LeetCode #2807 — MEDIUM

Insert Greatest Common Divisors in Linked List

Move from brute-force thinking to an efficient approach using linked list strategy.

Solve on LeetCode

The Problem

Problem Statement

Given the head of a linked list head, in which each node contains an integer value.

Between every pair of adjacent nodes, insert a new node with a value equal to the greatest common divisor of them.

Return the linked list after insertion.

The greatest common divisor of two numbers is the largest positive integer that evenly divides both numbers.

Example 1:

Input: head = [18,6,10,3]
Output: [18,6,6,2,10,1,3]
Explanation: The 1^st diagram denotes the initial linked list and the 2^nd diagram denotes the linked list after inserting the new nodes (nodes in blue are the inserted nodes).
- We insert the greatest common divisor of 18 and 6 = 6 between the 1^st and the 2^nd nodes.
- We insert the greatest common divisor of 6 and 10 = 2 between the 2^nd and the 3^rd nodes.
- We insert the greatest common divisor of 10 and 3 = 1 between the 3^rd and the 4^th nodes.
There are no more adjacent nodes, so we return the linked list.

Example 2:

Input: head = [7]
Output: [7]
Explanation: The 1^st diagram denotes the initial linked list and the 2^nd diagram denotes the linked list after inserting the new nodes.
There are no pairs of adjacent nodes, so we return the initial linked list.

Constraints:

The number of nodes in the list is in the range [1, 5000].
1 <= Node.val <= 1000

Step 01

Brute Force Baseline

Problem summary: Given the head of a linked list head, in which each node contains an integer value. Between every pair of adjacent nodes, insert a new node with a value equal to the greatest common divisor of them. Return the linked list after insertion. The greatest common divisor of two numbers is the largest positive integer that evenly divides both numbers.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Linked List · Math

Example 1

[18,6,10,3]

Example 2

[7]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2807: Insert Greatest Common Divisors in Linked List
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode insertGreatestCommonDivisors(ListNode head) {
        for (ListNode pre = head, cur = head.next; cur != null; cur = cur.next) {
            int x = gcd(pre.val, cur.val);
            pre.next = new ListNode(x, cur);
            pre = cur;
        }
        return head;
    }

    private int gcd(int a, int b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }
}

// Accepted solution for LeetCode #2807: Insert Greatest Common Divisors in Linked List
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func insertGreatestCommonDivisors(head *ListNode) *ListNode {
	for pre, cur := head, head.Next; cur != nil; cur = cur.Next {
		x := gcd(pre.Val, cur.Val)
		pre.Next = &ListNode{x, cur}
		pre = cur
	}
	return head
}

func gcd(a, b int) int {
	if b == 0 {
		return a
	}
	return gcd(b, a%b)
}

# Accepted solution for LeetCode #2807: Insert Greatest Common Divisors in Linked List
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def insertGreatestCommonDivisors(
        self, head: Optional[ListNode]
    ) -> Optional[ListNode]:
        pre, cur = head, head.next
        while cur:
            x = gcd(pre.val, cur.val)
            pre.next = ListNode(x, cur)
            pre, cur = cur, cur.next
        return head

// Accepted solution for LeetCode #2807: Insert Greatest Common Divisors in Linked List
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2807: Insert Greatest Common Divisors in Linked List
// /**
//  * Definition for singly-linked list.
//  * public class ListNode {
//  *     int val;
//  *     ListNode next;
//  *     ListNode() {}
//  *     ListNode(int val) { this.val = val; }
//  *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
//  * }
//  */
// class Solution {
//     public ListNode insertGreatestCommonDivisors(ListNode head) {
//         for (ListNode pre = head, cur = head.next; cur != null; cur = cur.next) {
//             int x = gcd(pre.val, cur.val);
//             pre.next = new ListNode(x, cur);
//             pre = cur;
//         }
//         return head;
//     }
// 
//     private int gcd(int a, int b) {
//         if (b == 0) {
//             return a;
//         }
//         return gcd(b, a % b);
//     }
// }

// Accepted solution for LeetCode #2807: Insert Greatest Common Divisors in Linked List
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function insertGreatestCommonDivisors(head: ListNode | null): ListNode | null {
    for (let pre = head, cur = head.next; cur; cur = cur.next) {
        const x = gcd(pre.val, cur.val);
        pre.next = new ListNode(x, cur);
        pre = cur;
    }
    return head;
}

function gcd(a: number, b: number): number {
    if (b === 0) {
        return a;
    }
    return gcd(b, a % b);
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

COPY TO ARRAY

O(n) time

O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS

O(n) time

O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.