr;
            (*fr)->prev = news->next;
            *fr = news;
        }
 
        // If p is more rear node's priority,
        // then insert after the rear.
        else if (p > (*rr)->priority) {
            news->next = NULL;
            (*rr)->next = news;
            news->prev = (*rr)->next;
            *rr = news;
        }
 
        // Handle other cases
        else {
 
            // Find position where we need to
            // insert.
            struct Node* start = (*fr)->next;
            while (start->priority > p)
                start = start->next;
            (start->prev)->next = news;
            news->next = start->prev;
            news->prev = (start->prev)->next;
            start->prev = news->next;
        }
    }
}
 
// Return the value at rear
int peek(struct Node* fr) { return fr->info; }
 
int isEmpty(struct Node* fr) { return (fr == NULL); }
 
// Removes the element with the
// least priority value from the list
int pop(struct Node** fr, struct Node** rr)
{
    struct Node* temp = *fr;
    int res = temp->info;
    (*fr) = (*fr)->next;
    free(temp);
    if (*fr == NULL)
        *rr = NULL;
    return res;
}
 
// Driver code
int main()
{
    struct Node *front = NULL, *rear = NULL;
    push(&front, &rear, 2, 3);
    push(&front, &rear, 3, 4);
    push(&front, &rear, 4, 5);
    push(&front, &rear, 5, 6);
    push(&front, &rear, 6, 7);
    push(&front, &rear, 1, 2);
 
    printf("%d\n", pop(&front, &rear));
    printf("%d\n", peek(front));
    return 0;
}

Java

// Java code to implement priority
// queue using doubly linked list
 
import java.util.*;
 
class Solution {
 
    static Node front, rear;
 
    // Linked List Node
    static class Node {
        int info;
        int priority;
        Node prev, next;
    }
 
    // Function to insert a new Node
    static void push(Node fr, Node rr, int n, int p)
    {
        Node news = new Node();
        news.info = n;
        news.priority = p;
 
        // If linked list is empty
        if (fr == null) {
            fr = news;
            rr = news;
            news.next = null;
        }
        else {
            // If p is less than or equal front
            // node's priority, then insert at
            // the front.
            if (p <= (fr).priority) {
                news.next = fr;
                (fr).prev = news.next;
                fr = news;
            }
 
            // If p is more rear node's priority,
            // then insert after the rear.
            else if (p > (rr).priority) {
                news.next = null;
                (rr).next = news;
                news.prev = (rr).next;
                rr = news;
            }
 
            // Handle other cases
            else {
 
                // Find position where we need to
                // insert.
                Node start = (fr).next;
                while (start.priority > p)
                    start = start.next;
                (start.prev).next = news;
                news.next = start.prev;
                news.prev = (start.prev).next;
                start.prev = news.next;
            }
        }
        front = fr;
        rear = rr;
    }
 
    // Return the value at rear
    static int peek(Node fr) { return fr.info; }
 
    static boolean isEmpty(Node fr) { return (fr == null); }
 
    // Removes the element with the
    // least priority value from the list
    static int pop(Node fr, Node rr)
    {
        Node temp = fr;
        int res = temp.info;
        (fr) = (fr).next;
        if (fr == null)
            rr = null;
 
        front = fr;
        rear = rr;
        return res;
    }
 
    // Driver code
    public static void main(String args[])
    {
 
        push(front, rear, 2, 3);
        push(front, rear, 3, 4);
        push(front, rear, 4, 5);
        push(front, rear, 5, 6);
        push(front, rear, 6, 7);
        push(front, rear, 1, 2);
 
        System.out.println(pop(front, rear));
        System.out.println(peek(front));
    }
}
 
// This code is contributed
// by Arnab Kundu

Python3

# Python3 code to implement priority
# queue using doubly linked list
 
# Linked List Node
class Node:
     
    def __init__(self):
         
        self.info = 0
        self.priority = 0
        self.next = None
        self.prev = None
 
front = None
rear = None
 
# Function to insert a new Node
def push(fr, rr, n, p):
     
    global front, rear
     
    news = Node()
    news.info = n
    news.priority = p
     
    # If linked list is empty
    if (fr == None):
        fr = news
        rr = news
        news.next = None
     
    else:
         
        # If p is less than or equal fr
        # node's priority, then insert at
        # the fr.
        if (p <= (fr).priority):
            news.next = fr
            (fr).prev = news.next
            fr = news
 
        # If p is more rr node's priority,
        # then insert after the rr.
        elif (p > (rr).priority):
            news.next = None
            (rr).next = news
            news.prev = (rr).next
            rr = news
         
        # Handle other cases
        else:
 
            # Find position where we need to
            # insert.
            start = (fr).next
             
            while (start.priority > p):
                start = start.next
                 
            (start.prev).next = news
            news.next = start.prev
            news.prev = (start.prev).next
            start.prev = news.next
             
    front = fr
    rear = rr
     
# Return the value at rr
def peek(fr):
     
    return fr.info
             
def isEmpty(fr):
     
    return fr == None
 
# Removes the element with the
# least priority value from the list
def pop(fr, rr):
     
    global front , rear
    temp = fr
    res = temp.info
    (fr) = (fr).next
     
    if (fr == None):
        rr = None
         
    front = fr
    rear = rr
    return res
 
# Driver code
if __name__=='__main__':
     
    push( front, rear, 2, 3)
    push( front, rear, 3, 4)
    push( front, rear, 4, 5)
    push( front, rear, 5, 6)
    push( front, rear, 6, 7)
    push( front, rear, 1, 2)
     
    print(pop(front, rear))
    print(peek(front))
 
