Master Heaps for Coding Interviews: Top LeetCode Problems Explained with Optimized Solutions
Master Heaps for Coding Interviews: Top LeetCode Problems Explained with Optimized Solutions
⏱️ Estimated
reading time: 18 minutes
In the landscape of data
structures, heaps are very powerful but quietly operating within the wish to
maintain perfect order and focus on prioritizing what is important. A heap is a
type of tree structure that is designed to ensure that the most important item
in the heap will always be at the root (top of the tree) either by having the
minimum value (in the case of a Min Heap) or maximum value (in the case of a
Max Heap). This assurance allows developers to use heaps to get the best
candidate as needed very quickly and efficiently. The `heapq` module of Python
allows developers to efficiently manage priorities by utilizing heaps without
having to construct tree structures manually.
There are many real-life applications for heaps because many problems faced by organizations require the ability to select the most important items associated with that organization and not necessarily all of the items in order. The following are all examples of how heaps are used in real life: CPU task scheduling in operating systems; route optimization using Dijkstra's Algorithm; real-time event processing; leaderboard systems on gaming platforms; streaming median calculations in analysis dashboards; top K queries in search engines, etc. When an organization needs to continuously determine the highest or lowest-priority items from a dataset that is always changing, a heap will typically be used behind the scenes for making those decisions as efficiently and quickly as possible.
🟢 EASY (Foundational
+ Pattern Building)
1.
Kth Largest Element in a Stream
🧠 Explanation
Many people tend to make
the mistake of sorting all scores each time they are added when they are
solving this problem, but while that method is sufficient for small data sets,
it won't work as efficiently with larger data sets. Rather than sorting the
scores repeatedly, we will maintain only the k highest scores in a Min Heap.
Why use a Min Heap?
Because the lowest
element in the Min Heap will represent the kth highest score as a whole. By
limiting the size of the Heap to k, we minimize both the amount of storage
required for our data and the cost of performing a sort operation. Thus, the
solution allows for efficient real-time streaming of data from the source.
⚙️ Logic in Steps
1. Use a Min Heap to
maintain the k largest values of the stream.
2. During initialization
of the heap, pass in the entire array and call `heapify()` to convert it into a
heap.
3. If the size of the
heap exceeds k, continue to pop the smallest element until the size is k. This
will guarantee that only the k largest values are stored in the heap.
4. Each time a new number
is added to the heap, push it into the heap.
5. If the size of the
heap exceeds k after the new number has been inserted, pop the smallest number
from the heap.
6. Thus, ‘heap[0]’ will
always contain the kth largest element.
📝 Key Takeaways from This
Problem and Solution
1.
Improving Efficiency by Keeping Only The
Most Important Items/Components.
2.
We Can Use The Min Heap To Help Us Keep
Track Of The Kth Largest Element.
3.
Do Not Continuously Re-Sort The Entire
Dataset.
4.
Using The Fixed-Size Heap Is A Very Common
Pattern In Streaming Applications.
5.
The MinHeap Approach Has Been Widely Used
As A Way To Rank/Track Things In Real-Time, Ranking Sites, Leaderboards, And
Cut-Off Rank Systems.
🧠 Explanation
One of the most common
mistakes made in solving this problem is to continually sort the array after
performing a smash operation. While sorting the array over and over again works
when you have small amounts of data, it is an inefficient approach for data of
larger quantities. Instead of having to sort the entire array each time, you
can utilize a max heap to give yourself quick access to the two largest
(heaviest) stones in your array.
Because python only
offers you a min heap, we can simulate a max heap by inserting negative numbers
into the min heap so that when we extract the element from the min heap, it
will be the max value (negative value).By doing this, the max heap maintains
that we can pick the two largest stones, smash them together, and then insert
the remaining weight (if needed) in log(n) time. Therefore, at every point in
time, we can quickly identify and smash the two largest stones in our array
using a max heap.
⚙️ Logic in Steps
1.
Convert all weights of stones into
negative value, to simulate Max Heap.
2.
Use heapify() to create the heap in O(n)
time.
3.
Continue looping until there is more than
one stone left in the heap:
a. Remove
the heaviest stone and convert to a positive value.
b. Remove
the next heaviest stone, converting to a positive value.
c. If
the stones are not equal push the difference of the two stones back into the
heap, as a negative value.
4.
Repeat this process until there are at
most one stone left.
5.
Return the value of the remaining stone as
a positive value if one stone remains.
6.
If there are no stones left, return 0.
⏱ Time and Space Complexity
1.
Time Complexity
O(n
log n): Each smash involves two pop operations and possibly one push, each
taking O(log n). In the worst case, this happens n times.
2.
