Limited Time Offer:30% off Hello Interview Premium 🎉

Back to Main

Learn DSA

Introduction

Two Pointers

Overview

Container With Most Water

3-Sum

Triangle Numbers

Move Zeroes

Sort Colors

Trapping Rain Water

Sliding Window

Overview

Longest Substring Without Repeating Characters

Longest Repeating Character Replacement

Overview

Maximum Sum of Subarrays of Size K

Max Points You Can Obtain From Cards

Max Sum of Distinct Subarrays Length k

Intervals

Overview

Can Attend Meetings

Insert Interval

Non-Overlapping Intervals

Merge Intervals

Employee Free Time

Stack

Overview

Valid Parentheses

Decode String

Longest Valid Parentheses

Overview

Daily Temperatures

Largest Rectangle in Histogram

Linked List

Overview

Linked List Cycle

Palindrome Linked List

Remove Nth Node From End of List

Reorder List

Swap Nodes in Pairs

Binary Search

Overview

Apple Harvest (Koko Eating Bananas)

Search in Rotated Sorted Array

Heap

Overview

Kth Largest Element in an Array

K Closest Points to Origin

Find K Closest Elements

Merge K Sorted Lists

Breadth-First Search

Introduction

Overview

Level Order Sum

Rightmost Node

Zigzag Level Order

Maximum Width of Binary Tree

Graphs Overview

Minimum Knight Moves

Rotting Oranges

01-Matrix

Bus Routes

Backtracking

Overview

Word Search

Solution Space Trees

Subsets

Generate Parentheses

Combination Sum

Graphs

Overview

Course Schedule

Course Schedule II

Dynamic Programming

Fundamentals

Solving a Question with Dynamic Programming

Counting Bits

Decode Ways

Maximal Square

Unique Paths

Longest Increasing Subsequence

Word Break

Maximum Profit in Job Scheduling

Greedy Algorithms

Overview

Best Time to Buy and Sell Stock

Gas Station

Jump Game

Trie

Overview

Implement Trie Methods

Prefix Matching

Prefix Sum

Overview

Count Vowels in Substrings

Subarray Sum Equals K

Matrices

Spiral Matrix

Rotate Image

Set Matrix Zeroes

Pricing

Become a Coach

Pricing Become a Coach

Get Premium

Heap

Merge K Sorted Lists

hard

Watch Video Walkthrough

Watch a video walking through the coding problem step-by-step

Watch Video Walkthrough

Watch a video walking through the coding problem step-by-step

DESCRIPTION (credit Leetcode.com)

Given k linked lists, each sorted in ascending order, in a list lists, write a function to merge the input lists into one sorted linked list.

Example 1:

Inputs:


lists = [[3,4,6],[2,3,5],[-1,6]]
3 -> 4 -> 6
2 -> 3 -> 5
-1 -> 6

Output:


[-1,2,3,3,4,5,6,6]

-1 -> 2 -> 3 -> 3 -> 4 -> 5 -> 6 -> 6

Explanation

This problem can be solved using a min-heap of size k.

The min-heap will always store the first unmerged node from each of the k sorted linked lists, and the node with the smallest value sits at the root of the heap. We build the merged list we will return at the end by repeatedly popping the root from the heap. Each time we pop from the heap, we will also push the next element from the linked list where the popped element came from onto the heap.

When the heap is empty, all the elements from the linked lists have been merged, and we can return the merged list.

