Kth Smallest Element Using Min Heap in DSA Typescript - Time & Space Complexity

Time Complexity: Kth Smallest Element Using Min Heap

O(n log n + k log n)

Understanding Time Complexity

We want to understand how the time needed to find the kth smallest element using a min heap changes as the input size grows.

Specifically, how does the number of steps grow when we add more numbers or change k?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function kthSmallest(arr: number[], k: number): number {
  const minHeap = new MinHeap();
  for (const num of arr) {
    minHeap.insert(num);
  }
  let result = -1;
  for (let i = 0; i < k; i++) {
    result = minHeap.extractMin();
  }
  return result;
}
    

This code builds a min heap from the array, then extracts the smallest element k times to find the kth smallest.

Identify Repeating Operations

• Primary operation: Inserting all n elements into the min heap and extracting the minimum k times.
• How many times: Insert runs n times; extractMin runs k times.

How Execution Grows With Input

As the input size n grows, building the heap takes more steps, and extracting k times adds more steps depending on k.

Input Size (n)	Approx. Operations
10, k=3	About 10 insertions + 3 extractions
100, k=10	About 100 insertions + 10 extractions
1000, k=100	About 1000 insertions + 100 extractions

Pattern observation: The total steps grow roughly with n plus k, but each insertion and extraction takes a small extra time that grows with the heap size.

Final Time Complexity

Time Complexity: O(n log n + k log n)

This means the time grows mostly by inserting all n elements into the heap and then extracting k times, each operation taking time related to the heap size.

Common Mistake

[X] Wrong: "Building the heap is just O(n) so the whole process is O(n + k)."

[OK] Correct: While heap construction can be done in O(n), each insertion here is done one by one, costing O(log n) each, so total insertion is O(n log n). Also, each extraction costs O(log n), so k extractions add up.

Interview Connect

Understanding this time complexity helps you explain why heaps are useful for selection problems and shows your grasp of how data structures affect performance.

Self-Check

"What if we used a max heap of size k instead of a min heap of size n? How would the time complexity change?"

Practice

1. What is the main idea behind finding the kth smallest element using a min heap?

easy

A. Use a hash map to count frequencies and find the kth smallest

B. Build a min heap from the array and extract the minimum element k times

C. Build a max heap and extract the maximum element k times

D. Sort the array and pick the element at index k

Solution

1. Step 1: Understand min heap property

   A min heap always keeps the smallest element at the root, so extracting the root gives the smallest element.

2. Step 2: Extract minimum k times

   By extracting the root k times, we get the kth smallest element.

3. Final Answer:

   Build a min heap and extract minimum k times -> Option B

4. Quick Check:

   Min heap extraction = kth smallest element [OK]

Hint: Extract min from min heap k times to get kth smallest [OK]

Common Mistakes:

• Using max heap instead of min heap
• Sorting array instead of using heap
• Extracting less or more than k times

2. Which of the following TypeScript code snippets correctly creates a min heap using a priority queue for numbers?

easy

A. const minHeap = new PriorityQueue<number>((a, b) => a - b);

B. const minHeap = new MinPriorityQueue<number>();

C. const minHeap = new PriorityQueue<number>((a, b) => b - a);

D. const minHeap = new MaxPriorityQueue<number>();

Solution

1. Step 1: Understand comparator for min heap

   A min heap comparator returns negative if first argument is smaller, so (a, b) => a - b is correct.

2. Step 2: Check code options

   const minHeap = new PriorityQueue<number>((a, b) => a - b); uses (a, b) => a - b, which creates a min heap. const minHeap = new PriorityQueue<number>((a, b) => b - a); reverses order, making max heap.

3. Final Answer:

   const minHeap = new PriorityQueue<number>((a, b) => a - b); -> Option A

4. Quick Check:

   Comparator a-b creates min heap [OK]

Hint: Use comparator (a, b) => a - b for min heap [OK]

Common Mistakes:

• Using max heap comparator for min heap
• Using wrong class name for min heap
• Confusing comparator logic

3. What will be the output of the following TypeScript code snippet?

const arr = [7, 10, 4, 3, 20, 15];
const k = 3;

// Build min heap
const minHeap = new MinPriorityQueue<number>();
arr.forEach(num => minHeap.enqueue(num));

let kthSmallest = -1;
for (let i = 0; i < k; i++) {
  kthSmallest = minHeap.dequeue().element;
}
console.log(kthSmallest);

medium

A. 7

B. 10

C. 3

D. 4

Solution

1. Step 1: Insert all elements into min heap

   Min heap will organize elements so smallest is at root: 3, 4, 7, 10, 15, 20.

2. Step 2: Extract min element k=3 times

   Extract 1: 3, Extract 2: 4, Extract 3: 7. So kth smallest is 7.

3. Final Answer:

   7 -> Option A

4. Quick Check:

   3rd smallest in array = 7 [OK]

Hint: Extract min k times from min heap to get kth smallest [OK]

Common Mistakes:

• Confusing kth smallest with kth largest
• Off-by-one error in loop
• Not building min heap correctly

4. Identify the bug in the following TypeScript code that tries to find the kth smallest element using a min heap:

function kthSmallest(arr: number[], k: number): number {
  const minHeap = new MinPriorityQueue<number>();
  for (let i = 0; i <= arr.length; i++) {
    minHeap.enqueue(arr[i]);
  }
  let result = -1;
  for (let j = 0; j < k; j++) {
    result = minHeap.dequeue().element;
  }
  return result;
}

medium

A. The dequeue loop should run k-1 times

B. MinPriorityQueue should be MaxPriorityQueue

C. Loop condition should be i < arr.length to avoid undefined enqueue

D. Return statement should be inside the loop

Solution

1. Step 1: Check array loop bounds

   The loop uses i <= arr.length, which accesses arr[arr.length] undefined, causing error.

2. Step 2: Correct loop condition

   Change to i < arr.length to avoid out-of-bounds access.

3. Final Answer:

   Loop condition should be i < arr.length to avoid undefined enqueue -> Option C

4. Quick Check:

   Array index out of bounds causes error [OK]

Hint: Use i < arr.length to avoid out-of-bounds errors [OK]

Common Mistakes:

• Using <= instead of < in loops
• Confusing min heap with max heap
• Incorrect dequeue loop count

5. Given a large array with many duplicate elements, which approach using a min heap is best to efficiently find the kth smallest unique element?

hard

A. Build a min heap from the array and extract k times ignoring duplicates during extraction

B. Sort the array and pick the kth element directly

C. Build a max heap of size k to track unique smallest elements

D. Use a hash set to filter unique elements first, then build a min heap

Solution

1. Step 1: Remove duplicates using a hash set

   Duplicates can cause incorrect counting of unique elements, so first filter unique elements.

2. Step 2: Build min heap from unique elements and extract kth smallest

   After filtering, build min heap and extract k times to get kth smallest unique element.

3. Final Answer:

   Use a hash set to filter unique elements first, then build a min heap -> Option D

4. Quick Check:

   Filter duplicates before heap for unique kth smallest [OK]

Hint: Filter duplicates first, then use min heap for kth unique smallest [OK]

Common Mistakes:

• Ignoring duplicates causing wrong kth element
• Using max heap incorrectly
• Extracting without filtering unique elements