Count Of Range Sum

To solve this coding challenge, we need to understand how to efficiently count the number of range sums that lie within a given inclusive range

[lower, upper]

for an array of integers. Here's a detailed explanation and pseudocode for the solution.

Explanation

The primary task is to find all possible subarray sums and then check if they lie within the given range. A brute-force solution would involve calculating sums for all possible subarrays, but this can be too slow for large arrays (e.g., size 10^5). Hence, we need a more efficient approach. We use a combination of prefix sums and a data structure that allows efficient range queries to solve the problem. Here, we use a balanced binary search tree (like

std::multiset

in C++ or

SortedList

from the

sortedcontainers

library in Python) to help with efficient range queries.

Key steps:

1. Prefix Sum Array : This helps in quick sum calculations for any subarray.
2. Balanced BST : This allows us to count how many prefix sums fall within the required range efficiently.
The idea is to maintain a running sum and for each sum, count the number of times prefix sums fall within a specified range using a BST. For each element in the array:
3. Update the running sum.
4. Determine the range of prefix sums that would make the current sum fall within [lower, upper].
5. Count those prefix sums using the BST.
6. Insert the current running sum into the BST for future queries.

Pseudocode

Let's break down the detailed pseudocode along with comments to clarify each step.

                                            
# Define a function to count range sums
function countRangeSum(nums, lower, upper):
# Initialize count of valid range sums to 0
validRangeSumCount = 0

# Initialize current running sum to 0
runningSum = 0

# Use a multiset (or balanced BST) to store prefix sums
prefixSums = BalancedBST()

# Insert an initial prefix sum of 0 to handle the sum from the start
prefixSums.insert(0)

# Iterate through each number in the given nums array
for num in nums:
# Update the running sum
runningSum += num

# Determine the lower bound for valid prefix sums
lowerBound = runningSum - upper
# Determine the upper bound for valid prefix sums
upperBound = runningSum - lower

# Count how many prefix sums lie within the calculated bounds
validPrefixCount = prefixSums.countRange(lowerBound, upperBound)

# Add this count to the total count
validRangeSumCount += validPrefixCount

# Insert the current running sum to the prefix sums for future queries
prefixSums.insert(runningSum)

# Return the total count of valid range sums
return validRangeSumCount

# BalancedBST is a conceptual representation of a balanced binary search tree
# Define a class to represent the balanced BST with necessary operations
class BalancedBST:
# Initialize the BST
function __init__():
# Use a list to store elements (conceptual representation)
self.elements = []

# Function to insert a new element into the BST
function insert(value):
# Insert the value while maintaining the order
self.elements.insertSorted(value)

# Function to count elements within a specific range [low, high]
function countRange(low, high):
# Perform a binary search to determine the start and end indices
start = self.binarySearchFirstGreaterEqual(low)
end = self.binarySearchFirstGreater(high)

# Return the count of elements within the range
return end - start

# Note: The `insertSorted`, `binarySearchFirstGreaterEqual`, and `binarySearchFirstGreater`
# functions are not explicitly defined here. They need to be implemented to perform the 
# specific tasks described in the comments.

                                        

Detailed Steps in Pseudocode

7. Initialize the variables:
• validRangeSumCount
  to 0 (to store the count of valid range sums).
• runningSum
  to 0 (to keep track of the current sum of elements).
• prefixSums
  as an empty balanced binary search tree (BST).
8. Initial Prefix Sum:
• Insert 0 into
  prefixSums
  to handle cases where the subarray starts at index 0.
9. Iterate through the array:
• For each element
  num
  in
  nums
  :
• Update
  runningSum
  by adding
  num
  .
• Calculate the lower and upper bounds (
  lowerBound
  and
  upperBound
  ) for valid prefix sums.
• Count the number of prefix sums within the calculated range using
  prefixSums.countRange(lowerBound, upperBound)
  .
• Add this count to
  validRangeSumCount
  .
• Insert the current
  runningSum
  into
  prefixSums
  .
10. Return the Result:
• Return
  validRangeSumCount
  as the final count of subarray sums within the range
  [lower, upper]
  .

This approach ensures that each step is handled efficiently, making the solution feasible for large input sizes. The use of prefix sums and a balanced BST helps in achieving the required performance.