Space Complexity
O(n):
The heap stores at most n elements.
📝 Key Takeaways from
This Problem and Solution
1.
If you need to access the largest value,
utilise a Max Heap data structure.
2.
You can create a Max Heap from Python’s
built-in Min Heap by converting all values into their negative counterparts.
3.
When you need to repeatedly remove an item
by its priority level, do not sort the data each time.
4.
In general, problems involving greedy
selection can be solved more efficiently with Heaps compared to other types of
data structures.
5.
This pattern exists in virtually all
scheduling applications, resource management and competitive simulation
systems.
6.
This illustrates how heaps can be used to
efficiently simulate repeated "highest priority removal" operations.
🧠
Explanation
Sorting an array multiple times to figure out which
value is the next largest is a common error when solving a ranking problem.
Although it is true that sorting can be done many times, as the size of the
input grows, it will become less and less efficient. Rather than having to sort
again to find the next highest score every time, we use what's called a Max
Heap. The properties of a Max Heap make it possible to extract the highest
value from it in O(log n) time.
Python provides Min Heaps via the heapq package, but
there is no built-in way to create a Max Heap. Instead, we simulating a Max
Heap with negative values so that the smallest element (i.e. least negative
value) corresponds to the largest original value that we are ranking. By
stashing both of these items in the Max Heap:
1. Negative
value of the score
2. Original
index of the score
We are able to rank the scores and place medals
directly into their respective spots within a given result list. In summary:
1. We
can immediately access the next highest score at all times.
2. We
can assign ranks while making only a single pass through the scores.
3. We
can avoid unnecessary sorting.
⚙️
Logic in Steps
1.
To simulate a Max Heap using Python's Min
Heap, run `(-score, index)` conversion on each score (tuple).
2.
By doing this, you can build a Heap in
O(n) time with `heapify()`.
3.
Create result array of size (n).
4.
Create a rank counter that starts from 1.
5.
Repeat the following until the heap has
been drained of elements:
6.
Remove (pop) the maximum score (the
minimum of the negative scores) from the Heap.
7.
Use the removed element's original index
to record its current rank value in the result array.
8.
If the rank is 1, set the corresponding
result array element to "Gold Medal".
9.
If the rank is 2; set the corresponding
result array element to "Silver Medal".
10. If
the rank is 3; set the corresponding result array element to "Bronze
Medal".
11. Otherwise
set the rank's value (conversion of rank to a string) in the corresponding
result array index.
12. Increment
the rank counter.
13. When
all of the ranks have been processed, return the result array.
⏱
Time and Space Complexity
1. Time
Complexity O(n log n):
a.
Heapify takes O(n)
b.
Each of the n pops takes O(log n)
2. Space
Complexity O(n):
a.
Heap stores n elements
b.
Result array stores n elements
📝
Key Takeaways from This Problem
1. Use
a max heap if you need to obtain the maximum value from your data multiple
times.
2. Negative
values are used to create a max heap from python's min heap.
3. Prioritised
removal of elements does not require sorting of your data multiple times.
4. Store
the indices into the heap instead of the values if the original location of the
value is important.
5. Heaps
are very useful for ranking, scheduling, creating priority queues and greedy
type problems.
6. This
pattern can be found in leaderboards, scheduling tasks and scoring engine for
competition.
4. Minimum Cost of Connecting Sticks
🧠
Explanation
In order to find the maximum pile after every second
in this problem, the most prevalent mistake is to sort the array over and over
again. Even though sorting can be done quickly for small datasets, because
sorting will be required to be completed O(n log n) every time, this will
become exponentially slow as we increase the volume of data. Rather than
sorting the array and performing a search for the maximum pile at the end of
each second, we can maintain a max heap (binary tree) which will allow for
O(log n) time complexity when removing the largest pile.
Python does not have a max heap implementation;
therefore, we will use a min heap and enter negative values into the heap to
simulate a max heap. By inserting into a min heap using negative values, the
largest original value will be at the top of the heap (i.e., the smallest
negative value). So, our operation to maintain the largest available pile every
second is as follows:
1. Remove
the largest pile from the heap, and replace it with the floor of the square
root of the largest pile.
2. Push
the newly created pile back into the heap.
⚙️
Logic in Steps
1. Make
all the values inside the gift piles negative, this will create a Max Heap of
all gift piles.
2. To
create the heap, use heapify() to turn the entire list of negative gift piles
into a Max Heap in O(n) time.
3. Then
you will repeat the following process 'k' amount of times:
a. Remove
the largest pile (smallest negative gift pile) from the Max Heap and convert
the value back to positive.
b. Compute
floor(sqrt(value)).
c. Push
the negative of the new value into the Max Heap.
