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Finding Median in a Sorted Linked List

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Given A sorted linked list of N              elements. The task is to find the median in the given Sorted Linked List.
We know that median in a sorted array is the middle element.

Procedure to find median of N sorted numbers:  

if N is odd:
median is N/2th element
else
median is N/2th element + (N/2+1)th element

Examples: 

Input : 1->2->3->4->5->NULL
Output : 3
Input : 1->2->3->4->5->6->NULL
Output : 3.5

Simple approach  

  1. Traverse the linked list and count all elements.
  2. if count is odd then again traverse the linked list and find n/2th element.
  3. if count is even then again traverse the linked list and find: 
    (n/2th element+ (n/2+1)th element)/2

Note: The above solution traverse the linked list two times.

Efficient Approach: an efficient approach is to traverse the list using two pointers to find the number of elements. See method 2 of this post.
We can use the above algorithm for finding the median of the linked list. Using this algorithm we won’t need to count the number of element: 

  1. if the fast_ptr is Not NULL then it means linked list contain odd element we simply print the data of the slow_ptr.
  2. else if fast_ptr reach to NULL its means linked list contain even element we create backup of the previous node of slow_ptr and print (previous node of slow_ptr+ slow_ptr->data)/2

Below is the implementation of the above approach:  

C++

// C++ program to find median
// of a linked list
#include <bits/stdc++.h>
using namespace std;
 
// Link list node
struct Node {
    int data;
    struct Node* next;
};
 
/* Function to get the median of the linked list */
void printMidean(Node* head)
{
    Node* slow_ptr = head;
    Node* fast_ptr = head;
    Node* pre_of_slow = head;
 
    if (head != NULL) {
        while (fast_ptr != NULL && fast_ptr->next != NULL) {
 
            fast_ptr = fast_ptr->next->next;
 
            // previous of slow_ptr
            pre_of_slow = slow_ptr;
            slow_ptr = slow_ptr->next;
        }
 
        // if the below condition is true linked list
        // contain odd Node
        // simply return middle element
        if (fast_ptr != NULL)
            cout << "Median is : " << slow_ptr->data;
 
        // else linked list contain even element
        else
            cout << "Median is : "
                 << float(slow_ptr->data + pre_of_slow->data) / 2;
    }
}
 
/* Given a reference (pointer to
    pointer) to the head of a list
    and an int, push a new node on
    the front of the list. */
void push(struct Node** head_ref, int new_data)
{
    // allocate node
    Node* new_node = new Node;
 
    // put in the data
    new_node->data = new_data;
 
    // link the old list
    // off the new node
    new_node->next = (*head_ref);
 
    // move the head to point
    // to the new node
    (*head_ref) = new_node;
}
 
// Driver Code
int main()
{
    // Start with the
    // empty list
    struct Node* head = NULL;
 
    // Use push() to construct
    // below list
    // 1->2->3->4->5->6
    push(&head, 6);
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
 
    // Check the count
    // function
    printMidean(head);
 
    return 0;
}

                    

Java

// Java program to find median
// of a linked list
class GFG
{
 
    // Link list node
    static class Node
    {
 
        int data;
        Node next;
    };
 
    /* Function to get the median of the linked list */
    static void printMidean(Node head)
    {
        Node slow_ptr = head;
        Node fast_ptr = head;
        Node pre_of_slow = head;
 
        if (head != null)
        {
            while (fast_ptr != null && fast_ptr.next != null)
            {
 
                fast_ptr = fast_ptr.next.next;
 
                // previous of slow_ptr
                pre_of_slow = slow_ptr;
                slow_ptr = slow_ptr.next;
            }
 
            // if the below condition is true linked list
            // contain odd Node
            // simply return middle element
            if (fast_ptr != null)
            {
                System.out.print("Median is : " + slow_ptr.data);
            }
             
            // else linked list contain even element
            else
            {
                System.out.print("Median is : "
                        + (float) (slow_ptr.data + pre_of_slow.data) / 2);
            }
        }
    }
 
    /* Given a reference (pointer to
    pointer) to the head of a list
    and an int, push a new node on
    the front of the list. */
    static Node push(Node head_ref, int new_data)
    {
        // allocate node
        Node new_node = new Node();
 
