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DSA Goprogramming~10 mins

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

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to create a min heap from the given array.

DSA Go
h := &IntHeap[1]
heap.Init(h)
Drag options to blanks, or click blank then click option'
A[]int{10, 20, 30}
B[]int{}
C[]int{3, 1, 4, 1, 5}
D[]int{5, 4, 3, 2, 1}
Attempts:
3 left
💡 Hint
Common Mistakes
Using an empty slice will create an empty heap.
Using a sorted slice descending will not create a min heap automatically.
2fill in blank
medium

Complete the code to pop the smallest element from the min heap.

DSA Go
smallest := heap.[1](h).(int)
Drag options to blanks, or click blank then click option'
ARemove
BPop
CPush
DFix
Attempts:
3 left
💡 Hint
Common Mistakes
Using Push instead of Pop will add an element instead of removing.
Remove requires an index, not used here.
3fill in blank
hard

Fix the error in the loop that pops k-1 smallest elements from the heap.

DSA Go
for i := 0; i < [1]-1; i++ {
    heap.Pop(h)
}
Drag options to blanks, or click blank then click option'
Ak
Blen(h)
Ccap(h)
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Using len(h) will pop too many or too few elements.
Using 0 will not run the loop at all.
4fill in blank
hard

Fill both blanks to define the IntHeap type and implement the Len method.

DSA Go
type IntHeap []int

func (h IntHeap) [1]() int {
    return len([2])
}
Drag options to blanks, or click blank then click option'
ALen
Bh
CPush
DPop
Attempts:
3 left
💡 Hint
Common Mistakes
Using Push or Pop as method name is incorrect here.
Using a different receiver name than h will cause errors.
5fill in blank
hard

Fill all three blanks to implement the Less method for the min heap.

DSA Go
func (h IntHeap) [1](i, j int) bool {
    return h[[2]] [3] h[j]
}
Drag options to blanks, or click blank then click option'
ALess
Bi
C<
Dj
Attempts:
3 left
💡 Hint
Common Mistakes
Using > instead of < will create a max heap.
Swapping i and j will invert the comparison.

Practice

(1/5)
1. What does a min heap help you find efficiently when looking for the kth smallest element in a list?
easy
A. The smallest elements in order
B. The largest elements in order
C. The middle element only
D. The sum of all elements

Solution

  1. Step 1: Understand min heap property

    A min heap always keeps the smallest element at the top (root).
  2. Step 2: Use min heap to find kth smallest

    By extracting the root k times, you get the kth smallest element.
  3. Final Answer:

    The smallest elements in order -> Option A
  4. Quick Check:

    Min heap = smallest elements first [OK]
Hint: Min heap root is always the smallest element [OK]
Common Mistakes:
  • Confusing min heap with max heap
  • Thinking min heap gives largest elements first
  • Assuming kth smallest is the middle element
2. Which of the following Go code snippets correctly creates a min heap from a slice of integers nums using container/heap package?
easy
A. heap.Init(IntHeap(&nums))
B. heap.Init(nums)
C. heap.Init(&nums)
D. heap.Init(&IntHeap(nums))

Solution

  1. Step 1: Understand heap.Init usage

    heap.Init requires a pointer to a type implementing heap.Interface.
  2. Step 2: Convert slice to IntHeap and pass pointer

    We must convert nums to IntHeap and pass its pointer: &IntHeap(nums).
  3. Final Answer:

    heap.Init(&IntHeap(nums)) -> Option D
  4. Quick Check:

    heap.Init needs pointer to heap.Interface [OK]
Hint: Pass pointer to heap.Interface type for heap.Init [OK]
Common Mistakes:
  • Passing slice directly without conversion
  • Passing value instead of pointer
  • Using wrong type conversion syntax
3. What is the output of the following Go code that finds the 3rd smallest element using a min heap?
package main

import (
  "container/heap"
  "fmt"
)