4. Repeat
this verse for 'k' total iterations.
5. After
completing 'k' iterations, sum all the remaining values in the Max Heap.
6. Finally,
take the total sum of all the remaining values in the Max Heap and return the
negative of it (because they were originally stored as negative values to begin
with).
⏱
Time and Space Complexity
1. Time
Complexity O(n + k log n):
1.
Heapify takes O(n)
2.
Each of the k operations involves one pop
and one push, each taking O(log n)
2. Space Complexity O(n):
The heap stores at most n elements.
📝
Key Takeaways from This Problem
1. To
access the largest item multiple times, you would want to consider a max heap
data structure.
2. If
you are working in Python and you have a min-heap, you can simulate a max-heap
by simply inserting the negative of each item that you want to store within the
heap.
3. To
reduce repeated sorting on priority based selection use a heap.
4. Greedy
problems that involve selecting a maximum item multiple times can use a heap.
5. The
use of that pattern is very common for scheduling systems, priority queues, and
resource optimization type problems.
🟡 MEDIUM
(Industry-Relevant, Interview Favourite)
5. Kth Largest Element in an Array
🧠
Explanation
An efficient solution to
the Kth largest element issue is to use the heap libraries built in to the
Python programming language. The first step in this process is to use the
heapify() function to convert a given list to a heap data structure. A min-heap
is the type of heap that will be created when you do this. A min-heap will
always contain its minimum value at the "root" or first node. The
second step is to use the nlargest(k, nums) built in function to retrieve the k
largest items from the list. This built in function allows for efficient
retrieval of k largest items from the min-heap without the need for performing
a full sort of the entire list.
The return value from
nlargest(k, nums) will be a list of the k largest items from the original list
sorted in descending order (k largest value at index 0). Therefore, since the
returned value from nlargest(k, nums) is sorted in reverse order from largest
to smallest, if you want the k-th largest item from the input array, you can
simply return "largest[k - 1]".
⚙️
Logic in Steps
1.
Use the heapify() method to transform the
input array into Min Heap format at O(n) time.
2.
Utilize the builtin module
heapq.nlargest(k, nums) to return the k largest elements from an array.
3.
Make a list with these k elements.
4.
Because the list is already sorted from
largest to smallest, use the kth element from the list (the kth largest
element) at index k-1.
⏱
Time and Space Complexity
1. Time
Complexity O(n log k):
a. heapify()
takes O(n)
b. nlargest(k,
nums) takes O(n log k)
c. Overall
complexity is O(n log k)
2. Space
Complexity O(k):
nlargest() stores k
largest elements in a separate list.
📝
Key Takeaways from This Problem
1. Heap
utilities can be applied whenever you are looking for the top k elements, as
opposed to doing a complete sort of a dataset.
2. When
k is much smaller than n, heapq.nlargest() will be more efficient than sorting.
3. Even
for largest element problems, using Min Heaps with helper functions may lead to
a somewhat more optimized solution.
4. In
cases where only a partial ordering is required, avoid unnecessary complete
ordering (i.e., sort()), and simply use heap-based selection.
5. Heap-based
selection has been a frequent optimization method in some interview type
questions.
6. Top
K Frequent Elements
🧠 Explanation
The task at hand is to
retrieve the k most commonly occurring items in an array of integers. In
practice, one common way to solve this is through sorting by frequency; this
works fine as an approach to the problem but is more than we need given that we
only require the top k most frequently occurring items (and thus we can
accomplish our goal in less than O(n log n)).
1.
Instead, what we will do is combine:
a. A
map of frequencies - count each integer's frequency.
b. A
min heap containing k items, this will allow us to keep track of the items with
the largest k frequency counts.
2.
The overall steps of the algorithm are
simple:
a.
Create your frequency counter.
b.
Create your min heap which stores only the
top k integer frequency counts.
c.
For every integer frequency in the
counter, if the size of the heap is greater than k - remove the smallest item
(min heap property enforces this).
By following the above
procedure, when you are finished, your heap will contain the k items with the
highest frequency counts.
⚙️
Logic in Steps
1.
First, utilize the built-in Counter
function to construct a dictionary with the frequency of each number in
"nums".
2.
Create an empty Minimum Heap using
"heapq".
3.
Loop through each of the (num, count)
pairs in the frequency dictionary, inserting the tuple (count, num) into the
Minimum Heap.
4.
If the size of the Minimum Heap is greater
than k, pop the smallest element from it using the "heappop" method.
5.
After looping through all of the input,
the Minimum Heap will contain only the k most frequently appearing elements.
6.
Finally, extract and return the numbers
from the Minimum Heap.