        // put in the data
        new_node.data = new_data;
 
        // link the old list
        // off the new node
        new_node.next = head_ref;
 
        // move the head to point
        // to the new node
        head_ref = new_node;
        return head_ref;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Start with the
        // empty list
        Node head = null;
 
        // Use push() to construct
        // below list
        // 1.2.3.4.5.6
        head = push(head, 6);
        head = push(head, 5);
        head = push(head, 4);
        head = push(head, 3);
        head = push(head, 2);
        head = push(head, 1);
 
        // Check the count
        // function
        printMidean(head);
    }
}
 
// This code is contributed by PrinciRaj1992

                    

Python3

# Python3 program to find median
# of a linked list
class Node:
     
    def __init__(self, value):
         
        self.data = value
        self.next = None
     
class LinkedList:
 
    def __init__(self):
         
        self.head = None
 
    # Create Node and make linked list
    def push(self, new_data):
         
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node
         
    # Function to get the median
    # of the linked list   
    def printMedian(self):
         
        slow_ptr = self.head
        fast_ptr = self.head
        pre_of_show = self.head
        count = 0
         
        while (fast_ptr != None and
               fast_ptr.next != None):
            fast_ptr = fast_ptr.next.next
             
            # Previous of slow_ptr
            pre_of_slow = slow_ptr
            slow_ptr = slow_ptr.next
        # If the below condition is true
        # linked list contain odd Node
        # simply return middle element   
        if (fast_ptr):
            print("Median is :", (slow_ptr.data))
             
        # Else linked list contain even element
        else:
            print("Median is :", (slow_ptr.data +
                                  pre_of_slow.data) / 2)
                                   
# Driver code
llist = LinkedList()
 
# Use push() to construct
# below list
# 1->2->3->4->5->6
llist.push(6)
llist.push(5)
llist.push(4)
llist.push(3)
llist.push(2)
llist.push(1)
 
# Check the count
# function
llist.printMedian()
 
# This code is contributed by grand_master

                    

C#

// C# program to find median
// of a linked list
using System;
 
class GFG
{
 
    // Link list node
    class Node
    {
 
        public int data;
        public Node next;
    };
 
    /* Function to get the median
    of the linked list */
    static void printMidean(Node head)
    {
        Node slow_ptr = head;
        Node fast_ptr = head;
        Node pre_of_slow = head;
 
        if (head != null)
        {
            while (fast_ptr != null &&
                   fast_ptr.next != null)
            {
                fast_ptr = fast_ptr.next.next;
 
                // previous of slow_ptr
                pre_of_slow = slow_ptr;
                slow_ptr = slow_ptr.next;
            }
 
            // if the below condition is true linked list
            // contain odd Node
            // simply return middle element
            if (fast_ptr != null)
            {
                Console.Write("Median is : " +
                               slow_ptr.data);
            }
             
            // else linked list contain even element
            else
            {
                Console.Write("Median is : " +
                       (float)(slow_ptr.data +
                               pre_of_slow.data) / 2);
            }
        }
    }
 
    /* Given a reference (pointer to
    pointer) to the head of a list
    and an int, push a new node on
    the front of the list. */
    static Node push(Node head_ref, int new_data)
    {
        // allocate node
        Node new_node = new Node();
 
        // put in the data
        new_node.data = new_data;
 
        // link the old list
        // off the new node
        new_node.next = head_ref;
 
        // move the head to point
        // to the new node
        head_ref = new_node;
        return head_ref;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        // Start with the
        // empty list
        Node head = null;
 
        // Use push() to construct
        // below list
        // 1->2->3->4->5->6
        head = push(head, 6);
        head = push(head, 5);
        head = push(head, 4);
        head = push(head, 3);
        head = push(head, 2);
        head = push(head, 1);
 
        // Check the count
        // function
        printMidean(head);
    }
}
 
// This code is contributed by Rajput-Ji

                    

Javascript

<script>
 
// Javascript program to find median
// of a linked list
 
// A linked list node
class Node {
        constructor() {
                this.data = 0;
                this.next = null;
             }
        }
         
