Kth Smallest Element Using Min Heap in DSA C++ - Execution Trace

Concept Flow - Kth Smallest Element Using Min Heap

Build Min Heap from array

↓

Extract min element k-1 times

↓

Heap root is kth smallest

↓

Return root value

Build a min heap from the array, then remove the smallest element k-1 times. The root after these removals is the kth smallest.

Execution Sample

DSA C++

int kthSmallest(vector<int>& arr, int k) {
  priority_queue<int, vector<int>, greater<int>> minHeap(arr.begin(), arr.end());
  for (int i = 1; i < k; i++) {
    minHeap.pop();
  }
  return minHeap.top();
}

Builds a min heap from the array and pops the smallest element k-1 times to find the kth smallest.

Execution Table

Step	Operation	Heap Size	Heap Root	Heap Visual State
1	Build min heap from [7, 10, 4, 3, 20, 15]	6	3	[3, 7, 4, 10, 20, 15]
2	Pop min element (1st pop)	5	4	[4, 7, 15, 10, 20]
3	Pop min element (2nd pop)	4	7	[7, 10, 15, 20]
4	Return heap root as 3rd smallest	4	7	[7, 10, 15, 20]

💡 After popping k-1=2 times, the root is the kth smallest element.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
minHeap Size	0	6	5	4	4
minHeap Root	N/A	3	4	7	7
k	3	3	3	3	3

Key Moments - 3 Insights

Why do we pop k-1 times instead of k times?

Because after popping k-1 smallest elements, the root of the heap is the kth smallest. See execution_table rows 2 and 3 where pops happen, and row 4 where root is returned.

Why is the heap root always the smallest element?

A min heap keeps the smallest element at the root by design. After building the heap (row 1), the root is 3, the smallest in the array.

What happens to the heap size after each pop?

Each pop removes one element, so the heap size decreases by 1 each time. See variable_tracker for minHeap Size changes after steps.

Visual Quiz - 3 Questions

Look at the execution_table, what is the heap root after the first pop?

A 3

B 7

C 4

D 10

💡 Hint

Check execution_table row 2 under 'Heap Root'

At which step does the heap size become 4?

A Step 3

B Step 2

C Step 1

D Step 4

💡 Hint

Look at variable_tracker row for minHeap Size after each step

If k was 1, what would the returned element be?

A 4

B 3

C 7

D 10

💡 Hint

If k=1, no pops happen, so root after building heap is returned (see execution_table row 1)

Concept Snapshot

Kth Smallest Element Using Min Heap:
- Build a min heap from the array
- Pop the smallest element k-1 times
- The root after these pops is the kth smallest
- Time: O(n + k log n)
- Space: O(n) for heap

Full Transcript

To find the kth smallest element using a min heap, first build a min heap from the input array. The min heap keeps the smallest element at the root. Then, pop the smallest element k-1 times from the heap. Each pop removes the current smallest element. After these removals, the root of the heap is the kth smallest element. Return this root value. The heap size decreases by one after each pop. This method efficiently finds the kth smallest element by leveraging the heap's properties.