# This code is contributed by rutvik_56

C#

// C# code to implement priority
// queue using doubly linked list
using System;
 
class GFG {
    public static Node front, rear;
 
    // Linked List Node
    public class Node {
        public int info;
        public int priority;
        public Node prev, next;
    }
 
    // Function to insert a new Node
    public static void push(Node fr, Node rr, int n, int p)
    {
        Node news = new Node();
        news.info = n;
        news.priority = p;
 
        // If linked list is empty
        if (fr == null) {
            fr = news;
            rr = news;
            news.next = null;
        }
        else {
            // If p is less than or equal front
            // node's priority, then insert at
            // the front.
            if (p <= (fr).priority) {
                news.next = fr;
                (fr).prev = news.next;
                fr = news;
            }
 
            // If p is more rear node's priority,
            // then insert after the rear.
            else if (p > (rr).priority) {
                news.next = null;
                (rr).next = news;
                news.prev = (rr).next;
                rr = news;
            }
 
            // Handle other cases
            else {
 
                // Find position where we
                // need to insert.
                Node start = (fr).next;
                while (start.priority > p) {
                    start = start.next;
                }
                (start.prev).next = news;
                news.next = start.prev;
                news.prev = (start.prev).next;
                start.prev = news.next;
            }
        }
        front = fr;
        rear = rr;
    }
 
    // Return the value at rear
    public static int peek(Node fr) { return fr.info; }
 
    public static bool isEmpty(Node fr)
    {
        return (fr == null);
    }
 
    // Removes the element with the
    // least priority value from the list
    public static int pop(Node fr, Node rr)
    {
        Node temp = fr;
        int res = temp.info;
        (fr) = (fr).next;
        if (fr == null) {
            rr = null;
        }
 
        front = fr;
        rear = rr;
        return res;
    }
 
    // Driver code
    public static void Main(string[] args)
    {
        push(front, rear, 2, 3);
        push(front, rear, 3, 4);
        push(front, rear, 4, 5);
        push(front, rear, 5, 6);
        push(front, rear, 6, 7);
        push(front, rear, 1, 2);
 
        Console.WriteLine(pop(front, rear));
        Console.WriteLine(peek(front));
    }
}
 
// This code is contributed by Shrikant13

Javascript

<script>
// javascript code to implement priority
// queue using doubly linked list
var front, rear;
 
    // Linked List Node
     class Node {
         constructor(){
        this.info = 0;
        this.priority = 0;
this.prev = null;
this.next = null;
}
    }
 
    // Function to insert a new Node
    function push(fr,  rr , n , p) {
var news = new Node();
        news.info = n;
        news.priority = p;
 
        // If linked list is empty
        if (fr == null) {
            fr = news;
            rr = news;
            news.next = null;
        } else {
            // If p is less than or equal front
            // node's priority, then insert at
            // the front.
            if (p <= (fr).priority) {
                news.next = fr;
                (fr).prev = news.next;
                fr = news;
            }
 
            // If p is more rear node's priority,
            // then insert after the rear.
            else if (p > (rr).priority) {
                news.next = null;
                (rr).next = news;
                news.prev = (rr).next;
                rr = news;
            }
 
            // Handle other cases
            else {
 
                // Find position where we need to
                // insert.
        var start = (fr).next;
                while (start.priority > p)
                    start = start.next;
                (start.prev).next = news;
                news.next = start.prev;
                news.prev = (start.prev).next;
                start.prev = news.next;
            }
        }
        front = fr;
        rear = rr;
    }
 
    // Return the value at rear
    function peek(fr) {
        return fr.info;
    }
 
    function isEmpty(fr) {
        return (fr == null);
    }
 
    // Removes the element with the
    // least priority value from the list
    function pop(fr,  rr) {
var temp = fr;
        var res = temp.info;
        (fr) = (fr).next;
        if (fr == null)
            rr = null;
 
        front = fr;
        rear = rr;
        return res;
    }
 
    // Driver code
     
 
        push(front, rear, 2, 3);
        push(front, rear, 3, 4);
        push(front, rear, 4, 5);
        push(front, rear, 5, 6);
        push(front, rear, 6, 7);
        push(front, rear, 1, 2);
 
        document.write(pop(front, rear)+"<br/>");
        document.write(peek(front));
 
// This code contributed by aashish1995
</script>

Output:

1
2

Related Article :
Priority Queue using Singly Linked List

Time Complexities and Comparison with Binary Heap:

               peek()    push()    pop()
-----------------------------------------
Linked List |   O(1)      O(n)      O(1)
            |
Binary Heap |   O(1)    O(Log n)   O(Log n)

Space complexity :- O(N) The space complexity of the given code is O(n), where n is the number of elements in the priority queue. This is because we are using a doubly linked list to store the elements, and each element requires a node that contains its value, priority, and pointers to its previous and next nodes in the list. The space required for the linked list grows linearly with the number of elements. Additionally, we are using some constant extra space for temporary variables used in the push and pop operations.

Suggest improvement

Convert given Binary Tree to Doubly Linked List in Linear time

Reverse a doubly linked list in groups of given size

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