Common Heap Patterns - Learning Module

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Merge K Sorted Lists

When Sorted Data Comes in Pieces

Imagine you're a search engine processing results from a hundred different index servers, each returning a sorted list of relevant documents. Or you're a database merging sorted runs from an external sort operation. Or you're a logging system combining timestamp-ordered events from dozens of microservices.

The common thread: you have k sorted sequences and need to produce one unified sorted output.

Merging two sorted lists is simple—the classic merge operation from MergeSort runs in O(n) time. But what happens when k grows large? Naively merging pairs iteratively becomes increasingly inefficient. This is where heaps provide an elegant, optimal solution.

What You Will Master

By the end of this page, you will understand why the naive pairwise merge approach is suboptimal, master the heap-based k-way merge that achieves O(n log k) time, explore real-world applications from external sorting to distributed systems, and implement solutions for both linked list and array-based variations.

Problem Definition: K-Way Merge

The K-Way Merge Problem:

Given k sorted sequences (lists, arrays, or streams), produce a single sorted sequence containing all elements from all inputs.

Formal Statement:

Input: k sorted lists L₁, L₂, ..., Lₖ with sizes n₁, n₂, ..., nₖ
Output: A single sorted list containing all N = Σnᵢ elements
Constraint: Maintain sorted order in the output

Example:

Input:
  List 1: [1, 4, 5]
  List 2: [1, 3, 4]
  List 3: [2, 6]

Output: [1, 1, 2, 3, 4, 4, 5, 6]

Total Element Count

Throughout this page, N refers to the total number of elements across all k lists (N = n₁ + n₂ + ... + nₖ). This distinction matters for complexity analysis—we want algorithms that scale well with both N and k.

Why This Problem Matters:

The k-way merge is not just a textbook exercise—it's a fundamental operation in:

External Sorting: When data exceeds memory, we sort chunks and merge them
Distributed Query Processing: Merging results from multiple database shards
Log Aggregation: Combining timestamped events from many sources
MapReduce/Shuffle Phase: Merging sorted outputs from multiple mappers
LSM-Tree Compaction: Merging sorted string tables in databases like RocksDB
Multi-Stream Processing: Combining sensor readings from multiple devices

The Two-Pointer Baseline: Merging Two Lists

Before tackling k lists, let's revisit the simpler case: merging exactly two sorted lists. This is the merge step from MergeSort.

merge-two-lists
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from typing import Optional
 
class ListNode:
                                "Definition for singly-linked list."
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
 
def merge_two_lists(l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
    """
    Merge two sorted linked lists.
    
    Time: O(n + m) where n, m are list lengths
    Space: O(1) extra (just rewiring pointers)
    
    The classic two-pointer merge technique.
    """
    # Dummy head simplifies edge cases
    dummy = ListNode(-1)
    current = dummy
    
    while l1 and l2:
        if l1.val <= l2.val:
            current.next = l1
            l1 = l1.next
        else:
            current.next = l2
            l2 = l2.next
        current = current.next
    
    # Attach remaining elements (one list exhausted)
    current.next = l1 if l1 else l2
    
    return dummy.next
 
 
def merge_two_arrays(arr1: list[int], arr2: list[int]) -> list[int]:
    """
    Merge two sorted arrays.
    
    Time: O(n + m)
    Space: O(n + m) for the result array
    """
    result = []
    i = j = 0
    
    while i < len(arr1) and j < len(arr2):
        if arr1[i] <= arr2[j]:
            result.append(arr1[i])
            i += 1
        else:
            result.append(arr2[j])
            j += 1
    
    # Append remaining elements
    result.extend(arr1[i:])
    result.extend(arr2[j:])
    
    return result
 
 
# Example
print(merge_two_arrays([1, 4, 5], [1, 3, 4]))
# Output: [1, 1, 3, 4, 4, 5]

Key Properties of Two-Way Merge:

Time: O(n + m) — each element is processed exactly once
Comparisons: At most (n + m - 1) comparisons
Decisions: At each step, we compare two elements and pick the smaller

This is optimal for two lists. But how do we extend to k lists?

