2807. Insert Greatest Common Divisors in Linked List - Explanation

Description

You are 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 head of 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 = [12,3,4,6]

Output: [12,3,3,1,4,2,6]

Example 2:

Input: head = [2,1]

Output: [2,1,1]

Constraints:

1 <= The length of the list <= 5000.
1 <= Node.val <= 1000

Topics

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Prerequisites

Before attempting this problem, you should be comfortable with:

Linked Lists - Understanding node structure, traversal, and inserting nodes between existing nodes
GCD (Greatest Common Divisor) - Knowing the Euclidean algorithm for efficiently computing the GCD of two numbers
Pointer Manipulation - Safely adjusting next pointers without losing references or creating infinite loops

1. Simulation

Intuition

The problem asks us to insert a new node between every pair of adjacent nodes, where the new node's value is the GCD of its neighbors. We traverse the list and for each pair of consecutive nodes, compute their gcd and create a new node with that value.
The Euclidean algorithm efficiently computes the gcd: repeatedly replace the larger number with the remainder of dividing the two numbers until one becomes zero. The other number is the gcd.
Since we're inserting nodes as we traverse, we need to be careful to advance the pointer past the newly inserted node to avoid processing it again.

Algorithm

Start with cur pointing to the head of the list.
While cur.next exists:
- Get the values of cur and cur.next as n1 and n2.
- Compute their gcd using the Euclidean algorithm.
- Create a new node with the gcd value.
- Insert it between cur and cur.next by adjusting pointers.
- Move cur to cur.next.next to skip over the newly inserted node.
Return the head of the modified 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]:
        def gcd(a, b):
            while b > 0:
                a, b = b, a % b
            return a

        cur = head
        while cur.next:
            n1, n2 = cur.val, cur.next.val
            cur.next = ListNode(gcd(n1, n2), cur.next)
            cur = cur.next.next

        return head

/**
 * 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; }
 * }
 */
public class Solution {
    public ListNode insertGreatestCommonDivisors(ListNode head) {
        if (head == null) return null;

        ListNode cur = head;

        while (cur.next != null) {
            int n1 = cur.val, n2 = cur.next.val;
            int gcdValue = gcd(n1, n2);
            ListNode newNode = new ListNode(gcdValue, cur.next);
            cur.next = newNode;
            cur = newNode.next;
        }

        return head;
    }

    private int gcd(int a, int b) {
        while (b > 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* insertGreatestCommonDivisors(ListNode* head) {
        if (!head) return nullptr;

        ListNode* cur = head;

        while (cur->next) {
            int n1 = cur->val, n2 = cur->next->val;
            int gcdValue = gcd(n1, n2);
            ListNode* newNode = new ListNode(gcdValue, cur->next);
            cur->next = newNode;
            cur = newNode->next;
        }

        return head;
    }

private:
    int gcd(int a, int b) {
        while (b > 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
};

/**
 * Definition for singly-linked list.
 * class ListNode {
 *     constructor(val = 0, next = null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
class Solution {
    /**
     * @param {ListNode} head
     * @return {ListNode}
     */
    insertGreatestCommonDivisors(head) {
        if (!head) return null;
        const gcd = (a, b) => {
            while (b > 0) {
                [a, b] = [b, a % b];
            }
            return a;
        };

        let cur = head;

        while (cur.next) {
            const n1 = cur.val,
                n2 = cur.next.val;
            const gcdValue = gcd(n1, n2);
            const newNode = new ListNode(gcdValue, cur.next);
            cur.next = newNode;
            cur = newNode.next;
        }

        return head;
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int val=0, ListNode next=null) {
 *         this.val = val;
 *         this.next = next;
 *     }
 * }
 */
public class Solution {
    public ListNode InsertGreatestCommonDivisors(ListNode head) {
        ListNode cur = head;

        while (cur.next != null) {
            int gcdVal = GCD(cur.val, cur.next.val);
            ListNode newNode = new ListNode(gcdVal, cur.next);
            cur.next = newNode;
            cur = newNode.next;
        }

        return head;
    }

    private int GCD(int a, int b) {
        while (b > 0) {
            int temp = b;
            b = a % b;
            a = temp;
        }
        return a;
    }
}

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func insertGreatestCommonDivisors(head *ListNode) *ListNode {
    gcd := func(a, b int) int {
        for b > 0 {
            a, b = b, a%b
        }
        return a
    }

    cur := head
    for cur.Next != nil {
        n1, n2 := cur.Val, cur.Next.Val
        newNode := &ListNode{Val: gcd(n1, n2), Next: cur.Next}
        cur.Next = newNode
        cur = newNode.Next
    }

    return head
}

/**
 * Example:
 * var li = ListNode(5)
 * var v = li.`val`
 * Definition for singly-linked list.
 * class ListNode(var `val`: Int) {
 *     var next: ListNode? = null
 * }
 */
class Solution {
    fun insertGreatestCommonDivisors(head: ListNode?): ListNode? {
        fun gcd(a: Int, b: Int): Int {
            var x = a
            var y = b
            while (y > 0) {
                val temp = y
                y = x % y
                x = temp
            }
            return x
        }

        var cur = head
        while (cur?.next != null) {
            val n1 = cur.`val`
            val n2 = cur.next!!.`val`
            val newNode = ListNode(gcd(n1, n2))
            newNode.next = cur.next
            cur.next = newNode
            cur = newNode.next
        }

        return head
    }
}

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public var val: Int
 *     public var next: ListNode?
 *     public init() { self.val = 0; self.next = nil; }
 *     public init(_ val: Int) { self.val = val; self.next = nil; }
 *     public init(_ val: Int, _ next: ListNode?) { self.val = val; self.next = next; }
 * }
 */
class Solution {
    func insertGreatestCommonDivisors(_ head: ListNode?) -> ListNode? {
        func gcd(_ a: Int, _ b: Int) -> Int {
            var a = a, b = b
            while b > 0 {
                let temp = b
                b = a % b
                a = temp
            }
            return a
        }

        var cur = head
        while cur?.next != nil {
            let n1 = cur!.val
            let n2 = cur!.next!.val
            let newNode = ListNode(gcd(n1, n2), cur!.next)
            cur!.next = newNode
            cur = newNode.next
        }

        return head
    }
}

Time & Space Complexity

Time complexity: $O(n * \log (min(a, b)))$
Space complexity:
- $O(n)$ space for the gcd ListNodes.
- $O(1)$ extra space.

Where $n$ is the length of the given list, and $a$ and $b$ are two numbers passed to the $gcd()$ function.

Common Pitfalls

Processing Newly Inserted Nodes

After inserting a GCD node between cur and cur.next, forgetting to skip over the newly inserted node causes an infinite loop. The pointer must advance by two positions (cur = cur.next.next) to move past both the inserted node and reach the next original node that needs processing.

Incorrect GCD Implementation

Implementing the Euclidean algorithm incorrectly by not handling the base case properly or swapping values in the wrong order. The algorithm should continue until one value becomes zero, and the order of operands in the modulo operation matters. Using a % b when a < b returns a, which is correct, but some implementations incorrectly assume a > b.

Off-by-One at List End

Attempting to access cur.next.val when cur.next is null causes a null pointer exception. The loop condition must be while cur.next (not while cur) to ensure there is always a valid pair of adjacent nodes to compute the GCD between.