LeetCode #1669 — MEDIUM

Merge In Between Linked Lists

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

Solve on LeetCode

The Problem

Problem Statement

You are given two linked lists: list1 and list2 of sizes n and m respectively.

Remove list1's nodes from the a^th node to the b^th node, and put list2 in their place.

The blue edges and nodes in the following figure indicate the result:

Build the result list and return its head.

Example 1:

Input: list1 = [10,1,13,6,9,5], a = 3, b = 4, list2 = [1000000,1000001,1000002]
Output: [10,1,13,1000000,1000001,1000002,5]
Explanation: We remove the nodes 3 and 4 and put the entire list2 in their place. The blue edges and nodes in the above figure indicate the result.

Example 2:

Input: list1 = [0,1,2,3,4,5,6], a = 2, b = 5, list2 = [1000000,1000001,1000002,1000003,1000004]
Output: [0,1,1000000,1000001,1000002,1000003,1000004,6]
Explanation: The blue edges and nodes in the above figure indicate the result.

Constraints:

3 <= list1.length <= 10⁴
1 <= a <= b < list1.length - 1
1 <= list2.length <= 10⁴

Step 01

Brute Force Baseline

Problem summary: You are given two linked lists: list1 and list2 of sizes n and m respectively. Remove list1's nodes from the ath node to the bth node, and put list2 in their place. The blue edges and nodes in the following figure indicate the result: Build the result list and return its head.

Baseline thinking

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

Pattern signal: Linked List

Example 1

[10,1,13,6,9,5]
3
4
[1000000,1000001,1000002]

Example 2

[0,1,2,3,4,5,6]
2
5
[1000000,1000001,1000002,1000003,1000004]

Step 02

Core Insight

What unlocks the optimal approach

Check which edges need to be changed.
Let the next node of the (a-1)th node of list1 be the 0-th node in list 2.
Let the next node of the last node of list2 be the (b+1)-th node in list 1.

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 #1669: Merge In Between Linked Lists
/**
 * 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 mergeInBetween(ListNode list1, int a, int b, ListNode list2) {
        ListNode p = list1, q = list1;
        while (--a > 0) {
            p = p.next;
        }
        while (b-- > 0) {
            q = q.next;
        }
        p.next = list2;
        while (p.next != null) {
            p = p.next;
        }
        p.next = q.next;
        q.next = null;
        return list1;
    }
}

// Accepted solution for LeetCode #1669: Merge In Between Linked Lists
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func mergeInBetween(list1 *ListNode, a int, b int, list2 *ListNode) *ListNode {
	p, q := list1, list1
	for ; a > 1; a-- {
		p = p.Next
	}
	for ; b > 0; b-- {
		q = q.Next
	}
	p.Next = list2
	for p.Next != nil {
		p = p.Next
	}
	p.Next = q.Next
	q.Next = nil
	return list1
}

# Accepted solution for LeetCode #1669: Merge In Between Linked Lists
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def mergeInBetween(
        self, list1: ListNode, a: int, b: int, list2: ListNode
    ) -> ListNode:
        p = q = list1
        for _ in range(a - 1):
            p = p.next
        for _ in range(b):
            q = q.next
        p.next = list2
        while p.next:
            p = p.next
        p.next = q.next
        q.next = None
        return list1

// Accepted solution for LeetCode #1669: Merge In Between Linked Lists
struct Solution;

use rustgym_util::*;

impl Solution {
    fn merge_in_between(list1: ListLink, a: i32, b: i32, list2: ListLink) -> ListLink {
        let mut l1 = vec![];
        let mut l2 = vec![];
        let mut p1 = list1;
        let start = a as usize;
        let end = (b + 1) as usize;
        while let Some(mut node) = p1 {
            p1 = node.next.take();
            l1.push(node);
        }
        let mut p2 = list2;
        while let Some(mut node) = p2 {
            p2 = node.next.take();
            l2.push(node);
        }
        let mut prev = None;
        for _ in (end..l1.len()).rev() {
            let mut node = l1.pop().unwrap();
            node.next = prev;
            prev = Some(node);
        }
        for _ in (start..end).rev() {
            l1.pop();
        }
        for _ in (0..l2.len()).rev() {
            let mut node = l2.pop().unwrap();
            node.next = prev;
            prev = Some(node);
        }
        for _ in (0..start).rev() {
            let mut node = l1.pop().unwrap();
            node.next = prev;
            prev = Some(node);
        }
        prev
    }
}

#[test]
fn test() {
    let list1 = list![0, 1, 2, 3, 4, 5];
    let a = 3;
    let b = 4;
    let list2 = list![1000000, 1000001, 1000002];
    let res = list![0, 1, 2, 1000000, 1000001, 1000002, 5];
    assert_eq!(Solution::merge_in_between(list1, a, b, list2), res);
    let list1 = list![0, 1, 2, 3, 4, 5, 6];
    let a = 2;
    let b = 5;
    let list2 = list![1000000, 1000001, 1000002, 1000003, 1000004];
    let res = list![0, 1, 1000000, 1000001, 1000002, 1000003, 1000004, 6];
    assert_eq!(Solution::merge_in_between(list1, a, b, list2), res);
}

// Accepted solution for LeetCode #1669: Merge In Between Linked Lists
/**
 * 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 mergeInBetween(
    list1: ListNode | null,
    a: number,
    b: number,
    list2: ListNode | null,
): ListNode | null {
    let p = list1;
    let q = list1;
    while (--a > 0) {
        p = p.next;
    }
    while (b-- > 0) {
        q = q.next;
    }
    p.next = list2;
    while (p.next) {
        p = p.next;
    }
    p.next = q.next;
    q.next = null;
    return list1;
}

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(m + 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.