⏱
Time and Space Complexity
1. Time
Complexity O(n log k):
a. Counting
frequencies → O(n)
b. Heap
operations for each unique element → O(log k)
2. Space
Complexity O(n):
a. Frequency
dictionary can store up to n unique elements.
b. Heap
stores at most k elements.
7. K
Closest Points to Origin
🧠 Explanation
To retrieve the k nearest
points to the origin, use a k-sized Max Heap solution. Instead of sorting all
points (takes O(n log n)), we will use a Max Heap to store only the k least
distant points we've seen thus far. The main idea is:
1.
Compute the squared distance using the
formula: x² + y²
2.
Store this squared distance as a negative
number (to provide us with a Max Heap functionality)
3.
Restrict the size of the Max Heap to k
points
Any time we try to add a
point resulting in a total size of the heap exceeding k, we will remove the
point with the greatest squared distance. This provides us with an efficient
way to maintain only the k nearest points to the origin.
⚙️ Logic in Steps
1.
You should start by creating an empty
heap.
2.
For every point, [x,y] calculate the
square of the distance by using the formula X^2 + Y^2.
3.
To simulate a Max heap, you will push the
negative of the distance along with the point to into the heap.
4.
Once the heap becomes larger than k you'll
need to pull the top (which is farthest from the origin) out of the heap.
5.
At the end of processing all your points,
you can pull out and return the points still remaining in the heap.
⏱ Time and Space Complexity
1.
Time Complexity O(n log k):
a.
For each of n points, we perform a push
operation (log k).
b.
If size exceeds k, we perform one pop (log
k).
2.
Space Complexity O(k):
Heap
stores at most k elements.
📝 Key Takeaways
1.
When comparing distances mathematically we
can compute by using square distances rather than calculating both square or
distance values separately. Using heap of fixed size k, , if we want only k
more than that.
2.
If using max heap to simulate, for example
we need to use negative values stored in an array as keys in order simulate a
max heap using min heap functions in Python.
3.
To keep memory usage optimal remove all
elements from heap when total heap size exceeds k.
4.
This technique is often found in nearest
neighbor or top-k selection algorithm implementations.
8. Top K Frequent Words
🧠 Explanation
This program takes a list
of words and finds the k most frequently used. To do this, it first counts the
word frequencies with a Counter. It then sorts the list of unique words using
two rules:
1. The
most frequently used words first.
2. If
two words are used the same number of times, order them alphabetically by their
value.
To accomplish this
sorting, the program will use the sorting key:
key=lambda w: (-count[w],
w)
This key will:
1.
Use -count[w] to ensure that counts are
sorted in descending order.
2.
Use w to ensure alphabetical ordering for
multiple instances of the same count.
Finally, the program will
return only the first k values from the sorted list.
⚙️ Logic in Steps
1.
Apply the Counter(words) function to
identify how many times each word appears in the list.
2.
Use count.keys() to obtain a list
containing each distinct word in the list.
3.
Sort the list of Distinct Words by
(1)`Negative Frequency (Largest to Smallest)` and (2) alphabetically.
4.
Return the first k entries from the sorted
list.
⏱ Time and Space Complexity
1.
Time Complexity O(n log n):
a.
Counting frequencies → O(n)
b.
Sorting unique words → O(m log m)
1.
where m = number of unique words
2.
worst case m = n
2. Space Complexity O(n):
a.
Frequency dictionary stores up to n unique
words.
b.
Sorted list stores up to n words.
📝 Key Takeaways
1.
Leverage Counter for fast & efficient
frequency counting.
2.
Sort based on multiple conditions using
tuple keys.
3.
Negative number sorts in reverse order.
4.
Lexicographic sort will sort automatically
and resolve ties.
5.
Simple sorts do not compare as well
against heap sort where k is much smaller than n.
Conclusion
These eight challenges
would sneakily relay to you the importance of heaps when it comes to priority. From
crushing rocks to ranking athletes, from recording streams, to finding nearest
points; the cycle continues to repeat itself. Don't sort all of the data you
receive. Keep your structure smart and retrieve what is essential quickly! By
mastering these patterns, you will not only be able to answer heap-related
problems, but also develop a gut feeling for Top K, Greedily Choosing and
Priority-focused Problems. Heaps can be utilized as not only a data structure,
but also as an engine of decision-making.
If you missed the
Previous Part
If you missed the
previous part on Master Queue & Deque Problems in Python, make sure to check it out to strengthen
your foundation before diving deeper into heap patterns.
Those concepts quietly power many of the problems discussed here.
Coming Up
Next, we dive into Strings
& String Matching, the art of decoding patterns, taming text, and turning
raw characters into powerful logic. Get ready to master the algorithms that
quietly power search engines, validations, and real-world text processing.
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