LeetCode #725 — MEDIUM

Split Linked List in Parts

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

The Problem

Problem Statement

Given the head of a singly linked list and an integer k, split the linked list into k consecutive linked list parts.

The length of each part should be as equal as possible: no two parts should have a size differing by more than one. This may lead to some parts being null.

The parts should be in the order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal to parts occurring later.

Return an array of the k parts.

Example 1:

Input: head = [1,2,3], k = 5
Output: [[1],[2],[3],[],[]]
Explanation:
The first element output[0] has output[0].val = 1, output[0].next = null.
The last element output[4] is null, but its string representation as a ListNode is [].

Example 2:

Input: head = [1,2,3,4,5,6,7,8,9,10], k = 3
Output: [[1,2,3,4],[5,6,7],[8,9,10]]
Explanation:
The input has been split into consecutive parts with size difference at most 1, and earlier parts are a larger size than the later parts.

Constraints:

The number of nodes in the list is in the range [0, 1000].
0 <= Node.val <= 1000
1 <= k <= 50

Step 01

Brute Force Baseline

Problem summary: Given the head of a singly linked list and an integer k, split the linked list into k consecutive linked list parts. The length of each part should be as equal as possible: no two parts should have a size differing by more than one. This may lead to some parts being null. The parts should be in the order of occurrence in the input list, and parts occurring earlier should always have a size greater than or equal to parts occurring later. Return an array of the k parts.

Baseline thinking

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

Pattern signal: Linked List

Example 1

[1,2,3]
5

Example 2

[1,2,3,4,5,6,7,8,9,10]
3

Core Insight

What unlocks the optimal approach

If there are N nodes in the list, and k parts, then every part has N/k elements, except the first N%k parts have an extra one.

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 #725: Split Linked List in Parts
/**
 * 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[] splitListToParts(ListNode head, int k) {
        int n = 0;
        for (ListNode cur = head; cur != null; cur = cur.next) {
            ++n;
        }
        int cnt = n / k, mod = n % k;
        ListNode[] ans = new ListNode[k];
        ListNode cur = head;
        for (int i = 0; i < k && cur != null; ++i) {
            ans[i] = cur;
            int m = cnt + (i < mod ? 1 : 0);
            for (int j = 1; j < m; ++j) {
                cur = cur.next;
            }
            ListNode nxt = cur.next;
            cur.next = null;
            cur = nxt;
        }
        return ans;
    }
}

// Accepted solution for LeetCode #725: Split Linked List in Parts
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func splitListToParts(head *ListNode, k int) []*ListNode {
	n := 0
	for cur := head; cur != nil; cur = cur.Next {
		n++
	}

	cnt := n / k
	mod := n % k
	ans := make([]*ListNode, k)
	cur := head

	for i := 0; i < k && cur != nil; i++ {
		ans[i] = cur
		m := cnt
		if i < mod {
			m++
		}
		for j := 1; j < m; j++ {
			cur = cur.Next
		}
		next := cur.Next
		cur.Next = nil
		cur = next
	}

	return ans
}

# Accepted solution for LeetCode #725: Split Linked List in Parts
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def splitListToParts(
        self, head: Optional[ListNode], k: int
    ) -> List[Optional[ListNode]]:
        n = 0
        cur = head
        while cur:
            n += 1
            cur = cur.next
        cnt, mod = divmod(n, k)
        ans = [None] * k
        cur = head
        for i in range(k):
            if cur is None:
                break
            ans[i] = cur
            m = cnt + int(i < mod)
            for _ in range(1, m):
                cur = cur.next
            nxt = cur.next
            cur.next = None
            cur = nxt
        return ans

// Accepted solution for LeetCode #725: Split Linked List in Parts
struct Solution;
use rustgym_util::*;

impl Solution {
    fn split_list_to_parts(head: ListLink, k: i32) -> Vec<ListLink> {
        let mut nodes = vec![];
        let mut p = head;
        while let Some(mut node) = p {
            p = node.next.take();
            nodes.push(node);
        }
        let n = nodes.len();
        let k = k as usize;
        let left = n % k;
        let right = k - left;
        let m = n / k;
        let mut res = vec![None; k];
        for i in 0..right {
            let mut prev = None;
            for _ in 0..m {
                let mut node = nodes.pop().unwrap();
                node.next = prev;
                prev = Some(node);
            }
            res[k - 1 - i] = prev;
        }
        for i in 0..left {
            let mut prev = None;
            for _ in 0..=m {
                let mut node = nodes.pop().unwrap();
                node.next = prev;
                prev = Some(node);
            }
            res[k - right - 1 - i] = prev;
        }
        res
    }
}

#[test]
fn test() {
    let head = list!(1, 2, 3);
    let k = 5;
    let res = vec![list!(1), list!(2), list!(3), None, None];
    assert_eq!(Solution::split_list_to_parts(head, k), res);
}

// Accepted solution for LeetCode #725: Split Linked List in Parts
/**
 * 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 splitListToParts(head: ListNode | null, k: number): Array<ListNode | null> {
    let n = 0;
    for (let cur = head; cur !== null; cur = cur.next) {
        n++;
    }
    const cnt = (n / k) | 0;
    const mod = n % k;
    const ans: Array<ListNode | null> = Array(k).fill(null);
    let cur = head;
    for (let i = 0; i < k && cur !== null; i++) {
        ans[i] = cur;
        let m = cnt + (i < mod ? 1 : 0);
        for (let j = 1; j < m; j++) {
            cur = cur.next!;
        }
        let next = cur.next;
        cur.next = null;
        cur = next;
    }
    return ans;
}

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(k)

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.