Naive Approaches for K Lists

Let's analyze two intuitive but suboptimal approaches before introducing the heap solution.

Approach 1: Sequential Pairwise Merge

•Algorithm: Merge list 1 and 2, then merge result with list 3, then with list 4, etc.
•Example: ((L₁ ⊕ L₂) ⊕ L₃) ⊕ L₄, where ⊕ denotes merge
•Time Analysis: First merge: O(n₁ + n₂), Second merge: O(n₁ + n₂ + n₃), ...
•Total Time: O(k × N) where N is total elements — each element potentially touched k times!
•Verdict: Highly inefficient when k is large

Why is this O(k × N)?

Consider k lists, each with n elements (N = k × n total):

Merge 1: n + n = 2n elements
Merge 2: 2n + n = 3n elements
Merge 3: 3n + n = 4n elements
...
Merge k-1: (k-1)n + n = kn elements

Total work: 2n + 3n + 4n + ... + kn = n(2 + 3 + ... + k) = n × O(k²) = O(k × N)

Approach 2: Collect All and Sort

•Algorithm: Collect all N elements into one array, then sort
•Time: O(N log N) — ignores the pre-sorted structure
•Verdict: Wastes the information that inputs are already sorted

The Smart Approach: Divide and Conquer

A better pairwise approach uses divide-and-conquer: pair up all k lists, merge each pair in parallel, repeat until one list remains. This gives O(N log k) time—already much better! But heaps offer an equally efficient solution with different trade-offs.

The Heap Solution: O(N log k)

The heap-based solution uses a fundamentally different approach: instead of merging pairs, we maintain a min-heap of k elements representing the current smallest unprocessed element from each list.

The Key Insight:

At any moment, the global minimum among all remaining elements must be the minimum of the k 'frontier' elements—one from each list. A min-heap lets us find this global minimum in O(1) and update in O(log k).

Algorithm:

Initialize: Create a min-heap with the first element from each non-empty list (along with which list it came from and its index)
Extract-Insert Loop:
- Extract the minimum from the heap → this is the next output element
- If that element's list has more elements, insert the next element from that list
Terminate: When the heap is empty, all elements have been processed

Why This Works:

The heap always contains at most k elements (one frontier element per list). Since lists are sorted, the smallest remaining element is always at the frontier. The heap efficiently maintains access to the smallest frontier element.

merge-k-sorted-lists
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import heapq
from typing import List, Optional
 
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
 
def merge_k_lists(lists: List[Optional[ListNode]]) -> Optional[ListNode]:
    """
    LeetCode 23: Merge k Sorted Lists
    
    Merge k sorted linked lists using a min-heap.
    
    Time: O(N log k) where N = total elements, k = number of lists
    Space: O(k) for the heap
    
    The heap contains at most k elements (one from each list).
    Each of N elements is pushed and popped exactly once.
    """
    # Min-heap storing (value, list_index, node)
    # list_index breaks ties to avoid comparing ListNode objects
    min_heap = []
    
    # Initialize: add first element from each non-empty list
    for i, head in enumerate(lists):
        if head:
            heapq.heappush(min_heap, (head.val, i, head))
    
    # Dummy head for result list
    dummy = ListNode(-1)
    current = dummy
    
    while min_heap:
        # Extract minimum
        val, list_idx, node = heapq.heappop(min_heap)
        
        # Add to result
        current.next = node
        current = current.next
        
        # If this list has more elements, add the next one
        if node.next:
            heapq.heappush(min_heap, (node.next.val, list_idx, node.next))
    
    return dummy.next
 
 
def merge_k_arrays(arrays: List[List[int]]) -> List[int]:
    """
    Merge k sorted arrays using a min-heap.
    
