Count of Range Sum
The given problem requires us to find the number of range sums that lie within a given range [lower, upper]. A naive solution would be to iterate through all possible subarrays, calculate their sum, and check if the sum lies within the given range. However, this approach is highly inefficient and will lead to a time complexity of (O(n^3)), where (n) is the length of the array.
We can significantly optimize the solution by using a prefix sum array and sorting.
Here’s a more optimized solution using merge sort:
 Calculate the prefix sum array,
prefixSum
.  Use a divideandconquer approach similar to merge sort to count the valid ranges.
 In the merge step, count the ranges where
lower <= right  left <= upper
and then merge the two sorted halves.
Here is the implementation of the above approach:


Key Takeaways
 By using a prefix sum array, we simplify the problem to finding ranges in a onedimensional array.
 Utilizing a merge sortlike divideandconquer approach helps us efficiently count the ranges.
The time complexity of this solution is (O(n \log n)), and the space complexity is also (O(n)).


Identifying Problem Isomorphism
“Count of Range Sum” can be mapped to “Count of Smaller Numbers After Self”.
Both problems deal with an array of integers and both require us to make certain calculations based on the values and their relative positions within the array.
In “Count of Range Sum”, we’re interested in finding the number of range sums that lie in the inclusive range [lower, upper]. A range sum is simply the sum of a range of numbers in the array.
In “Count of Smaller Numbers After Self”, we’re required to calculate how many numbers to the right of the current number are smaller than it.
The similarity here is that both problems are about doing some kind of range query on the array and return the count based on the conditions.
“Count of Smaller Numbers After Self” is a simpler problem because it’s only about comparing individual numbers, while “Count of Range Sum” needs to calculate sums over a range, making it a more complex problem.
The problem “327. Count of Range Sum” involves techniques related to prefix sum and divide and conquer. Here are some simpler problems to prepare for this:
LeetCode 53. Maximum Subarray
 This problem introduces you to the concept of subarray sums.
LeetCode 209. Minimum Size Subarray Sum
 This problem introduces the concept of a sliding window to achieve a target sum.
LeetCode 303. Range Sum Query  Immutable
 This problem introduces the concept of precomputing sums for faster queries, which is a key concept for handling range sum queries efficiently.
LeetCode 304. Range Sum Query 2D  Immutable
 This problem extends the previous concept to two dimensions.
LeetCode 560. Subarray Sum Equals K
 This problem involves counting subarrays that sum to a certain target, which may be helpful for understanding how to count range sums that fall within a certain range.
LeetCode 974. Subarray Sums Divisible by K
 This problem involves using a hash map to count prefix sums, which is a technique that can also be useful for counting range sums.
LeetCode 1480. Running Sum of 1D Array
 This problem is about calculating prefix sums, which is a simple but important concept for the problem “Count of Range Sum”.
LeetCode 1524. Number of Subarrays With Odd Sum
 This problem also involves counting certain subarrays based on their sums.
LeetCode 930. Binary Subarrays With Sum
 This problem involves a specific application of prefix sums, which is to count the number of subarrays with a certain sum.
LeetCode 1248. Count Number of Nice Subarrays
 This problem involves counting the number of subarrays with a certain property, which is similar to the “Count of Range Sum” problem.
These cover subarray sums and prefix sums, which are the key concepts for tackling the “327. Count of Range Sum” problem.


