Palindrome Linked List - Practice Coding Problems

Palindrome Linked List - Problem

Given the head of a singly linked list, return true if it is a palindrome or false otherwise.

A palindrome is a sequence that reads the same forward and backward.

Example: The linked list 1 → 2 → 2 → 1 is a palindrome because it reads the same forwards and backwards.

Input & Output

Example 1 — Basic Palindrome

$ Input: head = [1,2,2,1]

› Output: true

💡 Note: The linked list reads the same forwards (1→2→2→1) and backwards (1→2→2→1), so it's a palindrome.

Example 2 — Not a Palindrome

$ Input: head = [1,2]

› Output: false

💡 Note: The linked list reads 1→2 forwards but 2→1 backwards, which are different, so it's not a palindrome.

Example 3 — Single Node

$ Input: head = [1]

› Output: true

💡 Note: A single node is always a palindrome since it reads the same forwards and backwards.

Constraints

The number of nodes in the list is in the range [1, 10⁵]
0 ≤ Node.val ≤ 9

Visualization

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Understanding the Visualization

Input

Linked list: 1→2→2→1

Check

Compare forwards vs backwards reading

Output

Return true if palindrome

Key Takeaway

🎯 Key Insight: Find the middle, reverse second half, then compare both halves for optimal O(1) space solution

Asked in

F Facebook 45 A Amazon 38 M Microsoft 32 G Google 28

The key insight is to compare the first and second half of the linked list. The optimal reverse second half approach finds the middle using slow/fast pointers, reverses the second half in-place, then compares both halves. Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Array Conversion	O(n)	O(n)	Convert linked list to array, then check palindrome with two pointers
Stack-based Approach	O(n)	O(n/2)	Use stack to store first half, then compare with second half
Reverse Second Half	O(n)	O(1)	Find middle, reverse second half, compare both halves

Array Conversion — Algorithm Steps

Step 1: Traverse linked list and store all values in array
Step 2: Use two pointers to compare values from both ends of array

Visualization

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Step-by-Step Walkthrough

Convert to Array

Traverse linked list and store values

Two Pointers

Compare from both ends

Result

Return true if all pairs match

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

struct ListNode {
    int val;
    struct ListNode* next;
};

bool solution(struct ListNode* head) {
    if (!head) return true;
    
    // Convert to array
    int values[100000];
    int count = 0;
    struct ListNode* current = head;
    while (current) {
        values[count++] = current->val;
        current = current->next;
    }
    
    // Check palindrome using two pointers
    int left = 0, right = count - 1;
    while (left < right) {
        if (values[left] != values[right]) {
            return false;
        }
        left++;
        right--;
    }
    
    return true;
}

struct ListNode* arrayToLinkedList(int* arr, int size) {
    if (size == 0) return NULL;
    struct ListNode* head = malloc(sizeof(struct ListNode));
    head->val = arr[0];
    head->next = NULL;
    
    struct ListNode* current = head;
    for (int i = 1; i < size; i++) {
        current->next = malloc(sizeof(struct ListNode));
        current = current->next;
        current->val = arr[i];
        current->next = NULL;
    }
    return head;
}

int main() {
    char line[1000000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int arr[100000];
    int count = 0;
    char* token = strtok(line, "[],\n ");
    while (token) {
        arr[count++] = atoi(token);
        token = strtok(NULL, "[],\n ");
    }
    
    struct ListNode* head = arrayToLinkedList(arr, count);
    bool result = solution(head);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

One pass to build array + one pass to check palindrome

✓ Linear Growth

Space Complexity

O(n)

Array stores all n node values

⚡ Linearithmic Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Array Conversion — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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