    Time: O(N log k)
    Space: O(k) for heap + O(N) for result
    """
    result = []
    
    # Heap stores (value, array_index, element_index)
    min_heap = []
    
    # Initialize with first element from each non-empty array
    for i, arr in enumerate(arrays):
        if arr:
            heapq.heappush(min_heap, (arr[0], i, 0))
    
    while min_heap:
        val, arr_idx, elem_idx = heapq.heappop(min_heap)
        result.append(val)
        
        # If this array has more elements
        if elem_idx + 1 < len(arrays[arr_idx]):
            next_elem = arrays[arr_idx][elem_idx + 1]
            heapq.heappush(min_heap, (next_elem, arr_idx, elem_idx + 1))
    
    return result
 
 
# Example usage
arrays = [[1, 4, 5], [1, 3, 4], [2, 6]]
print(merge_k_arrays(arrays))
# Output: [1, 1, 2, 3, 4, 4, 5, 6]
 
# Step-by-step trace:
# Initial heap: [(1,0,0), (1,1,0), (2,2,0)]
# Extract (1,0,0): result=[1], push (4,0,1)
# Extract (1,1,0): result=[1,1], push (3,1,1)
# Extract (2,2,0): result=[1,1,2], push (6,2,1)
# Extract (3,1,1): result=[1,1,2,3], push (4,1,2)
# Extract (4,0,1): result=[1,1,2,3,4], push (5,0,2)
# Extract (4,1,2): result=[1,1,2,3,4,4], no more in list 1
# Extract (5,0,2): result=[1,1,2,3,4,4,5], no more in list 0
# Extract (6,2,1): result=[1,1,2,3,4,4,5,6], no more in list 2
# Heap empty, done!

Complexity Analysis: Why O(N log k)

Let's rigorously analyze the heap-based k-way merge and compare it to alternatives.

Time Complexity: O(N log k)

Breakdown:

Each of the N elements is pushed to the heap exactly once: N pushes
Each of the N elements is popped from the heap exactly once: N pops
Each heap operation (push or pop) takes O(log k) since heap size ≤ k
Total: O(N log k)

Why the heap size is always ≤ k:

Initially, we insert one element per non-empty list (at most k elements)
Each pop is followed by at most one push (from the same list)
Therefore, heap size never exceeds k

Space Complexity: O(k)

The heap contains at most k elements
For linked list version, no additional space needed (rewiring pointers)
For array version, O(N) for the result array

Comparison of K-Way Merge Approaches
Approach	Time Complexity	Space Complexity	Notes
Sequential Pairwise	O(k × N)	O(N)	Each element touched up to k times
Collect and Sort	O(N log N)	O(N)	Ignores pre-sorted structure
Divide & Conquer Pairwise	O(N log k)	O(N)	log k merge rounds, N work each
Heap-Based	O(N log k)	O(k)	Optimal time, minimal space

When O(N log k) Beats O(N log N)

When k is small compared to N, the difference is significant. For 10 million elements across 100 lists: O(N log N) ≈ 230 million ops vs O(N log k) ≈ 67 million ops—a 3.4x improvement. As the number of elements per list grows (N/k increases), the advantage grows.

Is O(N log k) Optimal?

Yes, for comparison-based algorithms! Here's the information-theoretic argument:

The output is one of (N choose n₁, n₂, ..., nₖ) = N! / (n₁! × n₂! × ... × nₖ!) possible interleavings
Any comparison-based algorithm needs Ω(log(number of outcomes)) comparisons
log(N! / (n₁! × ... × nₖ!)) ≥ N × log k (using entropy arguments)

Therefore, O(N log k) is optimal for comparison-based k-way merge.

Alternative: Divide and Conquer Merge

While the heap solution is elegant, there's an equally efficient divide-and-conquer approach worth understanding. This method pairs up lists and merges them, repeating until one list remains.

divide-conquer-merge
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from typing import List, Optional
 
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
 
def merge_k_lists_divide_conquer(lists: List[Optional[ListNode]]) -> Optional[ListNode]:
    """
    Merge k sorted lists using divide and conquer.
    
