Merge K Sorted Lists Using Min Heap in DSA Java - Time & Space Complexity

Time Complexity: Merge K Sorted Lists Using Min Heap

O(n log k)

Understanding Time Complexity

We want to understand how the time needed to merge multiple sorted lists grows as the number of lists and their sizes increase.

Specifically, how does using a min heap help manage this merging efficiently?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

import java.util.*;

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

This code merges k sorted linked lists into one sorted list using a min heap to always pick the smallest current node.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Removing the smallest node from the heap and adding the next node from that list.
• How many times: This happens once for every node across all lists, so total nodes = n.

How Execution Grows With Input

Each node is pushed and popped from the heap once. The heap size is at most k (number of lists).

Input Size (n)	Approx. Operations
10 (nodes total)	About 10 insertions and 10 removals from a heap of size ≤ k
100	About 100 insertions and removals, each costing log k steps
1000	About 1000 insertions and removals, each costing log k steps

Pattern observation: The total work grows linearly with the total number of nodes, multiplied by the cost of heap operations which grows slowly with the number of lists.

Final Time Complexity

Time Complexity: O(n log k)

This means the time grows mostly with the total number of nodes, but each step involves a quick operation on a small heap of size k.

Common Mistake

[X] Wrong: "Merging k lists is just like merging two lists, so it takes O(n) time."

[OK] Correct: Merging k lists involves managing multiple lists at once, so we must repeatedly find the smallest among k candidates, which adds a log k factor.

Interview Connect

Understanding this time complexity helps you explain why using a min heap is efficient for merging multiple sorted lists, a common real-world problem.

Self-Check

"What if we used a simple linear search to find the smallest node instead of a min heap? How would the time complexity change?"

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