Kth Smallest Element Using Min Heap in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to create a min heap using PriorityQueue.

DSA Java

PriorityQueue<Integer> minHeap = new PriorityQueue[1]();

Drag options to blanks, or click blank then click option'

A{}

B()

C<Integer>

D[]

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of angle brackets for the type.

Using curly braces or square brackets which are invalid here.

The PriorityQueue constructor requires the type parameter in angle brackets to create a min heap of integers.

2fill in blank

medium

Complete the code to add elements from the array to the min heap.

DSA Java

for (int num : nums) {
    minHeap.[1](num);
}

Drag options to blanks, or click blank then click option'

Aoffer

Bpop

Cpeek

Dpoll

Attempts:

3 left

💡 Hint

Common Mistakes

Using poll() which removes elements instead of adding.

Using peek() which only views the smallest element.

The offer() method adds an element to the PriorityQueue (min heap).

3fill in blank

hard

Fix the error in the code to extract the kth smallest element by removing elements from the min heap.

DSA Java

int kthSmallest = 0;
for (int i = 0; i < k; i++) {
    kthSmallest = minHeap.[1]();
}

Drag options to blanks, or click blank then click option'

Aremove

Boffer

Cpeek

Dpoll

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek() which does not remove elements.

Using offer() which adds elements instead of removing.

poll() removes and returns the smallest element from the min heap, which is needed to find the kth smallest.

4fill in blank

hard

Fill both blanks to complete the method that returns the kth smallest element using a min heap.

DSA Java

public int findKthSmallest(int[] nums, int k) {
    PriorityQueue<Integer> minHeap = new PriorityQueue[1]();
    for (int num : nums) {
        minHeap.[2](num);
    }
    for (int i = 0; i < k - 1; i++) {
        minHeap.poll();
    }
    return minHeap.peek();
}

Drag options to blanks, or click blank then click option'

A<Integer>

Boffer

Cadd

D()

Attempts:

3 left

💡 Hint

Common Mistakes

Omitting the type parameter in PriorityQueue.

Using add() instead of offer(), which also works but offer() is preferred here.

The PriorityQueue needs the type parameter and elements are added using offer() method.

5fill in blank

hard

Fill all three blanks to complete the method that returns the kth smallest element by removing elements from the min heap.

DSA Java

public int findKthSmallest(int[] nums, int k) {
    PriorityQueue<Integer> minHeap = new PriorityQueue[1]();
    for (int num : nums) {
        minHeap.[2](num);
    }
    int kthSmallest = 0;
    for (int i = 0; i < k; i++) {
        kthSmallest = minHeap.[3]();
    }
    return kthSmallest;
}

Drag options to blanks, or click blank then click option'

A<Integer>

Boffer

Cpoll

Dpeek

Attempts:

3 left

💡 Hint

Common Mistakes

Using peek() instead of poll() to remove elements.

Omitting the type parameter in PriorityQueue.

The PriorityQueue needs the type parameter , elements are added with offer(), and poll() removes the smallest element to find the kth smallest.

Practice

1. What does the kth smallest element mean in a list of numbers?

easy

A. The element that is the kth in ascending order

B. The largest element in the list

C. The element that appears kth from the end

D. The element with the kth highest frequency

Solution

1. Step 1: Understand the meaning of kth smallest

   The kth smallest element is the element that would appear at position k if the list was sorted in ascending order.

2. Step 2: Compare with other options

   Options B, C, and D describe different concepts: largest element, kth from end, and frequency-based element, which are not the kth smallest.

3. Final Answer:

   The element that is the kth in ascending order -> Option A

4. Quick Check:

   kth smallest = kth element in sorted list [OK]

Hint: Sort the list; kth smallest is element at index k-1 [OK]

Common Mistakes:

• Confusing kth smallest with kth largest
• Thinking kth smallest means kth from the end
• Mixing element frequency with position

2. Which Java code snippet correctly creates a min heap using PriorityQueue to find the kth smallest element?

easy

A. PriorityQueue<Integer> minHeap = new PriorityQueue<>(Comparator.reverseOrder());

B. PriorityQueue<Integer> minHeap = new PriorityQueue<>(Collections.reverseOrder());

C. PriorityQueue<Integer> minHeap = new PriorityQueue<>(Collections.reverseOrder(), k);

D. PriorityQueue<Integer> minHeap = new PriorityQueue<>();

Solution

1. Step 1: Identify min heap creation syntax

   In Java, new PriorityQueue<>() creates a min heap by default.