    /* Function to get the median of the linked list */
    function printMidean( head)
    {
        var slow_ptr = head;
        var fast_ptr = head;
        var pre_of_slow = head;
 
        if (head != null)
        {
            while (fast_ptr != null && fast_ptr.next != null)
            {
 
                fast_ptr = fast_ptr.next.next;
 
                // previous of slow_ptr
                pre_of_slow = slow_ptr;
                slow_ptr = slow_ptr.next;
            }
 
            // if the below condition is true linked list
            // contain odd Node
            // simply return middle element
            if (fast_ptr != null)
            {
                document.write("Median is : " + slow_ptr.data);
            }
             
            // else linked list contain even element
            else
            {
                document.write("Median is : "
                        +  (slow_ptr.data + pre_of_slow.data) / 2);
            }
        }
    }
 
    /* Given a reference (pointer to
    pointer) to the head of a list
    and an int, push a new node on
    the front of the list. */
    function push( head_ref,  new_data)
    {
        // allocate node
        var new_node = new Node();
 
        // put in the data
        new_node.data = new_data;
 
        // link the old list
        // off the new node
        new_node.next = head_ref;
 
        // move the head to point
        // to the new node
        head_ref = new_node;
        return head_ref;
    }
 
 
// Driver Code
 
// Start with the
// empty list
var head = null;
 
// Use push() to construct
// below list
// 1.2.3.4.5.6
head = push(head, 6);
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);
 
// Check the count
// function
printMidean(head);
 
// This code is contributed by jana_sayantan.
</script>

                    

Output
Median is : 3.5




Using Binary Search: The idea is simple we have to find the middle element of a linked list . 

As linked list is sorted then we can easily find middle element with the help of  binary search 

  • This approach involves dividing the linked list into two halves 
  • comparing the middle element with the median and making a recursive call on the appropriate half.

C++

#include <iostream>
 
// Definition for singly-linked list.
struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL) {}
};
 
class Solution {
public:
    double findMedian(ListNode *head) {
        int n = 0;
        ListNode *p = head;
         
        // Count the number of elements in the linked list
        while (p) {
            n++;
            p = p->next;
        }
         
        // Call the helper function to find the median
        return (n & 1) ? findKth(head, n / 2 ) :
               (findKth(head, n / 2) + findKth(head, n / 2 )) / 2.0;
    }
 
private:
    int findKth(ListNode *head, int k) {
        ListNode *p = head;
         
        // Iterate through the linked list until the kth element
        while (k-- > 0) {
            p = p->next;
        }
         
        // Return the kth element
        return p->val;
    }
};
 
int main() {
    ListNode *head = new ListNode(1);
    head->next = new ListNode(2);
    head->next->next = new ListNode(3);
    head->next->next->next = new ListNode(4);
    head->next->next->next->next = new ListNode(5);
     
    Solution solution;
    std::cout << "The median is:-> "<<solution.findMedian(head) << std::endl;
     
    return 0;
}
//This code is contributed by Veerendra Singh Rajpoot

                    

Java

class ListNode {
    int val;
    ListNode next;
    ListNode(int x)
    {
        val = x;
        next = null;
    }
}
 
class Solution {
    public double findMedian(ListNode head)
    {
        int n = 0;
        ListNode p = head;
 
        // Count the number of elements in the linked list
        while (p != null) {
            n++;
            p = p.next;
        }
 
        // Call the helper function to find the median
        return (n % 2 == 1) ? findKth(head, n / 2)
                            : (findKth(head, n / 2)
                               + findKth(head, n / 2 - 1))
                                  / 2.0;
    }
 
    private int findKth(ListNode head, int k)
    {
        ListNode p = head;
 
        // Iterate through the linked list until the kth
        // element
        while (k > 0) {
            p = p.next;
            k--;
        }
 
        // Return the kth element
        return p.val;
    }
}
 
public class GFG {
    public static void main(String[] args)
    {
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(3);
        head.next.next.next = new ListNode(4);
        head.next.next.next.next = new ListNode(5);
 
        Solution solution = new Solution();
        System.out.println("The median is: "
                           + solution.findMedian(head));
    }
}

                    