Step 1: Initialize the heap

The first step in the algorithm is to initialize the heap with the first element from each of the k linked lists. We iterate over each of the k linked lists, and for each linked list, we push a tuple containing the value of its head node, its index (in the input list), and the node itself onto the heap.


from heapq import heappush, heappop

def mergeKLists(lists):
  if not lists:
    return None

  heap = []
  for i, node in enumerate(lists):
    if node:
      heappush(heap, (node.val, i, node))

  dummy = ListNode(0)
  curr = dummy

  while heap:
    val, idx, node = heappop(heap)
    curr.next = node
    curr = curr.next

    if node.next:
      heappush(heap, (node.next.val, idx, node.next))

  return dummy.next

initialize heap

0 / 3

Step 2: Build the merged list

With the heap initialized, we can now build the merged linked list. We start by initializing a dummy node that acts as the head of the merged list, and a pointer curr that points to the last node in the merged list (initially the dummy node).


from heapq import heappush, heappop

def mergeKLists(lists):
  if not lists:
    return None

  heap = []
  for i, node in enumerate(lists):
    if node:
      heappush(heap, (node.val, i, node))

  dummy = ListNode(0)
  curr = dummy

  while heap:
    val, idx, node = heappop(heap)
    curr.next = node
    curr = curr.next

    if node.next:
      heappush(heap, (node.next.val, idx, node.next))

  return dummy.next

add -1 to heap

0 / 1

Now, we can repeatedly pop the root of the heap, which gives us a reference to the node with the smallest value among the first unmerged elements from the k linked lists. We append this node to the merged list and move the pointer curr to the node we just appended.


from heapq import heappush, heappop

def mergeKLists(lists):
  if not lists:
    return None

  heap = []
  for i, node in enumerate(lists):
    if node:
      heappush(heap, (node.val, i, node))

  dummy = ListNode(0)
  curr = dummy

  while heap:
    val, idx, node = heappop(heap)
    curr.next = node
    curr = curr.next

    if node.next:
      heappush(heap, (node.next.val, idx, node.next))

  return dummy.next

initialize dummy node

0 / 1

Last, to ensure that our heap always contains the first unmerged element from each of the k linked lists, we push the next node from the linked list where the popped node came from onto the heap, if it exists. This prepares the heap for the next iteration.


from heapq import heappush, heappop

def mergeKLists(lists):
  if not lists:
    return None

  heap = []
  for i, node in enumerate(lists):
    if node:
      heappush(heap, (node.val, i, node))

  dummy = ListNode(0)
  curr = dummy

  while heap:
    val, idx, node = heappop(heap)
    curr.next = node
    curr = curr.next

    if node.next:
      heappush(heap, (node.next.val, idx, node.next))

  return dummy.next

add -1 to result

0 / 1

When the heap is empty, all the elements from the linked lists have been merged, and we can return head of the the merged list by returning dummy.next.


from heapq import heappush, heappop

def mergeKLists(lists):
  if not lists:
    return None

  heap = []
  for i, node in enumerate(lists):
    if node:
      heappush(heap, (node.val, i, node))

  dummy = ListNode(0)
  curr = dummy

  while heap:
    val, idx, node = heappop(heap)
    curr.next = node
    curr = curr.next

    if node.next:
      heappush(heap, (node.next.val, idx, node.next))

  return dummy.next

0 / 1

Solution


from heapq import heappush, heappop

def mergeKLists(lists):
  if not lists:
    return None

  heap = []
  for i, node in enumerate(lists):
    if node:
      heappush(heap, (node.val, i, node))

  dummy = ListNode(0)
  curr = dummy

  while heap:
    val, idx, node = heappop(heap)
    curr.next = node
    curr = curr.next

    if node.next:
      heappush(heap, (node.next.val, idx, node.next))

  return dummy.next

merge k sorted lists

0 / 19

Complexity Analysis

Time Complexity: O(n * log k), where n is the total number of nodes in each of the input linked lists. We iterate over all the nodes in the linked lists. At each iteration, we push and pop elements from the heap, which takes O(log k) time.

Space Complexity: O(k). We use a min-heap of size k to store the first unmerged element from each of the k linked lists.

Next: Introduction

Your account is free and you can post anonymously if you choose.

Sort By

Heap

Merge K Sorted Lists

Watch Video Walkthrough

Watch Video Walkthrough

DESCRIPTION (credit Leetcode.com)

Explanation

Step 1: Initialize the heap

Step 2: Build the merged list

Solution

Complexity Analysis

Questions

Learn

Links

Legal

Contact