Heap Concept Structure and Properties in DSA Javascript - Time & Space Complexity

Time Complexity: Heap Concept Structure and Properties

O(log n)

Understanding Time Complexity

We want to understand how the time to work with a heap changes as the heap grows.

Specifically, how fast can we add or remove items in a heap as it gets bigger?

Scenario Under Consideration

Analyze the time complexity of inserting an element into a heap.

function insert(heap, value) {
  heap.push(value);
  let index = heap.length - 1;
  while (index > 0) {
    let parentIndex = Math.floor((index - 1) / 2);
    if (heap[parentIndex] <= heap[index]) break;
    [heap[parentIndex], heap[index]] = [heap[index], heap[parentIndex]];
    index = parentIndex;
  }
}
    

This code adds a new value to a min-heap and moves it up to keep the heap property.

Identify Repeating Operations

Look for loops or repeated steps that happen as the heap changes.

• Primary operation: The while loop that moves the new value up the heap.
• How many times: At most, it moves up the height of the heap, which depends on the number of elements.

How Execution Grows With Input

As the heap grows, the height grows slowly compared to the number of elements.

Input Size (n)	Approx. Operations (moves up)
10	About 3 or 4
100	About 6 or 7
1000	About 9 or 10

Pattern observation: The number of steps grows slowly, roughly with the height of the heap, which is much smaller than the total elements.

Final Time Complexity

Time Complexity: O(log n)

This means adding an element takes time proportional to the heap's height, which grows slowly as the heap gets bigger.

Common Mistake

[X] Wrong: "Inserting into a heap takes the same time as scanning all elements, so it is O(n)."

[OK] Correct: The heap keeps elements in a tree shape, so we only move up the height, not through all elements.

Interview Connect

Knowing heap operations run in logarithmic time shows you understand efficient data structures, a key skill in many coding challenges.

Self-Check

"What if we changed the heap to a linked list? How would the time complexity of insertion change?"

Practice

1. What is a key property of a max heap data structure?

easy

A. Every parent node is greater than or equal to its child nodes

B. Every parent node is less than or equal to its child nodes

C. All leaf nodes are greater than the root node

D. The tree is always a complete binary tree but unordered

Solution

1. Step 1: Understand max heap property

   A max heap requires that each parent node is greater than or equal to its children.

2. Step 2: Compare options with max heap definition

   Every parent node is greater than or equal to its child nodes matches the max heap property exactly, others describe min heap or incorrect properties.

3. Final Answer:

   Every parent node is greater than or equal to its child nodes -> Option A

4. Quick Check:

   Max heap = parent ≥ children [OK]

Hint: Max heap means parent is always bigger or equal [OK]

Common Mistakes:

• Confusing max heap with min heap property
• Thinking leaf nodes must be bigger than root
• Ignoring the complete binary tree requirement

2. Which of the following is the correct way to represent a heap in JavaScript using an array?

easy

A. Parent at index i, left child at i + 1, right child at i + 2

B. Parent at index i, left child at 2*i + 1, right child at 2*i + 2

C. Parent at index i, left child at 2*i, right child at 2*i + 1

D. Parent at index i, left child at i - 1, right child at i - 2

Solution

1. Step 1: Recall array indexing for heap

   In zero-based arrays, left child index = 2*i + 1, right child index = 2*i + 2.

2. Step 2: Match options with formula

   Parent at index i, left child at 2*i + 1, right child at 2*i + 2 matches the correct formula for zero-based indexing heaps.

3. Final Answer:

   Parent at index i, left child at 2*i + 1, right child at 2*i + 2 -> Option B

4. Quick Check:

   Heap array indexing = 2*i+1, 2*i+2 [OK]

Hint: Remember zero-based indexing: left=2i+1, right=2i+2 [OK]

Common Mistakes:

• Using one-based indexing formulas incorrectly
• Swapping left and right child indices
• Using i+1 or i+2 which is wrong for heap

3. Consider the following JavaScript code that builds a max heap from an array:

const arr = [3, 9, 2, 1, 4, 5];
function maxHeapify(arr, i, n) {
  let largest = i;
  let left = 2 * i + 1;
  let right = 2 * i + 2;
  if (left < n && arr[left] > arr[largest]) largest = left;
  if (right < n && arr[right] > arr[largest]) largest = right;
  if (largest !== i) {
    [arr[i], arr[largest]] = [arr[largest], arr[i]];
    maxHeapify(arr, largest, n);
  }
}

function buildMaxHeap(arr) {
  let n = arr.length;
  for (let i = Math.floor(n / 2) - 1; i >= 0; i--) {
    maxHeapify(arr, i, n);
  }
}

buildMaxHeap(arr);
console.log(arr);

What is the output printed?

medium

A. [3, 9, 2, 1, 4, 5]

B. [9, 3, 5, 1, 4, 2]

C. [5, 4, 3, 1, 9, 2]

D. [9, 4, 5, 1, 3, 2]

Solution

1. Step 1: Understand buildMaxHeap process

   Start heapifying from last non-leaf node at index 2 (value 2), then index 1 (value 9), then index 0 (value 3).

