DSA Visualizations | Hello Interview

Merge K Sorted Lists

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.

EXAMPLES

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.

Loading comments...

Back to Home

Eliminating Pairs

Overview

Container With Most Water

3-Sum

Triangle Numbers

Pointers as Regions

Move Zeroes

Sort Colors

Trapping Rain Water

Variable-Length

Overview

Longest Substring Without Repeating Characters

Longest Repeating Character Replacement

Fixed-Length

Overview

Maximum Sum of Subarrays of Size K

Max Points You Can Obtain From Cards

Max Sum of Distinct Subarrays Length k

Overview

Can Attend Meetings

Insert Interval

Non-Overlapping Intervals

Merge Intervals

Employee Free Time

Overview

Valid Parentheses

Decode String

Longest Valid Parentheses

Monotonic Stack

Overview

Daily Temperatures

Largest Rectangle in Histogram

Overview

Linked List Cycle

Palindrome Linked List

Remove Nth Node From End of List

Reorder List

Swap Nodes in Pairs

Overview

Apple Harvest (Koko Eating Bananas)

Search in Rotated Sorted Array

Overview

Kth Largest Element in an Array

K Closest Points to Origin

Find K Closest Elements

Merge K Sorted Lists

Introduction

Binary Trees

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

Overview

Word Search

Solution Space Trees

Subsets

Generate Parentheses

Combination Sum

Topological Sort

Overview

Course Schedule

Course Schedule II

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

Overview

Best Time to Buy and Sell Stock

Gas Station

Jump Game

Overview

Implement Trie Methods

Prefix Matching

Overview

Count Vowels in Substrings

Subarray Sum Equals K

Spiral Matrix

Rotate Image

Set Matrix Zeroes

Find K Closest Elements

Depth-First Search: Introduction

Merge K Sorted Lists

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.

EXAMPLES

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.

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


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


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


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


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

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.

Loading comments...

BlogBehavioral Interview Examples Interviewing for Experienced SWEs All Blog Posts

Compare UsCompare to Interviewing.io

LinksFAQ Schedule Mock Interviews Pricing Become a Coach Our Coaches Learn System Design Learn DSA

LegalTerms and Conditions Privacy Policy

Contactsupport@hellointerview.com7511 Greenwood Ave North
Unit #4238 Seattle, WA 98103