    Time: O(N log k) - same as heap approach
    Space: O(log k) - recursion stack for divide and conquer
    
    This approach may have better cache locality than heap-based.
    """
    if not lists:
        return None
    
    def merge_two(l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        """Standard two-list merge."""
        dummy = ListNode(-1)
        current = dummy
        
        while l1 and l2:
            if l1.val <= l2.val:
                current.next = l1
                l1 = l1.next
            else:
                current.next = l2
                l2 = l2.next
            current = current.next
        
        current.next = l1 if l1 else l2
        return dummy.next
    
    def merge_range(start: int, end: int) -> Optional[ListNode]:
        """Recursively merge lists[start:end+1]."""
        if start > end:
            return None
        if start == end:
            return lists[start]
        
        mid = (start + end) // 2
        left = merge_range(start, mid)
        right = merge_range(mid + 1, end)
        return merge_two(left, right)
    
    return merge_range(0, len(lists) - 1)
 
 
# Iterative version (avoids recursion)
def merge_k_lists_iterative(lists: List[Optional[ListNode]]) -> Optional[ListNode]:
    """
    Iterative divide and conquer - pairs up lists and merges.
    
    Round 1: Merge pairs -> k/2 lists
    Round 2: Merge pairs -> k/4 lists
    ...
    Final: 1 list
    
    Time: O(N log k)
    Space: O(1) extra (modifies input list)
    """
    if not lists:
        return None
    
    def merge_two(l1, l2):
        dummy = ListNode(-1)
        current = dummy
        while l1 and l2:
            if l1.val <= l2.val:
                current.next = l1
                l1 = l1.next
            else:
                current.next = l2
                l2 = l2.next
            current = current.next
        current.next = l1 or l2
        return dummy.next
    
    # Reduce lists by half each round
    while len(lists) > 1:
        merged = []
        for i in range(0, len(lists), 2):
            l1 = lists[i]
            l2 = lists[i + 1] if i + 1 < len(lists) else None
            merged.append(merge_two(l1, l2))
        lists = merged
    
    return lists[0]

Heap Advantages

•Lower memory: O(k) vs O(log k) stack + list storage
•Streaming friendly: can process infinite streams
•Simpler mental model for some

D&C Advantages

•Better cache locality (accessing adjacent memory)
•Parallelizable: merge pairs independently
•No heap overhead in languages without efficient heaps

Real-World Applications

The k-way merge pattern is fundamental to many critical systems. Understanding these applications helps you recognize when to apply this pattern.

Application: External Sorting

•Problem: Sort 1TB of data with only 8GB RAM
•Solution: Divide data into ~125 sorted 'runs' that fit in memory
•Merge Phase: K-way merge the runs, reading chunks into memory as needed
•Why Heaps: Heap-based merge minimizes disk I/O while maintaining optimal merge
•Used By: Database sort operations, MapReduce shuffle phase, Unix 'sort' command

Application: LSM-Tree Compaction

•Used By: RocksDB, LevelDB, Cassandra, HBase
•Structure: Multiple levels of sorted string tables (SSTables)
•Compaction: Merge overlapping SSTables within a level
•K-Way Merge: Core operation when merging multiple SSTables
•Critical Path: Compaction throughput directly affects write performance

Application: Distributed Query Processing

•Scenario: Database query across 100 shards
•Each Shard: Returns sorted results for ORDER BY clause
•Coordinator: Must merge k sorted streams into final result
•Streaming: K-way heap merge processes results as they arrive
•LIMIT Optimization: Can stop early once LIMIT is reached

The Log Aggregation Pattern

Modern microservice architectures generate logs from dozens or hundreds of services. Each service produces timestamp-ordered log entries. A centralized logging system must merge these streams in timestamp order for querying. This is exactly k-way merge—often operating on potentially infinite streams.

