Insert Greatest Common Divisors in Linked List

Insert Greatest Common Divisors in Linked List - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: head = [18,6,10,3]

› Output: [18,6,6,2,10,1,3]

💡 Note: GCD(18,6)=6, insert 6. GCD(6,10)=2, insert 2. GCD(10,3)=1, insert 1. Result: [18,6,6,2,10,1,3]

Example 2 — Two Nodes

$ Input: head = [7]

› Output: [7]

💡 Note: Single node, no adjacent pairs exist, so no insertions needed

Example 3 — Coprime Numbers

$ Input: head = [7,11,13]

› Output: [7,1,11,1,13]

💡 Note: GCD(7,11)=1, GCD(11,13)=1. Prime numbers are coprime, so GCD is always 1

Constraints

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

Visualization

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Asked in

G Google 25 M Microsoft 18 a Amazon 15

The key insight is to traverse the linked list once and insert GCD nodes between adjacent pairs. Best approach uses the Euclidean algorithm for efficient GCD calculation. Time: O(n × log(min(a,b))), Space: O(1)

Common Approaches

✓ Brute Force GCD

⏱️ Time: O(n × min(a,b)) Space: O(1)

Traverse the linked list and for each pair of adjacent nodes, calculate their GCD by testing all possible divisors from 1 to the minimum of the two values. Insert a new node with the GCD value between them.

Euclidean Algorithm

⏱️ Time: O(n × log(min(a,b))) Space: O(1)

Traverse the linked list once and for each pair of adjacent nodes, calculate their GCD using the Euclidean algorithm. This algorithm repeatedly applies the formula GCD(a,b) = GCD(b, a mod b) until one number becomes 0.

Brute Force GCD — Algorithm Steps

Traverse linked list node by node
For each adjacent pair, find GCD by testing all divisors
Create new node with GCD value and insert between them

Visualization

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Step-by-Step Walkthrough

Find Adjacent Pair

Locate two adjacent nodes (18, 6)

Test All Divisors

Check 1,2,3,4,5,6 - find GCD = 6

Insert GCD Node

Create new node with value 6 between them

Code -

solution.c — C

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

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

int gcdBrute(int a, int b) {
    int result = 1;
    int min = (a < b) ? a : b;
    for (int i = 1; i <= min; i++) {
        if (a % i == 0 && b % i == 0) {
            result = i;
        }
    }
    return result;
}

struct ListNode* solution(struct ListNode* head) {
    if (!head || !head->next) {
        return head;
    }
    
    struct ListNode* current = head;
    while (current && current->next) {
        int gcdVal = gcdBrute(current->val, current->next->val);
        struct ListNode* newNode = malloc(sizeof(struct ListNode));
        newNode->val = gcdVal;
        newNode->next = current->next;
        current->next = newNode;
        current = newNode->next;
    }
    
    return head;
}

struct ListNode* arrayToList(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->next->val = arr[i];
        current->next->next = NULL;
        current = current->next;
    }
    return head;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    int arr[1000], size = 0;
    char* ptr = strchr(line, '[');
    if (ptr) {
        ptr++;
        char* token = strtok(ptr, ",]");
        while (token) {
            arr[size++] = atoi(token);
            token = strtok(NULL, ",]");
        }
    }
    
    struct ListNode* head = arrayToList(arr, size);
    struct ListNode* result = solution(head);
    
    printf("[");
    struct ListNode* current = result;
    int first = 1;
    while (current) {
        if (!first) printf(",");
        printf("%d", current->val);
        first = 0;
        current = current->next;
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × min(a,b))

Visit n nodes, for each pair calculate GCD by testing up to min(a,b) divisors

✓ Linear Growth

Space Complexity

O(1)

Only use a few variables, modify list in place

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force GCD — Algorithm Steps

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Code -

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