2. Step 2: Analyze other options

   Options B, C, and D create max heaps using reverse order comparators.

3. Final Answer:

   PriorityQueue<Integer> minHeap = new PriorityQueue<>(); -> Option D

4. Quick Check:

   Default PriorityQueue = min heap [OK]

Hint: Use PriorityQueue without comparator for min heap [OK]

Common Mistakes:

• Using reverseOrder() creates max heap, not min heap
• Confusing initial capacity with heap type
• Not importing java.util.PriorityQueue

3. What is the output of the following Java code snippet?

int[] arr = {7, 10, 4, 3, 20, 15};
int k = 3;
PriorityQueue<Integer> minHeap = new PriorityQueue<>();
for (int num : arr) {
    minHeap.add(num);
}
int result = 0;
for (int i = 0; i < k; i++) {
    result = minHeap.poll();
}
System.out.println(result);

medium

A. 10

B. 4

C. 7

D. 3

Solution

1. Step 1: Add all elements to min heap

   Elements added: 7, 10, 4, 3, 20, 15. Min heap orders them internally as 3,4,7,10,15,20.

2. Step 2: Poll k=3 times from min heap

   First poll: 3, second poll: 4, third poll: 7. The variable result holds the last polled value, 7.

3. Final Answer:

   7 -> Option C

4. Quick Check:

   3rd smallest element = 7 [OK]

Hint: Poll k times from min heap; last polled is kth smallest [OK]

Common Mistakes:

• Returning first polled element instead of kth
• Confusing min heap order with insertion order
• Off-by-one error in loop count

4. Identify the error in this Java code to find kth smallest element using min heap:

int kthSmallest(int[] arr, int k) {
    PriorityQueue<Integer> minHeap = new PriorityQueue<>();
    for (int i = 0; i <= arr.length; i++) {
        minHeap.add(arr[i]);
    }
    int result = 0;
    for (int i = 0; i < k; i++) {
        result = minHeap.poll();
    }
    return result;
}

medium

A. Loop condition should be i < arr.length, not i <= arr.length

B. PriorityQueue should be max heap, not min heap

C. Poll loop should run k+1 times

D. Return statement should return minHeap.peek()

Solution

1. Step 1: Check array loop bounds

   The for loop uses i <= arr.length, which causes ArrayIndexOutOfBoundsException because valid indices are 0 to arr.length-1.

2. Step 2: Verify other parts

   Min heap is correct for kth smallest. Poll loop runs k times correctly. Return statement returns last polled element, which is correct.

3. Final Answer:

   Loop condition should be i < arr.length, not i <= arr.length -> Option A

4. Quick Check:

   Array index out of bounds due to loop condition [OK]

Hint: Use i < arr.length to avoid index errors [OK]

Common Mistakes:

• Using <= instead of < in array loops
• Changing heap type unnecessarily
• Polling wrong number of times

5. Given a large array with millions of elements, which approach is most efficient to find the kth smallest element?

hard

A. Add all elements to a min heap and poll k times

B. Add first k elements to a max heap, then for each remaining element, replace root if smaller

C. Sort the entire array and pick element at index k-1

D. Use a hash map to count frequencies and pick kth smallest key

Solution

1. Step 1: Understand efficiency of adding all elements to min heap

   Adding all millions of elements to a min heap and polling k times is O(n + k log n), which is costly for large n.

2. Step 2: Use max heap of size k for better efficiency

   Adding first k elements to a max heap keeps track of k smallest seen so far. For each new element, if smaller than max heap root, replace root. This is O(n log k), much better for large n.

3. Step 3: Compare other options

   Sorting entire array is O(n log n), slower than max heap approach. Hash map frequency counting is not suitable for kth smallest element.

4. Final Answer:

   Add first k elements to a max heap, then for each remaining element, replace root if smaller -> Option B

5. Quick Check:

   Max heap of size k = efficient kth smallest [OK]

Hint: Use max heap of size k for large arrays [OK]

Common Mistakes:

• Using min heap with all elements for large data
• Sorting entire array unnecessarily
• Confusing frequency counting with order statistics