Problem Variations

The k-way merge pattern appears in several disguised forms. Recognizing these variations is key to applying the pattern effectively.

K-Way Merge Variations
Problem	Input Structure	Heap Contents	Key Insight
Merge K Sorted Lists	k linked lists	(value, list_idx, node)	Standard k-way merge
Merge K Sorted Arrays	k arrays	(value, arr_idx, elem_idx)	Track element index
K-th Smallest in Matrix	n×n sorted matrix	(value, row, col)	Matrix = n sorted rows
Smallest Range Covering K Lists	k lists	(value, list_idx, elem_idx)	Track max too, minimize range
Find K Pairs with Smallest Sums	2 arrays	(sum, i, j)	Virtual k lists of pair sums

kth-smallest-matrix
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import heapq
from typing import List
 
def kth_smallest_in_matrix(matrix: List[List[int]], k: int) -> int:
    """
    LeetCode 378: Kth Smallest Element in a Sorted Matrix
    
    Matrix where each row and column is sorted.
    Find k-th smallest element overall.
    
    Approach: View as k-way merge of n sorted rows.
    Use min-heap with first element of each row.
    
    Time: O(k log n) where n = number of rows
    Space: O(n) for the heap
    """
    n = len(matrix)
    
    # Heap: (value, row, col)
    min_heap = [(matrix[i][0], i, 0) for i in range(n)]
    heapq.heapify(min_heap)
    
    # Extract k times
    for _ in range(k):
        val, row, col = heapq.heappop(min_heap)
        
        # This is the answer on k-th extraction
        if _ == k - 1:
            return val
        
        # Push next element from same row if exists
        if col + 1 < n:
            heapq.heappush(min_heap, (matrix[row][col + 1], row, col + 1))
    
    return -1  # Should never reach
 
 
def smallest_range_covering_k_lists(nums: List[List[int]]) -> List[int]:
    """
    LeetCode 632: Smallest Range Covering Elements from K Lists
    
    Find smallest range [a, b] that includes at least one number 
    from each of the k lists.
    
    Approach: K-way merge while tracking current max.
    Range = [heap_min, current_max]. Minimize (max - min).
    
    Time: O(N log k)
    Space: O(k)
    """
    # Heap: (value, list_idx, elem_idx)
    min_heap = []
    current_max = float('-inf')
    
    # Initialize with first element from each list
    for i, lst in enumerate(nums):
        heapq.heappush(min_heap, (lst[0], i, 0))
        current_max = max(current_max, lst[0])
    
    best_range = [min_heap[0][0], current_max]
    
    while True:
        min_val, list_idx, elem_idx = heapq.heappop(min_heap)
        
        # Current range: [min_val, current_max]
        if current_max - min_val < best_range[1] - best_range[0]:
            best_range = [min_val, current_max]
        
        # Try to advance in this list
        if elem_idx + 1 < len(nums[list_idx]):
            next_val = nums[list_idx][elem_idx + 1]
            heapq.heappush(min_heap, (next_val, list_idx, elem_idx + 1))
            current_max = max(current_max, next_val)
        else:
            # One list exhausted - cannot have a valid range anymore
            break
    
    return best_range
 
 
# Examples
matrix = [
    [1, 5, 9],
    [10, 11, 13],
    [12, 13, 15]
]
print(f"8th smallest: {kth_smallest_in_matrix(matrix, 8)}")  # 13
 
nums = [[4, 10, 15, 24, 26], [0, 9, 12, 20], [5, 18, 22, 30]]
print(f"Smallest range: {smallest_range_covering_k_lists(nums)}")  # [20, 24]

Implementation Tips and Pitfalls

Implementing k-way merge correctly requires attention to several subtle details that can cause bugs.

