Merge K Sorted Lists Using Min Heap in DSA Javascript

Mental Model

We combine many sorted lists into one sorted list by always picking the smallest current element from all lists using a min heap.

Analogy: Imagine you have K friends each reading a sorted book. You want to write down all the words in order. You ask each friend for their current word, pick the smallest word, write it down, then ask that friend for their next word. Repeat until all words are written.

Lists:
List1: 1 -> 4 -> 7 -> null
List2: 2 -> 5 -> 8 -> null
List3: 3 -> 6 -> 9 -> null

MinHeap (initial): [1, 2, 3]

Merged List: null

Dry Run Walkthrough

Input: lists: [1->4->7, 2->5->8, 3->6->9]

Goal: Merge all 3 sorted lists into one sorted list: 1->2->3->4->5->6->7->8->9

Step 1: Insert first nodes of all lists into min heap

MinHeap: [1, 2, 3]
Merged List: null

Why: We start by knowing the smallest elements from each list to pick the smallest overall

Step 2: Extract min (1) from heap and add to merged list; insert next node (4) from list1 into heap

MinHeap: [2, 3, 4]
Merged List: 1 -> null

Why: We add smallest element to merged list and bring next candidate from same list

Step 3: Extract min (2) from heap and add to merged list; insert next node (5) from list2 into heap

MinHeap: [3, 4, 5]
Merged List: 1 -> 2 -> null

Why: Continue picking smallest and replenishing heap from lists

Step 4: Extract min (3) from heap and add to merged list; insert next node (6) from list3 into heap

MinHeap: [4, 5, 6]
Merged List: 1 -> 2 -> 3 -> null

Why: Keep merging by always picking smallest current element

Step 5: Extract min (4) from heap and add to merged list; insert next node (7) from list1 into heap

MinHeap: [5, 6, 7]
Merged List: 1 -> 2 -> 3 -> 4 -> null

Why: Add next smallest and continue replenishing heap

Step 6: Extract min (5) from heap and add to merged list; insert next node (8) from list2 into heap

MinHeap: [6, 7, 8]
Merged List: 1 -> 2 -> 3 -> 4 -> 5 -> null

Why: Continue merging in sorted order

Step 7: Extract min (6) from heap and add to merged list; insert next node (9) from list3 into heap

MinHeap: [7, 8, 9]
Merged List: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null

Why: Keep picking smallest and adding next nodes

Step 8: Extract min (7) from heap and add to merged list; list1 has no more nodes

MinHeap: [8, 9]
Merged List: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> null

Why: Add remaining elements from heap as lists finish

Step 9: Extract min (8) from heap and add to merged list; list2 has no more nodes

MinHeap: [9]
Merged List: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> null

Why: Continue until all nodes are merged

Step 10: Extract min (9) from heap and add to merged list; list3 has no more nodes

MinHeap: []
Merged List: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

Why: All nodes merged, heap empty

Result:

1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

Annotated Code

DSA Javascript

class ListNode {
  constructor(val = 0, next = null) {
    this.val = val;
    this.next = next;
  }
}

class MinHeap {
  constructor() {
    this.heap = [];
  }

  insert(node) {
    this.heap.push(node);
    this.bubbleUp();
  }

  bubbleUp() {
    let index = this.heap.length - 1;
    while (index > 0) {
      let parentIndex = Math.floor((index - 1) / 2);
      if (this.heap[parentIndex].val <= this.heap[index].val) break;
      [this.heap[parentIndex], this.heap[index]] = [this.heap[index], this.heap[parentIndex]];
      index = parentIndex;
    }
  }

  extractMin() {
    if (this.heap.length === 0) return null;
    if (this.heap.length === 1) return this.heap.pop();
    const min = this.heap[0];
    this.heap[0] = this.heap.pop();
    this.bubbleDown();
    return min;
  }

  bubbleDown() {
    let index = 0;
    const length = this.heap.length;
    while (true) {
      let left = 2 * index + 1;
      let right = 2 * index + 2;
      let smallest = index;

      if (left < length && this.heap[left].val < this.heap[smallest].val) {
        smallest = left;
      }
      if (right < length && this.heap[right].val < this.heap[smallest].val) {
        smallest = right;
      }
      if (smallest === index) break;
      [this.heap[index], this.heap[smallest]] = [this.heap[smallest], this.heap[index]];
      index = smallest;
    }
  }

  isEmpty() {
    return this.heap.length === 0;
  }
}

function mergeKLists(lists) {
  const minHeap = new MinHeap();

  // Insert first node of each list into min heap
  for (const node of lists) {
    if (node !== null) {
      minHeap.insert(node);
    }
  }

  const dummy = new ListNode(0);
  let current = dummy;

  // Extract min and add next nodes until heap is empty
  while (!minHeap.isEmpty()) {
    const minNode = minHeap.extractMin();
    current.next = minNode;
    current = current.next;
    if (minNode.next !== null) {
      minHeap.insert(minNode.next);
    }
  }

  return dummy.next;
}

// Driver code using dry_run input
const list1 = new ListNode(1, new ListNode(4, new ListNode(7)));
const list2 = new ListNode(2, new ListNode(5, new ListNode(8)));
const list3 = new ListNode(3, new ListNode(6, new ListNode(9)));

const mergedHead = mergeKLists([list1, list2, list3]);

// Print merged list
let output = '';
let curr = mergedHead;
while (curr !== null) {
  output += curr.val + ' -> ';
  curr = curr.next;
}
output += 'null';
console.log(output);

for (const node of lists) { if (node !== null) { minHeap.insert(node); } }

Insert first nodes of all lists into min heap to start merging

while (!minHeap.isEmpty()) { const minNode = minHeap.extractMin(); current.next = minNode; current = current.next; if (minNode.next !== null) { minHeap.insert(minNode.next); } }

Repeatedly extract smallest node, add to merged list, and insert next node from same list

Output

1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null

Complexity Analysis

Time: O(N log K) because we process all N nodes and each heap operation takes O(log K) where K is number of lists

Space: O(K) for the heap storing one node from each list at a time

vs Alternative: Naive merging by repeatedly merging two lists at a time takes more time (O(KN)) compared to min heap approach which is more efficient

Edge Cases

Empty input lists array

Returns null as there are no nodes to merge

DSA Javascript

if (node !== null) { minHeap.insert(node); }

Some lists are empty (null)

Only non-empty lists contribute nodes; empty lists ignored

DSA Javascript

if (node !== null) { minHeap.insert(node); }

All lists empty

Returns null as no nodes exist

DSA Javascript

if (node !== null) { minHeap.insert(node); }

When to Use This Pattern

When you need to merge multiple sorted lists efficiently, use a min heap to always pick the smallest current element from all lists.

Common Mistakes

Mistake: Not inserting the next node of the extracted node back into the heap
Fix: After extracting minNode, check if minNode.next exists and insert it into the heap

Mistake: Using a max heap or sorting all nodes at once instead of incremental merging
Fix: Use a min heap to always pick the smallest node incrementally for efficiency

Summary

Merges K sorted linked lists into one sorted linked list using a min heap.

Use when you have multiple sorted lists and want to merge them efficiently.

The key is to always pick the smallest current node from all lists using a min heap.