type IntHeap []int

func (h IntHeap) Len() int           { return len(h) }
func (h IntHeap) Less(i, j int) bool { return h[i] < h[j] }
func (h IntHeap) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }

func (h *IntHeap) Push(x interface{}) {
  *h = append(*h, x.(int))
}

func (h *IntHeap) Pop() interface{} {
  old := *h
  n := len(old)
  x := old[n-1]
  *h = old[0 : n-1]
  return x
}

func main() {
  nums := []int{7, 10, 4, 3, 20, 15}
  h := IntHeap(nums)
  heap.Init(&h)
  k := 3
  var kthSmallest int
  for i := 0; i < k; i++ {
    kthSmallest = heap.Pop(&h).(int)
  }
  fmt.Println(kthSmallest)
}
medium
A. 10
B. 7
C. 4
D. 3

Solution

  1. Step 1: Build min heap from nums

    Heap after init: [3, 7, 4, 10, 20, 15] (smallest at root)
  2. Step 2: Pop 3 times to get 3rd smallest

    Pop 1: 3, Pop 2: 4, Pop 3: 7
  3. Final Answer:

    7 -> Option B
  4. Quick Check:

    3rd smallest in [7,10,4,3,20,15] is 7 [OK]
Hint: Pop k times from min heap to get kth smallest [OK]
Common Mistakes:
  • Not popping k times
  • Confusing min heap with max heap
  • Incorrect heap initialization
4. Identify the error in the following Go code snippet that tries to find the 2nd smallest element using a min heap:
nums := []int{5, 3, 8, 1}
h := IntHeap(nums)
heap.Init(h)
k := 2
for i := 0; i < k; i++ {
  val := heap.Pop(&h).(int)
  if i == k-1 {
    fmt.Println(val)
  }
}
medium
A. heap.Init should receive &h, not h
B. Pop should be called with h, not &h
C. k should be 3 to get 2nd smallest
D. nums slice should be sorted before heap.Init

Solution

  1. Step 1: Check heap.Init argument

    heap.Init requires a pointer to the heap, so it should be heap.Init(&h).
  2. Step 2: Confirm other calls are correct

    heap.Pop correctly uses &h, and k=2 is correct for 2nd smallest.
  3. Final Answer:

    heap.Init should receive &h, not h -> Option A
  4. Quick Check:

    heap.Init needs pointer to heap [OK]
Hint: Always pass pointer to heap in heap.Init [OK]
Common Mistakes:
  • Passing value instead of pointer to heap.Init
  • Confusing Pop argument
  • Incorrect k value for kth smallest
5. Given a large unsorted array of integers, which approach is most efficient to find the 5th smallest element using heaps in Go? A) Build a min heap of all elements and pop 5 times. B) Build a max heap of size 5 and iterate through array. C) Sort the array and pick the 5th element. D) Use a min heap but pop all elements until the 5th smallest is found.
hard
A. Build a min heap of all elements and pop 5 times.
B. Sort the array and pick the 5th element.
C. Build a max heap of size 5 and iterate through array.
D. Use a min heap but pop all elements until the 5th smallest is found.

Solution

  1. Step 1: Understand efficiency of building full min heap

    Building a min heap of all elements and popping 5 times is O(n + k log n), which can be costly for large n.
  2. Step 2: Use max heap of size k for better efficiency

    Maintaining a max heap of size 5 while iterating keeps track of 5 smallest elements efficiently in O(n log k).
  3. Step 3: Compare other options

    Sorting is O(n log n), popping all elements is inefficient, so max heap size k is best.
  4. Final Answer:

    Build a max heap of size 5 and iterate through array. -> Option C
  5. Quick Check:

    Max heap size k = efficient kth smallest [OK]
Hint: Use max heap of size k for kth smallest in large arrays [OK]
Common Mistakes:
  • Building full min heap for large arrays
  • Sorting entire array unnecessarily
  • Popping all elements from min heap
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10 questions~5 mineasy hard