Common Pitfalls

•Tie-Breaking in Heap: When values are equal, heaps may compare the second element. Storing non-comparable objects (like ListNode) can crash. Include an index as a tiebreaker.
•Empty List Handling: Some input lists may be empty. Skip them during initialization—don't push null/None to the heap.
•Index Management: Keep careful track of which list an element came from and its position within that list. Triple (value, list_idx, elem_idx) is the standard pattern.
•Exhausted Lists: When popping an element, check if its source list has more elements before pushing. Don't assume all lists have the same length.
•Off-By-One Errors: Is your list's next element at index elem_idx + 1 or node.next? Be consistent with your representation.

robust-merge-k
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import heapq
from typing import List, Optional
from dataclasses import dataclass, field
 
@dataclass(order=True)
class HeapEntry:
    """
    Properly ordered heap entry that avoids comparison issues.
    
    order=True makes dataclass generate comparison methods
    based on field order. 'node' is excluded from comparison.
    """
    value: int
    list_index: int
    elem_index: int
    node: object = field(compare=False)  # Excluded from comparisons
 
 
def merge_k_lists_robust(lists: List[Optional['ListNode']]) -> Optional['ListNode']:
    """
    Robust implementation handling all edge cases.
    """
    # Filter out empty lists upfront
    non_empty = [(head.val, i, head) for i, head in enumerate(lists) if head]
    
    if not non_empty:
        return None
    
    # Build initial heap
    min_heap = []
    for val, idx, node in non_empty:
        heapq.heappush(min_heap, HeapEntry(val, idx, 0, node))
    
    dummy = ListNode(-1)
    current = dummy
    
    while min_heap:
        entry = heapq.heappop(min_heap)
        node = entry.node
        
        current.next = node
        current = current.next
        
        # Check if source list has more elements
        if node.next:
            heapq.heappush(min_heap, HeapEntry(
                node.next.val,
                entry.list_index,
                entry.elem_index + 1,
                node.next
            ))
    
    return dummy.next
 
 
# Alternative: Use tuple with index as tiebreaker
def merge_k_lists_tuple(lists):
    """Using tuples with guaranteed ordering."""
    min_heap = []
    
    for i, head in enumerate(lists):
        if head:
            # (value, list_index, node) - list_index breaks ties
            heapq.heappush(min_heap, (head.val, i, head))
    
    dummy = ListNode(-1)
    current = dummy
    
    while min_heap:
        val, list_idx, node = heapq.heappop(min_heap)
        current.next = node
        current = current.next
        
        if node.next:
            # Same list_idx ensures proper tiebreaking
            heapq.heappush(min_heap, (node.next.val, list_idx, node.next))
    
    return dummy.next

Python's heapq Gotcha

Python's heapq compares tuples element-by-element. If first elements are equal, it compares second elements. If you store (value, node), and two values are equal, Python tries to compare nodes and crashes. Always include an index: (value, index, node).

Summary: The K-Way Merge Pattern

We've thoroughly explored the k-way merge pattern—a fundamental technique with broad applications. Let's consolidate the key insights:

Key Takeaways

•The Problem: Merge k sorted sequences into one sorted output efficiently—leveraging pre-sorted structure.
•The Solution: Maintain a min-heap of k frontier elements, always extracting the global minimum in O(log k).
•Complexity: O(N log k) time, O(k) space—optimal for comparison-based algorithms.
•Alternative: Divide-and-conquer pairwise merge achieves same time complexity with different trade-offs.
•Applications: External sorting, LSM-tree compaction, distributed databases, log aggregation.
•Variations: Sorted matrix k-th element, smallest range, k pairs with smallest sums—same core pattern.

What's Next:

The k-way merge and k-th element patterns are powerful, but heaps have even more tricks. In the next page, we'll explore median maintenance—a pattern where we use two heaps together to maintain a running median as data streams in. This technique builds directly on the heap intuition we've developed.

Pattern Mastered

You now understand the k-way merge pattern deeply—from the naive approaches and why they're suboptimal, through the elegant heap solution, to real-world applications and subtle implementation details. This pattern appears constantly in systems programming and distributed computing.