How do you solve Count of Range Sum on LeetCode?

Count of Range Sum (LeetCode #327) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Understanding prefix sums is crucial for efficiently solving range sum problems.

What is the brute force solution for Count of Range Sum?

The brute force approach for Count of Range Sum is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves calculating the sum of all possible subarrays and checking if they fall within the specified range. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What are the interview tips for Count of Range Sum?

Explain your thought process clearly while coding. Discuss potential optimizations even if you start with a brute-force solution. Practice writing clean and efficient code.

What are common mistakes when solving Count of Range Sum?

Not considering edge cases like empty arrays or single elements. Failing to optimize the brute-force solution when it is clearly inefficient.

#327

Count of Range Sum

Hard

ArrayBinary SearchDivide and ConquerBinary Indexed TreeSegment TreeMerge SortOrdered SetHash MapArray

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n)Space O(n)

The optimal approach uses a modified merge sort algorithm to count the range sums efficiently. By sorting the prefix sums while counting, we can leverage the divide-and-conquer strategy to achieve better performance.

⚙️

Algorithm

4 steps

1Step 1: Calculate the prefix sums for the array, storing them in a new array.
2Step 2: Use a recursive function to sort the prefix sums while counting how many of them fall within the specified range [lower, upper].
3Step 3: For each prefix sum, use binary search to find how many previous sums are within the range when combined with the current prefix sum.
4Step 4: Return the total count of valid range sums.

solution.py24 lines

1# Full working Python code
2
3def countRangeSum(nums, lower, upper):
4    n = len(nums)
5    prefix_sums = [0] * (n + 1)
6    for i in range(n):
7        prefix_sums[i + 1] = prefix_sums[i] + nums[i]
8    return merge_sort_count(prefix_sums, 0, n, lower, upper)
9
10
11def merge_sort_count(prefix_sums, left, right, lower, upper):
12    if left >= right:
13        return 0
14    mid = (left + right) // 2
15    count = merge_sort_count(prefix_sums, left, mid, lower, upper) + merge_sort_count(prefix_sums, mid + 1, right, lower, upper)
16    j = k = mid + 1
17    for i in range(left, mid + 1):
18        while j <= right and prefix_sums[j] - prefix_sums[i] < lower:
19            j += 1
20        while k <= right and prefix_sums[k] - prefix_sums[i] <= upper:
21            k += 1
22        count += k - j
23    prefix_sums[left:right + 1] = sorted(prefix_sums[left:right + 1])
24    return count

ℹ

Complexity note: The time complexity is O(n log n) due to the merge sort process, which involves dividing the array and merging it back while counting valid sums. The space complexity is O(n) because we store the prefix sums in a separate array.

1Understanding prefix sums is crucial for efficiently solving range sum problems.
2Using sorting and binary search can significantly reduce the time complexity.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.