Merge K Sorted Lists Using Min Heap in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of merging 3 sorted lists using a min heap

What is the printed merged list after running the following Java code that merges 3 sorted linked lists using a min heap?

DSA Java

import java.util.*;

class ListNode {
    int val;
    ListNode next;
    ListNode(int val) { this.val = val; }
}

public class MergeKLists {
    public static ListNode mergeKLists(ListNode[] lists) {
        PriorityQueue<ListNode> minHeap = new PriorityQueue<>(Comparator.comparingInt(a -> a.val));
        for (ListNode node : lists) {
            if (node != null) minHeap.offer(node);
        }
        ListNode dummy = new ListNode(0);
        ListNode current = dummy;
        while (!minHeap.isEmpty()) {
            ListNode smallest = minHeap.poll();
            current.next = smallest;
            current = current.next;
            if (smallest.next != null) minHeap.offer(smallest.next);
        }
        return dummy.next;
    }

    public static void printList(ListNode head) {
        while (head != null) {
            System.out.print(head.val + " -> ");
            head = head.next;
        }
        System.out.println("null");
    }

    public static void main(String[] args) {
        ListNode l1 = new ListNode(1);
        l1.next = new ListNode(4);
        l1.next.next = new ListNode(5);

        ListNode l2 = new ListNode(1);
        l2.next = new ListNode(3);
        l2.next.next = new ListNode(4);

        ListNode l3 = new ListNode(2);
        l3.next = new ListNode(6);

        ListNode[] lists = new ListNode[]{l1, l2, l3};
        ListNode merged = mergeKLists(lists);
        printList(merged);
    }
}

A 1 -> 1 -> 3 -> 4 -> 5 -> 6 -> null

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

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

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

💡 Hint

Trace the smallest elements picked from each list step-by-step using the min heap.

The min heap always picks the smallest current node among all lists. The merged list contains all elements in sorted order: 1,1,2,3,4,4,5,6.

🧠 Conceptual

intermediate

Understanding the role of the min heap in merging K sorted lists

Why is a min heap used when merging K sorted linked lists?

A To always extract the smallest current element among all lists efficiently

B To store all elements from all lists before sorting

C To reverse the order of elements in each list

D To merge lists by concatenating them without sorting

💡 Hint

Think about how to efficiently find the smallest element among multiple sorted lists at each step.

A min heap allows efficient extraction of the smallest element among the heads of all lists, enabling the merged list to be built in sorted order without sorting all elements at once.

🔧 Debug

advanced

Identify the runtime error in the mergeKLists method

What runtime error will occur when running the following code snippet for merging K sorted lists using a min heap?

DSA Java

PriorityQueue<ListNode> minHeap = new PriorityQueue<>();
for (ListNode node : lists) {
    if (node != null) minHeap.offer(node);
}
while (!minHeap.isEmpty()) {
    ListNode smallest = minHeap.poll();
    current.next = smallest;
    current = current.next;
    if (smallest.next != null) minHeap.offer(smallest.next);
}

A No error, code runs correctly

B NullPointerException when lists array is empty

C ClassCastException because ListNode does not implement Comparable

D IllegalArgumentException due to invalid comparator

💡 Hint

Check if ListNode has a natural ordering or if a comparator is provided to PriorityQueue.

PriorityQueue without a comparator requires elements to implement Comparable. ListNode does not implement Comparable, so ClassCastException occurs at runtime.

❓ Predict Output

advanced

Output after merging empty and non-empty lists

What is the output of the merged list after running this code with one empty list and two non-empty sorted lists?

DSA Java

ListNode l1 = null;
ListNode l2 = new ListNode(2);
l2.next = new ListNode(3);
ListNode l3 = new ListNode(1);
l3.next = new ListNode(4);
ListNode[] lists = new ListNode[]{l1, l2, l3};
ListNode merged = mergeKLists(lists);
printList(merged);

A 1 -> 4 -> 2 -> 3 -> null

B 1 -> 2 -> 3 -> 4 -> null

C null

D 2 -> 3 -> 1 -> 4 -> null

💡 Hint

Empty lists do not affect the merged output; the merged list is sorted from the non-empty lists.

