Find a peak element in Linked List

Given a linked list of integers, the task is to find a peak element in the given linked list. An element is a peak, if it is greater than or equals to its neighbors. For boundary elements, consider only one neighbor.

Examples:

Input : List = {1 -> 6 -> 8 -> 4 -> 12}
Output : 8
Explanation: In this list, 8 is greater than both its neighbors (6 and 4), making it a peak element. Other elements do not satisfy the peak condition.
Input : List = {1->2->3->4}
Output : 4
Explanation: The list is strictly increasing. In such cases, the last element (4) is always a peak, as it is greater than its only neighbor (3).

Approach:

To find a peak element in a linked list, we need to identify a node that is greater than or equal to its neighbors. Start from the beginning of the list and compare each node with the next one. If we find a node that is greater than or equal to the next node, it’s a peak, return it.
Notice, we only check the current node with next node, why not with previous node? Because, in the previouse iteration we have already checked the same and if prev was not peak it means it must be smaller than current element).
If we reach the end of the list without finding such a node, check if the last node is greater than or equal to the previous node; if so, it’s a peak.

Below is the implementation of the above approach:

C++

// C++ implementation to find the peak
// element in the Linked List
#include <bits/stdc++.h>
using namespace std;

struct Node {
    int data;  
    Node* next; 
    Node(int x) {
        data = x;
        next = nullptr;
    }
};

// Function to find a peak element in linked list
int findPeakElement(Node* head) {

     // If the list is empty, return -1
    if (!head) return -1;
    
    // Pointer to store the previous node
    Node* prev = nullptr; 
    
    // Traverse the list to find the peak element
    while (head->next) {
        // If the current node is a peak
        // then return its value
        if (head->data >= head->next->data)
          return head->data; 

        // Update the previous node pointer
        prev = head; 

        // Move to the next node
        head = head->next; 
    }
    
    // If the last node is greater than or equal
    // to the previous node, it's a peak
    int peakEle;
    if (prev && head->data >= prev->data) 
      peakEle = head->data;
    
    return peakEle; 
}

int main() {
    // Create a linked list with values:
    // 1 -> 2 -> 3 -> 4
    Node* head = new Node(1);
    head->next = new Node(2);
    head->next->next = new Node(3);
    head->next->next->next = new Node(4);

    // Find the peak element in the linked list
    int peakElement = findPeakElement(head);

    cout << peakElement << endl;

    return 0;
}

// C implementation to find the peak
// element in the Linked List
#include <stdio.h>
#include <stdlib.h>

typedef struct Node {
    int data;
    struct Node* next;
} Node;

// Function to find a peak element in linked list
int findPeakElement(Node* head) {
    // If the list is empty, return -1
    if (!head) return -1;

    // Pointer to store the previous node
    Node* prev = NULL;

    // Traverse the list to find the peak element
    while (head->next) {
        // If the current node is a peak
        // then return its value
        if (head->data >= head->next->data)
            return head->data;

        // Update the previous node pointer
        prev = head;

        // Move to the next node
        head = head->next;
    }

    // If the last node is greater than or equal
    // to the previous node, it's a peak
    int peakEle = head->data;
    if (prev && head->data >= prev->data)
        peakEle = head->data;

    return peakEle;
}

int main() {
    // Create a linked list with values:
    // 1 -> 2 -> 3 -> 4
    Node* head = (Node*)malloc(sizeof(Node));
    head->data = 1;
    head->next = (Node*)malloc(sizeof(Node));
    head->next->data = 2;
    head->next->next = (Node*)malloc(sizeof(Node));
    head->next->next->data = 3;
    head->next->next->next = (Node*)malloc(sizeof(Node));
    head->next->next->next->data = 4;
    head->next->next->next->next = NULL;

    // Find the peak element in the linked list
    int peakElement = findPeakElement(head);

    printf("%d\n", peakElement);

    return 0;
}

Java

// Java implementation to find the peak
// element in the Linked List
class Node {
    int data;
    Node next;
    Node(int x) {
        data = x;
        next = null;
    }
}

// Function to find a peak element in linked list
class GfG {
    static int findPeakElement(Node head) {
        // If the list is empty, return -1
        if (head == null) return -1;

