Master Prefix Sum for Coding Interviews: Must-Know LeetCode Subarray Problems Explained
Master Prefix Sum for
Coding Interviews: Must-Know LeetCode Subarray Problems Explained
⏱️ Estimated reading time: 16 minutes
In a production environment, being efficient does not come from performing clever, fancy calculations just once; rather, it's all about avoiding having to re-compute every time you want to do something large (this is what makes prefix sum such a great example of this concept). Whenever there are large numbers of requests to perform cumulative totals over a set of data, you will see this pattern. Analytical tools can calculate daily active users or revenue trends without having to read all the data each time a request comes in. The finance sector can rely on this design to provide faster balance calculations over time periods. In data engineering pipelines, cumulative aggregation of raw events into analysis is done very efficiently by using the cumulative sum style. Cumulative busy sums of several dimensions can be accomplished quickly in image processing and telemetry-related applications for returning answers to questions based on a defined region on the screen.
Prefix Sum demonstrates
Tradeoff Engineers are always Working To Minimize – High up front information
processing cost vs. a very large number of requests made later for the same or
similar data (from O(N) to O(1)). Thus, enabling the Engineer to create a
scalable system versus a system that fails under heavy requests.Prefix Sum is a
clear illustration of the overarching Engineering Principle – To move as much
as is possible from Running-time to Precomputation without degrading system
responsiveness.
🟢 EASY (Foundational
+ Pattern Building)
I.
Range Sum Query – Immutable
🧠 Explanation
We make a prefix sum
array before we start working with the array. Each index in this array holds
the sum of the elements from the start to that index. This lets us answer any
range sum question in constant time, without having to do the sums over and
over again. Instead of doing the work again, think of it as a "memory
shortcut."
⚙️ Logic in Steps
1.
Make an array of prefixes that looks like
this:
2.
Every element keeps track of the total sum
up to that index.
3.
Make a prefix array: prefix[i] =
prefix[i-1] plus nums[i]
4.
To get the sum from left to right:
5.
If left == 0, return prefix[right] right
away.
6.
Otherwise, take away the part you don't
want: prefix[right] - prefix[left - 1]
⏱ Time and Space Complexity
Preprocessing Time: O(n)
Query Time: O(1)
Space Complexity: O(n)
📝 Key Takeaways
1.
Preprocessing turns work that needs to be
done over and over into answers right away.
2.
Prefix sum is a classic way to "trade
space for speed."
3.
Very helpful for range queries problems
4.
Interview gold when there are more than
one question
II.
Find Pivot Index
🧠 Explanation
We
calculate the total sum of the array first.
Then
while iterating:
The
left_sum keeps getting bigger and bigger, but the right_sum isn't calculated
separately each time. Instead, it's calculated using:
right_sum
= total - left_sum - the current item
So
instead of scanning both sides repeatedly, we reuse information like a pro
⚙️ Logic in Steps
1.
Add up all the numbers in the array.
2.
Set left_sum to 0.
3.
Go through the array:
4.
Find right_sum by subtracting left_sum and
nums[i] from total.
5.
If left_sum is equal to right_sum, you
have found the pivot. Otherwise, add nums[i] to left_sum.
6.
If there is no pivot, return -1.
⏱ Time and Space Complexity
Time Complexity: O(n) →
single pass
Space Complexity: O(1) →
no extra space used
📝 Key Takeaways
1.
You don't have to do the math for the
right sum every time.
2.
The total sum works like a "global
reference."
3.
Smart rearranging keeps extra loops from
happening.
4.
A common way to optimize arrays
III.
Running Sum of 1D Array
🧠 Explanation
Rather than creating a
separate array for storing the running totals, we will use the original array
to hold these totals. The total of each of the elements will be equal to itself
plus all of the previous elements contained within. This allows for a gradual
conversion of the entire database into a prefix sum equivalent, or history
rewritten, as the arrow points forward.
⚙️ Logic in Steps
1.
Start from index 1 (since first element
stays same)
2.
For each index:
a. Add
previous value to current
b. nums[i]
+= nums[i - 1]
3.
Return modified array
⏱ Time and Space Complexity
- Time Complexity: O(n) → single pass
- Space Complexity: O(1) → no extra
space (in-place)
📝 Key Takeaways
1.
Updates in-place are also being memory
efficient
2.
Building prefix sums will not use an extra
array
3.
Always confirm in the interview process
that you can modify the input
4.
The same idea applies to lots of advanced
problems.
IV.
Subarray Sum equals K
🧠 Explanation
We use the prefix sum +
hashmap (keeping track of frequency) to compute how many subarrays sum to k. At
each point, we ask ourselves: Is there a previous prefix sum that yields k when
we take it away?
In terms of math:
1.
If our current sum is the prefix_sum, we
can say:
a. prefix_sum
- previous_sum = k
b. previous_sum
= prefix_sum – k
If that value exists in
our hashmap, then we have found one or more valid subarrays.
⚙️ Logic in Steps
1.
Start with:
a. Prefix
history = 0
b. Continuous
set = 0
2.