The merged list contains all elements from the non-empty lists in sorted order: 1,2,3,4.

🧠 Conceptual

expert

Time complexity of merging K sorted lists using a min heap

What is the time complexity of merging K sorted lists with a total of N elements using a min heap?

A O(N log K)

B O(K log N)

C O(N K)

D O(N log N)

💡 Hint

Consider how many times elements are pushed and popped from the min heap and the heap size.

Each of the N elements is pushed and popped from a min heap of size at most K, so each operation costs O(log K), resulting in O(N log K) total time.

Practice

1. What is the main purpose of using a min heap when merging K sorted lists?

easy

A. To efficiently get the smallest element among the heads of all lists

B. To sort each individual list before merging

C. To store all elements from all lists at once

D. To reverse the order of elements in each list

Solution

1. Step 1: Understand the role of min heap in merging

   A min heap helps quickly find the smallest element among multiple sorted lists by always keeping the smallest element at the top.

2. Step 2: Why min heap is used here

   When merging K sorted lists, we repeatedly pick the smallest current element from all lists. The min heap efficiently supports this operation.

3. Final Answer:

   To efficiently get the smallest element among the heads of all lists -> Option A

4. Quick Check:

   Min heap = smallest element retrieval [OK]

Hint: Min heap always gives smallest element fast [OK]

Common Mistakes:

• Thinking min heap sorts entire lists
• Using max heap instead of min heap
• Storing all elements at once in heap

2. Which of the following Java code snippets correctly initializes a min heap for integers?

easy

A. PriorityQueue<Integer> minHeap = new PriorityQueue<>(Comparator.reverseOrder());

B. PriorityQueue<Integer> minHeap = new PriorityQueue<>();

C. PriorityQueue<Integer> minHeap = new PriorityQueue<>(Collections.reverseOrder());

D. PriorityQueue<Integer> minHeap = new PriorityQueue<>(Comparator.naturalOrder().reversed());

Solution

1. Step 1: Check default behavior of PriorityQueue

   By default, Java's PriorityQueue is a min heap, so new PriorityQueue<>() creates a min heap.

2. Step 2: Analyze other options

   Options A, B, and D create max heaps using reverse order or equivalent. Only C uses the default min heap.

3. Final Answer:

   PriorityQueue<Integer> minHeap = new PriorityQueue<>(); -> Option B

4. Quick Check:

   Default PriorityQueue = min heap [OK]

Hint: Default PriorityQueue is min heap in Java [OK]

Common Mistakes:

• Using reverseOrder() creates max heap, not min heap
• Confusing Comparator.naturalOrder() usage
• Not importing java.util.PriorityQueue

3. What will be the output of the following code snippet that merges 3 sorted lists using a min heap?

List<Integer> l1 = List.of(1, 4, 5);
List<Integer> l2 = List.of(1, 3, 4);
List<Integer> l3 = List.of(2, 6);

PriorityQueue<int[]> minHeap = new PriorityQueue<>((a, b) -> a[0] - b[0]);

minHeap.offer(new int[]{l1.get(0), 0, 0});
minHeap.offer(new int[]{l2.get(0), 1, 0});
minHeap.offer(new int[]{l3.get(0), 2, 0});

List<Integer> result = new ArrayList<>();

while (!minHeap.isEmpty()) {
    int[] curr = minHeap.poll();
    result.add(curr[0]);
    int listIndex = curr[1];
    int elementIndex = curr[2];
    if (elementIndex + 1 < (listIndex == 0 ? l1.size() : listIndex == 1 ? l2.size() : l3.size())) {
        int nextVal = (listIndex == 0 ? l1 : listIndex == 1 ? l2 : l3).get(elementIndex + 1);
        minHeap.offer(new int[]{nextVal, listIndex, elementIndex + 1});
    }
}

System.out.println(result);

medium

A. [2, 3, 4, 5, 6]

B. [1, 2, 3, 4, 4, 5, 6]

C. [1, 1, 3, 4, 4, 5, 6]

D. [1, 1, 2, 3, 4, 4, 5, 6]

Solution

1. Step 1: Initialize min heap with first elements

   Heap starts with [1(l1), 1(l2), 2(l3)]. Poll smallest 1 from l1, add next 4 from l1. Then poll 1 from l2, add next 3 from l2, and so on.

