Delete N Nodes After M Nodes of a Linked List - Problem

Linked List Easy

You are given the head of a linked list and two integers m and n.

Traverse the linked list and remove some nodes in the following way:

Start with the head as the current node.
Keep the first m nodes starting with the current node.
Remove the next n nodes
Keep repeating steps 2 and 3 until you reach the end of the list.

Return the head of the modified list after removing the mentioned nodes.

Input & Output

Example 1 — Basic Pattern

$ Input: head = [1,2,3,4,5,6,7,8,9,10,11], m = 2, n = 3

› Output: [1,2,6,7,11]

💡 Note: Keep first 2 nodes (1,2), delete next 3 (3,4,5), keep next 2 (6,7), delete next 3 (8,9,10), keep remaining (11)

Example 2 — Short List

$ Input: head = [1,2,3,4], m = 1, n = 2

› Output: [1,4]

💡 Note: Keep 1 node (1), delete next 2 (2,3), keep remaining (4)

Example 3 — Delete All After Keep

$ Input: head = [1,2,3], m = 1, n = 3

› Output: [1]

💡 Note: Keep 1 node (1), delete next 3 but only 2,3 exist, so final result is [1]

Constraints

The number of nodes in the given linked list is in the range [1, 10⁴].
1 ≤ Node.val ≤ 10⁶
1 ≤ m, n ≤ 1000

Visualization

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Asked in

M Microsoft 15 a Amazon 12 f Facebook 8 G Google 6

The key insight is to process the linked list in repeating cycles of keep-then-delete. Instead of checking each node individually, we can process m nodes at once (keeping them), then n nodes at once (deleting them). The optimal approach uses chunk processing for better efficiency. Time: O(n), Space: O(1).

Common Approaches

✓ Brute Force - Node by Node Processing

⏱️ Time: O(n) Space: O(1)

Process each node one at a time, maintaining counters to track whether we're in a 'keep' phase or 'delete' phase. This approach handles the pattern by checking each node's position in the current cycle.

Optimized - Chunk Processing

⏱️ Time: O(n) Space: O(1)

Instead of checking state for each node, we process m nodes at once (keeping them), then process n nodes at once (deleting them). This reduces the number of conditional checks and makes the algorithm more efficient.

Brute Force - Node by Node Processing — Algorithm Steps

Traverse the linked list node by node
Use counters to track current position in m-keep, n-delete cycle
Either advance or delete current node based on cycle position

Visualization

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Step-by-Step Walkthrough

Initialize

Set up counters and phase flags

Process Each Node

Check current phase and act accordingly

Update State

Switch between keep and delete phases

Code -

solution.c — C

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

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

struct ListNode* solution(struct ListNode* head, int m, int n) {
    if (!head || m == 0) return NULL;
    
    struct ListNode* current = head;
    
    while (current) {
        // Keep m nodes
        for (int i = 1; i < m && current; i++) {
            current = current->next;
        }
        
        if (!current) break;
        
        // Delete n nodes
        for (int i = 0; i < n && current->next; i++) {
            struct ListNode* nodeToDelete = current->next;
            current->next = nodeToDelete->next;
            free(nodeToDelete);
        }
        
        // Move to next segment
        current = current->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[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int arr[100], size = 0;
    int i = 0;
    
    // Skip to '['
    while (line[i] && line[i] != '[') i++;
    i++; // Skip '['
    
    // Parse numbers
    while (line[i] && line[i] != ']') {
        if (line[i] >= '0' && line[i] <= '9') {
            int num = 0;
            while (line[i] >= '0' && line[i] <= '9') {
                num = num * 10 + (line[i] - '0');
                i++;
            }
            arr[size++] = num;
        } else {
            i++;
        }
    }
    
    int m, n;
    scanf("%d", &m);
    scanf("%d", &n);
    
    struct ListNode* head = arrayToList(arr, size);
    struct ListNode* resultHead = solution(head, m, n);
    
    printf("[");
    struct ListNode* current = resultHead;
    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)

Visit each node in the linked list exactly once

✓ Linear Growth

Space Complexity

O(1)

Only use a few variables for counters and pointers

✓ Linear Space

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Delete N Nodes After M Nodes of a Linked List - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Node by Node Processing — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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