Problem Classification
Language Agnostic Coding Drills
The given Python code is to count the number of range sums that lie in the range [lower, upper]
inclusive. A Range Sum S(i, j)
for an array is defined as the sum of all the numbers in the array from index i
to j
(both inclusive).
To breakdown this problem, here are the steps and concepts that are required:
Understanding the Problem Statement: This is the first and one of the most crucial steps. It’s important to understand what a Range Sum is, and what the constraints of the problem are.
Cumulative Sum Array: A cumulative sum is a sequence of partial sums of a given sequence. In the code, it is used to keep track of the cumulative sums of the array at each index. You need to know how to create and use a cumulative sum array.
Iterating over an Array: This problem requires iterating over the
cumsum
array, which is a fundamental concept in any programming language. Each value ofcsum
incumsum
is a potential end point of a valid subarray.Dictionaries (HashMaps) and DefaultDict: The problem uses a Python
defaultdict
to keep track of the frequency of cumulative sums. Thedefaultdict
is a dictionary subclass that calls a factory function to supply missing values. It’s used here because we want to return 0 as default value when the key is not found in the dictionary.Understanding Range Sum in Terms of Cumulative Sum: Given the cumulative sums
cumsum[i]
andcumsum[j]
, we know thatcumsum[j]  cumsum[i]
gives us the sum of elements fromi+1
toj
in the array. Here, we use this fact to find the number of valid range sums.Range Iteration and InMemory Counting: For each cumulative sum, we check every target value between
lower
andupper
(inclusive). Ifcsum  target
is inrecord
(which implies there exists a subarray ending at current index with sum betweenlower
andupper
), we add its count tores
.Nested Loop Complexity Understanding: The time complexity of this code is O(n^2) because for every
csum
, we are iterating overrange(lower, upper + 1)
. Understanding this can help you realize that this solution is not suitable for large inputs.
This problemsolving approach focuses on iterating over each potential end point of subarrays, using cumulative sum to calculate the subarray sum and dictionary to count valid sums.
Targeted Drills in Python
Let’s break down the problem and write coding drills for the key parts:
Understanding the Problem Statement: While this is not exactly a code, understanding a problem statement is an essential skill. Practice reading and comprehending problem statements, paying close attention to the inputs and the expected outputs.
Cumulative Sum Array: You need to understand how to construct a cumulative sum array. Here’s a simple drill:
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def cumulative_sum(nums): cumsum = [0] for n in nums: cumsum.append(cumsum[1] + n) return cumsum print(cumulative_sum([1, 2, 3, 4, 5])) # Output: [0, 1, 3, 6, 10, 15]
Iterating over an Array: Iterating over an array is a fundamental operation. You can practice it with the following drill:
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def iterate_array(arr): for i in arr: print(i) iterate_array([1, 2, 3, 4, 5]) # Output: 1 2 3 4 5 (each in new line)
Dictionaries (HashMaps) and DefaultDict: Practice using Python defaultdict to handle missing keys in dictionary.
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import collections def default_dict_usage(): record = collections.defaultdict(int) record['a'] += 1 print(record['a']) # Output: 1 print(record['b']) # Output: 0 default_dict_usage()
Understanding Range Sum in Terms of Cumulative Sum: You can practice this concept by writing a function that calculates the sum of elements between any two indices in an array using cumulative sum.
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def range_sum(cumsum, i, j): return cumsum[j + 1]  cumsum[i] cumsum = cumulative_sum([1, 2, 3, 4, 5]) print(range_sum(cumsum, 1, 3)) # Output: 9
Range Iteration and InMemory Counting: Practice range iteration and inmemory counting with a simple example:
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def range_iteration_and_counting(nums): count_dict = collections.defaultdict(int) for num in nums: for i in range(num  1, num + 2): count_dict[i] += 1 return count_dict print(range_iteration_and_counting([1, 2, 3])) # Output: defaultdict(<class 'int'>, {0: 1, 1: 2, 2: 2, 3: 2, 4: 1})
Nested Loop Complexity Understanding: This is more of a conceptual understanding than a coding drill. However, you can write a simple nested loop and time it to understand the time complexity.
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import time def nested_loop(n): start = time.time() count = 0 for i in range(n): for j in range(n): count += 1 end = time.time() print("Time taken: ", end  start) nested_loop(10000) # Note the time taken
These drills should provide a good practice and understanding of the individual concepts used in the problem. Remember that these are only practice drills, the real solution uses these concepts together to solve the problem.