2. Step 2: Continue polling and adding next elements

   Sequence of polled elements is 1(l1), 1(l2), 2(l3), 3(l2), 4(l1), 4(l2), 5(l1), 6(l3). This forms the merged sorted list.

3. Final Answer:

   [1, 1, 2, 3, 4, 4, 5, 6] -> Option D

4. Quick Check:

   Merged sorted list = [1,1,2,3,4,4,5,6] [OK]

Hint: Poll min, add next from same list until all merged [OK]

Common Mistakes:

• Forgetting to add next element after polling
• Mixing list indices when adding next elements
• Not handling empty lists properly

4. Identify the bug in the following code snippet that merges K sorted lists using a min heap:

PriorityQueue<int[]> minHeap = new PriorityQueue<>((a, b) -> b[0] - a[0]);

for (int i = 0; i < lists.size(); i++) {
    if (!lists.get(i).isEmpty()) {
        minHeap.offer(new int[]{lists.get(i).get(0), i, 0});
    }
}

List<Integer> result = new ArrayList<>();

while (!minHeap.isEmpty()) {
    int[] curr = minHeap.poll();
    result.add(curr[0]);
    int listIndex = curr[1];
    int elementIndex = curr[2];
    if (elementIndex + 1 < lists.get(listIndex).size()) {
        minHeap.offer(new int[]{lists.get(listIndex).get(elementIndex + 1), listIndex, elementIndex + 1});
    }
}

medium

A. The result list is not initialized before use

B. The loop does not check for empty lists before adding

C. Comparator reverses order, creating max heap instead of min heap

D. The elementIndex is not incremented correctly when adding next element

Solution

1. Step 1: Analyze the comparator used in PriorityQueue

   The comparator (b[0] - a[0]) reverses natural order, making it a max heap, not a min heap.

2. Step 2: Effect of max heap on merging sorted lists

   Using max heap causes the largest element to be polled first, breaking the merge logic which requires smallest element first.

3. Final Answer:

   Comparator reverses order, creating max heap instead of min heap -> Option C

4. Quick Check:

   Comparator direction = min heap needed [OK]

Hint: Comparator must order ascending for min heap [OK]

Common Mistakes:

• Using (b[0] - a[0]) instead of (a[0] - b[0])
• Ignoring empty list checks
• Not incrementing elementIndex properly

5. Given K sorted linked lists, which approach best optimizes the time complexity when merging them using a min heap? A) Insert all elements from all lists into a min heap and then extract all. B) Insert only the first element of each list into the min heap and then insert next elements as you extract. C) Merge lists two at a time repeatedly without using a heap. D) Use a max heap to store all elements and then reverse the result.

hard

A. Insert first elements only, add next after extraction

B. Insert all elements at once into min heap

C. Merge lists pairwise without heap

D. Use max heap and reverse output

Solution

1. Step 1: Compare time complexities of approaches

   Inserting all elements at once (A) costs O(N log N) where N is total elements. Pairwise merging (C) costs more. Using max heap and reversing (D) is inefficient.

2. Step 2: Explain min heap incremental insertion (B)

   Inserting only first elements and adding next after extraction keeps heap size at K, leading to O(N log K) time, which is optimal.

3. Final Answer:

   Insert first elements only, add next after extraction -> Option A

4. Quick Check:

   Optimal merge = incremental min heap insertion [OK]

Hint: Keep heap size K by adding next elements after polling [OK]

Common Mistakes:

• Inserting all elements at once causing slow performance
• Using max heap incorrectly
• Ignoring heap size impact on complexity