        // Pointer to store the previous node
        Node prev = null;

        // Traverse the list to find the peak element
        while (head.next != null) {
            // If the current node is a peak
            // then return its value
            if (head.data >= head.next.data)
                return head.data;

            // Update the previous node pointer
            prev = head;

            // Move to the next node
            head = head.next;
        }

        // If the last node is greater than or equal
        // to the previous node, it's a peak
        int peakEle = head.data;
        if (prev != null && head.data >= prev.data)
            peakEle = head.data;

        return peakEle;
    }

    public static void main(String[] args) {
        // Create a linked list with values:
        // 1 -> 2 -> 3 -> 4
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);

        // Find the peak element in the linked list
        int peakElement = findPeakElement(head);

        System.out.println(peakElement);
    }
}

Python

# Python implementation to find the peak
# element in the Linked List
class Node:
    def __init__(self, x):
        self.data = x
        self.next = None

# Function to find a peak element in linked list
def find_peak_element(head):
    # If the list is empty, return -1
    if not head:
        return -1

    # Pointer to store the previous node
    prev = None

    # Traverse the list to find the peak element
    while head.next:
        # If the current node is a peak
        # then return its value
        if head.data >= head.next.data:
            return head.data

        # Update the previous node pointer
        prev = head

        # Move to the next node
        head = head.next

    # If the last node is greater than or equal
    # to the previous node, it's a peak
    peak_ele = head.data
    if prev and head.data >= prev.data:
        peak_ele = head.data

    return peak_ele

# Create a linked list with values:
# 1 -> 2 -> 3 -> 4
head = Node(1)
head.next = Node(2)
head.next.next = Node(3)
head.next.next.next = Node(4)

# Find the peak element in the linked list
peak_element = find_peak_element(head)

print(peak_element)

// C# implementation to find the peak
// element in the Linked List
using System;

public class Node {
    public int data;
    public Node next;
    public Node(int x) {
        data = x;
        next = null;
    }
}

// Function to find a peak element in linked list
class GfG {
    static int FindPeakElement(Node head) {
        // If the list is empty, return -1
        if (head == null) return -1;

        // Pointer to store the previous node
        Node prev = null;

        // Traverse the list to find the peak element
        while (head.next != null) {
            // If the current node is a peak
            // then return its value
            if (head.data >= head.next.data)
                return head.data;

            // Update the previous node pointer
            prev = head;

            // Move to the next node
            head = head.next;
        }

        // If the last node is greater than or equal
        // to the previous node, it's a peak
        int peakEle = head.data;
        if (prev != null && head.data >= prev.data)
            peakEle = head.data;

        return peakEle;
    }

    public static void Main() {
        // Create a linked list with values:
        // 1 -> 2 -> 3 -> 4
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        head.next.next.next = new Node(4);

        // Find the peak element in the linked list
        int peakElement = FindPeakElement(head);

        Console.WriteLine(peakElement);
    }
}

JavaScript

// JavaScript implementation to find the peak
// element in the Linked List
class Node {
    constructor(x) {
        this.data = x;
        this.next = null;
    }
}

// Function to find a peak element in linked list
function findPeakElement(head) {
    // If the list is empty, return -1
    if (!head) return -1;

    // Pointer to store the previous node
    let prev = null;

    // Traverse the list to find the peak element
    while (head.next) {
        // If the current node is a peak
        // then return its value
        if (head.data >= head.next.data)
            return head.data;

        // Update the previous node pointer
        prev = head;

        // Move to the next node
        head = head.next;
    }

    // If the last node is greater than or equal
    // to the previous node, it's a peak
    let peakEle = head.data;
    if (prev && head.data >= prev.data)
        peakEle = head.data;

    return peakEle;
}

// Create a linked list with values:
// 1 -> 2 -> 3 -> 4
let head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(3);
head.next.next.next = new Node(4);

// Find the peak element in the linked list
let peakElement = findPeakElement(head);

console.log(peakElement);

Output

Time Complexity: O(N) where N is the number of elements in the linked list.
Auxiliary Space: O(1)

Find a peak element in Linked List

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