Leetcode #327: Count of Range Sum
In this guide, we solve Leetcode #327 Count of Range Sum in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given an integer array nums and two integers lower and upper, return the number of range sums that lie in [lower, upper] inclusive. Range sum S(i, j) is defined as the sum of the elements in nums between indices i and j inclusive, where i <= j.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Binary Indexed Tree, Segment Tree, Array, Binary Search, Divide and Conquer, Ordered Set, Merge Sort
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
Input: nums = [-2,5,-1], lower = -2, upper = 2
Output: 3
Explanation: The three ranges are: [0,0], [2,2], and [0,2] and their respective sums are: -2, -1, 2.
Python Solution
class BinaryIndexedTree:
def __init__(self, n):
self.n = n
self.c = [0] * (n + 1)
def update(self, x, v):
while x <= self.n:
self.c[x] += v
x += x & -x
def query(self, x):
s = 0
while x > 0:
s += self.c[x]
x -= x & -x
return s
class Solution:
def countRangeSum(self, nums: List[int], lower: int, upper: int) -> int:
s = list(accumulate(nums, initial=0))
arr = sorted(set(v for x in s for v in (x, x - lower, x - upper)))
tree = BinaryIndexedTree(len(arr))
ans = 0
for x in s:
l = bisect_left(arr, x - upper) + 1
r = bisect_left(arr, x - lower) + 1
ans += tree.query(r) - tree.query(l - 1)
tree.update(bisect_left(arr, x) + 1, 1)
return ans
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.