2. Step 2: Trace heapify calls

   Initial: [3,9,2,1,4,5]
   Heapify i=2: left=5 (5>2), largest=5, swap 2-5: [3,9,5,1,4,2]. Heapify 5: no children.
   Heapify i=1: left=3(1<9), right=4(4<9), no swap.
   Heapify i=0: left=1(9>3), largest=1; right=2(5<9), swap 0-1: [9,3,5,1,4,2]. Heapify 1 (val=3): left=3(1<3), right=4(4>3), largest=4, swap 1-4: [9,4,5,1,3,2]. Heapify 4: no children.

3. Final Answer:

   [9, 4, 5, 1, 3, 2] -> Option D

4. Quick Check:

   Max heap array after build = [9,4,5,1,3,2] [OK]

Hint: Heapify from bottom non-leaf nodes up [OK]

Common Mistakes:

• Not heapifying recursively after swap
• Confusing indices for children
• Assuming original array order remains

4. The following JavaScript function is intended to insert a new value into a max heap represented as an array. Identify the bug:

function insertHeap(arr, value) {
  arr.push(value);
  let i = arr.length - 1;
  while (i > 0) {
    let parent = Math.floor(i / 2);
    if (arr[parent] < arr[i]) {
      [arr[parent], arr[i]] = [arr[i], arr[parent]];
      i = parent;
    } else {
      break;
    }
  }
}

medium

A. Parent index calculation should be Math.floor((i - 1) / 2)

B. The loop should run while i >= 0

C. The swap should happen only if arr[parent] > arr[i]

D. The function should remove the last element before inserting

Solution

1. Step 1: Check parent index formula in zero-based heap

   In zero-based arrays, parent index = Math.floor((i - 1) / 2), not Math.floor(i / 2).

2. Step 2: Understand impact of wrong parent index

   Using Math.floor(i / 2) causes incorrect parent calculation, breaking heap property restoration.

3. Final Answer:

   Parent index calculation should be Math.floor((i - 1) / 2) -> Option A

4. Quick Check:

   Parent index = (i-1)//2 in zero-based heap [OK]

Hint: Parent index in zero-based heap = (i-1)//2 [OK]

Common Mistakes:

• Using i/2 instead of (i-1)/2 for parent
• Loop condition allowing i=0 to cause errors
• Swapping when parent is larger, not smaller

5. You have a min heap represented as an array: [2, 5, 8, 10, 15, 20]. You want to insert the value 1 and maintain the min heap property. Which sequence of array states correctly shows the heap after each step of insertion and heapify up?

hard

A. [2, 5, 8, 10, 15, 20, 1] -> [2, 5, 8, 10, 15, 20, 1] -> [1, 5, 8, 10, 15, 20, 2]

B. [2, 5, 8, 10, 15, 20, 1] -> [1, 5, 8, 10, 15, 20, 2]

C. [2, 5, 8, 10, 15, 20, 1] -> [2, 5, 1, 10, 15, 20, 8] -> [1, 5, 2, 10, 15, 20, 8]

D. [2, 5, 8, 10, 15, 20, 1] -> [1, 2, 5, 10, 15, 20, 8]

Solution

1. Step 1: Insert value at end

   Insert 1 at the end: [2, 5, 8, 10, 15, 20, 1]

2. Step 2: Heapify up swaps

   Compare 1 with parent at index 2 (value 8), swap: [2, 5, 1, 10, 15, 20, 8]
   Compare 1 with new parent at index 0 (value 2), swap: [1, 5, 2, 10, 15, 20, 8]

3. Step 3: Final heap state

   Heap property restored with 1 at root.

4. Final Answer:

   [2, 5, 8, 10, 15, 20, 1] -> [2, 5, 1, 10, 15, 20, 8] -> [1, 5, 2, 10, 15, 20, 8] -> Option C

5. Quick Check:

   Heapify up swaps child with parent until order restored [OK]

Hint: Heapify up swaps child with parent until order restored [OK]

Common Mistakes:

• Not swapping with correct parent index
• Stopping heapify too early
• Incorrect final heap array after swaps