Frequency of subarrays that sum to 0,
where Indexation = 0; frequency = 1
3.
From here on out:
a. Proceed
through the integer list sequentially
b. Take
current integer, append it to prefix sum
c. Determine
the frequency of prefix history - Sum(K) exists in the prefix history (HashMap)
d. If
it does exist increment set
4.
Add/update prefix sum to the HashMap
5.
Return total frequency of prefix sums less
than or equal to K
⏱ Time and Space Complexity
Time Complexity: O(n) →
single traversal
Space Complexity: O(n) →
hashmap storage
📝 Key Takeaways
1.
The golden formula:
prefix_sum
- k
2.
Hashmap stores frequency, not just
existence
3.
Handles negative numbers (unlike sliding
window)
4.
This pattern repeats in many advanced
problems
🟡 Medium
(Interview-Level Skills)
V.
Continous Subarray Sum
🧠 Explanation
Use prefix sum +
remainder map.If two prefix sums have the same remainder when divided by k,
their difference (subarray) is divisible by k.
⚙️ Logic in Steps
1.
Track prefix_sum
2.
Compute remainder = prefix_sum % k
3.
If same remainder seen before and subarray
length ≥ 2 → return True
4.
Else store remainder with index
⏱ Time and Space Complexity
1.
Time: O(n)
2.
Space: O(n)
📝 Key Takeaways
1.
Same remainder ⇒ subarray sum divisible
by k
2.
Store first occurrence only to maximize
length
3.
Length condition (≥ 2) is essential
4.
Handle k = 0 separately
VI.
Binary Subarrays with Sum
🧠 Explanation
In order to determine how
many subarrays there are that equal our goal, we will use a prefix sum with a
hashmap (for frequency counting). Here's the process for each step:
1.
The current sum will be assigned to the
prefix sum.
2.
You then search for a past sum that equals
"current sum - goal." If you have found such sums, the instances of
those will then count as valid subarrays.
⚙️ Logic in Steps
1.
Start with:
a. prefix_sum
= 0, count = 0
b. freq
= {0: 1}
2.
For each number:
a. Update
prefix_sum
b. Add
freq[prefix_sum - goal] to count
c. Update
freq[prefix_sum]
⏱ Time and Space Complexity
1.
Time: O(n)
2.
Space: O(n)
📝 Key Takeaways
1.
Core idea → prefix_sum - goal
2.
Hashmap stores frequency, not just
presence
3.
Works for all integers (even negatives)
4.
Classic upgrade from brute force
VII.
Contiguous Array
🧠 Explanation
Convert the following:
1.
0 → -1
2.
1 → +1
Thus equalizing all zeros
and ones ⇒
total sum = 0
Using a prefix sum and a
hashmap allows us to get the first time a prefix sum occurs. If the same prefix
sum occurs, this means that a subarray containing these two prefix sums has a
sum of 0.
⚙️ Logic in Steps
1.
Initialize:
a. prefix_sum
= 0
b. hashmap
= {0: -1}
c. max_len
= 0
2.
Traverse array:
a. Add
+1 for 1, -1 for 0
b. If
prefix_sum seen before: → update max_len = i - first_index
c. Else
store index
3.
Return max_len
⏱ Time and Space Complexity
1.
Time: O(n)
2.
Space: O(n)
📝 Key Takeaways
1. Transform
problem (0 → -1 trick)
2. Same
prefix sum ⇒
subarray sum = 0
3. Store
first occurrence only for max length
4. Classic
prefix sum pattern
VIII.
Product of Array Except Self
🧠 Explanation
By performing left and
right passes, we can bypass separating and computing the product of all items. Each
index has the following:
The total of all items
before it (left) multiplied by the total of all items after it (right).
⚙️ Logic in Steps
1.
Initialize result with 1s
2.
Left pass:
Store
product of elements before index
3.
Right pass:
Multiply
with product of elements after index
⏱ Time and Space Complexity
1.
Time: O(n)
2.
Space: O(1) (excluding output)
📝 Key Takeaways
1.
No
division → handles zeros safely
2.
Two-pass approach = efficient
3.
Reuse result array to save space
4.
Classic pattern: prefix + suffix
Conclusion
The real strength of the
Prefix Sum is that it can not only speed up how quickly you can perform range
queries on an array of items, but it allows you to take a more complex problem
and break it down into simpler components so that you can solve them using
linear time algorithms. For example, by taking arrays and balancing them with a
simple formula or counting subarrays with clever hashmaps, the key principle is
to use previous calculations instead of having to re-compute everything again.
Understanding this pattern successfully helps you become less reliant on brute
force approaches and start thinking like a designer who has built something for
10x growth
If you Missed the
Previous Part
Missed the hashing part?
Go back—because Prefix Sum truly shines when combined with hash maps for O(1)
lookups, a pattern heavily used in real interviews Master Hashing for Coding Interviews: Top LeetCode Problems Explained
Coming Up
Coming up next,
Sorting—the backbone of efficient algorithms, turning messy data into
structured power for smarter problem-solving.
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