LeetCode #1721 — MEDIUM

Swapping Nodes in a Linked List

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

The Problem

Problem Statement

You are given the head of a linked list, and an integer k.

Return the head of the linked list after swapping the values of the k^th node from the beginning and the k^th node from the end (the list is 1-indexed).

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [1,4,3,2,5]

Example 2:

Input: head = [7,9,6,6,7,8,3,0,9,5], k = 5
Output: [7,9,6,6,8,7,3,0,9,5]

Constraints:

The number of nodes in the list is n.
1 <= k <= n <= 10⁵
0 <= Node.val <= 100

Step 01

Brute Force Baseline

Problem summary: You are given the head of a linked list, and an integer k. Return the head of the linked list after swapping the values of the kth node from the beginning and the kth node from the end (the list is 1-indexed).

Baseline thinking

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

Pattern signal: Linked List · Two Pointers

Example 1

[1,2,3,4,5]
2

Example 2

[7,9,6,6,7,8,3,0,9,5]
5

Core Insight

What unlocks the optimal approach

We can traverse the linked list and store the elements in an array.
Upon conversion to an array, we can swap the required elements by indexing the array.
We can rebuild the linked list using the order of the elements in the array.

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 #1721: Swapping Nodes in a 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 swapNodes(ListNode head, int k) {
        ListNode fast = head;
        while (--k > 0) {
            fast = fast.next;
        }
        ListNode p = fast;
        ListNode slow = head;
        while (fast.next != null) {
            fast = fast.next;
            slow = slow.next;
        }
        ListNode q = slow;
        int t = p.val;
        p.val = q.val;
        q.val = t;
        return head;
    }
}

// Accepted solution for LeetCode #1721: Swapping Nodes in a Linked List
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func swapNodes(head *ListNode, k int) *ListNode {
	fast := head
	for ; k > 1; k-- {
		fast = fast.Next
	}
	p := fast
	slow := head
	for fast.Next != nil {
		fast, slow = fast.Next, slow.Next
	}
	q := slow
	p.Val, q.Val = q.Val, p.Val
	return head
}

# Accepted solution for LeetCode #1721: Swapping Nodes in a 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 swapNodes(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:
        fast = slow = head
        for _ in range(k - 1):
            fast = fast.next
        p = fast
        while fast.next:
            fast, slow = fast.next, slow.next
        q = slow
        p.val, q.val = q.val, p.val
        return head

// Accepted solution for LeetCode #1721: Swapping Nodes in a Linked List
struct Solution;

use rustgym_util::*;

impl Solution {
    fn swap_nodes(head: ListLink, k: i32) -> ListLink {
        let mut p = head;
        let mut nodes: Vec<Box<ListNode>> = vec![];
        let k = k as usize;
        while let Some(mut node) = p {
            p = node.next.take();
            nodes.push(node);
        }
        let n = nodes.len();
        nodes.swap(k - 1, n - k);
        let mut prev = None;
        while let Some(mut node) = nodes.pop() {
            node.next = prev;
            prev = Some(node);
        }
        prev
    }
}

#[test]
fn test() {
    let head = list!(1, 2, 3, 4, 5);
    let k = 2;
    let res = list!(1, 4, 3, 2, 5);
    assert_eq!(Solution::swap_nodes(head, k), res);
    let head = list!(7, 9, 6, 6, 7, 8, 3, 0, 9, 5);
    let k = 5;
    let res = list!(7, 9, 6, 6, 8, 7, 3, 0, 9, 5);
    assert_eq!(Solution::swap_nodes(head, k), res);
    let head = list!(1);
    let k = 1;
    let res = list!(1);
    assert_eq!(Solution::swap_nodes(head, k), res);
    let head = list!(1, 2);
    let k = 1;
    let res = list!(2, 1);
    assert_eq!(Solution::swap_nodes(head, k), res);
    let head = list!(1, 2, 3);
    let k = 2;
    let res = list!(1, 2, 3);
    assert_eq!(Solution::swap_nodes(head, k), res);
}

// Accepted solution for LeetCode #1721: Swapping Nodes in a 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 swapNodes(head: ListNode | null, k: number): ListNode | null {
    let [fast, slow] = [head, head];
    while (--k) {
        fast = fast.next;
    }
    const p = fast;
    while (fast.next) {
        fast = fast.next;
        slow = slow.next;
    }
    const q = slow;
    [p.val, q.val] = [q.val, p.val];
    return head;
}

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.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.