Python

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
class GFG:
    def findMedian(self, head):
        n = 0
        p = head
        # Count the number of elements in linked list
        while p:
            n += 1
            p = p.next
        # Call the helper function to the find the median
        return self.find_kth(head, n // 2) if n % 2 == 1 else (self.find_kth(head, n // 2) + self.find_kth(head, n // 2 - 1)) / 2.0
    def find_kth(self, head, k):
        p = head
        # Iterate through the linked list until
        # kth element
        while k > 0:
            p = p.next
            k -= 1
        return p.val
# Main driver code
if __name__ == "__main__":
    # Creating a linked list: 1 -> 2 -> 3 -> 4 -> 5
    head = ListNode(1)
    head.next = ListNode(2)
    head.next.next = ListNode(3)
    head.next.next.next = ListNode(4)
    head.next.next.next.next = ListNode(5)
    # Creating an instance of Solution class
    solution = GFG()
    # Finding and printing the median of linked list
    print "The median is:", solution.findMedian(head)

                    

C#

using System;
public class ListNode
{
    public int val;
    public ListNode next;
 
    public ListNode(int x)
    {
        val = x;
        next = null;
    }
}
 
public class Solution
{
    public double FindMedian(ListNode head)
    {
        int n = 0;
        ListNode p = head;
 
        // Count the number of elements in the linked list
        while (p != null)
        {
            n++;
            p = p.next;
        }
 
        // Call the helper function to find the median
        return (n % 2 == 1) ? FindKth(head, n / 2)
                            : (FindKth(head, n / 2)
                               + FindKth(head, n / 2 - 1))
                                  / 2.0;
    }
 
    private int FindKth(ListNode head, int k)
    {
        ListNode p = head;
 
        // Iterate through the linked list until the kth
        // element
        while (k > 0)
        {
            p = p.next;
            k--;
        }
 
        // Return the kth element
        return p.val;
    }
}
 
public class GFG
{
    public static void Main(string[] args)
    {
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(3);
        head.next.next.next = new ListNode(4);
        head.next.next.next.next = new ListNode(5);
 
        Solution solution = new Solution();
        Console.WriteLine("The median is:-> "
                           + solution.FindMedian(head));
    }
}

                    

Javascript

class ListNode {
    constructor(val) {
        this.val = val;
        this.next = null;
    }
}
 
class Solution {
    findMedian(head) {
        let n = 0;
        let p = head;
 
        // Count the number of elements in the linked list
        while (p) {
            n++;
            p = p.next;
        }
 
        // Call the helper function to find the median
        return (n % 2 === 1) ? this.findKth(head, Math.floor(n / 2)) :
            (this.findKth(head, n / 2) + this.findKth(head, n / 2 - 1)) / 2.0;
    }
 
    findKth(head, k) {
        let p = head;
 
        // Iterate through the linked list until the kth element
        while (k > 0) {
            p = p.next;
            k--;
        }
 
        // Return the kth element's value
        return p.val;
    }
}
 
const head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(3);
head.next.next.next = new ListNode(4);
head.next.next.next.next = new ListNode(5);
 
const solution = new Solution();
console.log("The median is: " + solution.findMedian(head));

                    

Output
The median is:-> 3




Explanation:

In this code, the findMedian function first counts the number of elements in the linked list using a while loop. After that, it calls the findKth function to find the kth element in the linked list, which can be used to find the median. If the number of elements is odd, the median is the middle element, and if it is even, the median is the average of the two middle elements. The findKth function uses another while loop to iterate through the linked list until the kth element, and returns the value of that element.

Time Complexity O(n): The time complexity of this approach is O(n), where n is the number of elements in the linked list. This is because in the worst case, the linked list needs to be traversed twice – once to count the number of elements and once to find the median. The first while loop in the findMedian function takes O(n) time, and the second while loop in the findKth function takes O(k) time, where k is the index of the median. In the worst case, k can be n/2, so the overall time complexity is O(n).
Auxiliary Space:  O(1):The space complexity of this approach is O(1), because only a few variables are used, and no extra data structures are needed. The linked list is traversed in place, so no additional memory is used to store the elements.

This approach is contributed by Veerendra Singh Rajpoot

If you find anything wrong or incorrect please let  us know.



Last Updated : 07 Dec, 2023
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