Leetcode #307: Range Sum Query - Mutable
In this guide, we solve Leetcode #307 Range Sum Query - Mutable 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, handle multiple queries of the following types: Update the value of an element in nums. Calculate the sum of the elements of nums between indices left and right inclusive where left <= right.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Design, Binary Indexed Tree, Segment Tree, Array, Divide and Conquer
Intuition
The problem splits naturally into smaller subproblems.
Solving each part independently makes the whole problem manageable.
Approach
Divide the input, solve recursively, then merge results.
This structure often yields clean and efficient code.
Steps:
- Divide into subproblems.
- Solve recursively.
- Merge to get final answer.
Example
Input
["NumArray", "sumRange", "update", "sumRange"]
[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]
Output
[null, 9, null, 8]
Explanation
NumArray numArray = new NumArray([1, 3, 5]);
numArray.sumRange(0, 2); // return 1 + 3 + 5 = 9
numArray.update(1, 2); // nums = [1, 2, 5]
numArray.sumRange(0, 2); // return 1 + 2 + 5 = 8
Python Solution
class BinaryIndexedTree:
__slots__ = ["n", "c"]
def __init__(self, n):
self.n = n
self.c = [0] * (n + 1)
def update(self, x: int, delta: int):
while x <= self.n:
self.c[x] += delta
x += x & -x
def query(self, x: int) -> int:
s = 0
while x > 0:
s += self.c[x]
x -= x & -x
return s
class NumArray:
__slots__ = ["tree"]
def __init__(self, nums: List[int]):
self.tree = BinaryIndexedTree(len(nums))
for i, v in enumerate(nums, 1):
self.tree.update(i, v)
def update(self, index: int, val: int) -> None:
prev = self.sumRange(index, index)
self.tree.update(index + 1, val - prev)
def sumRange(self, left: int, right: int) -> int:
return self.tree.query(right + 1) - self.tree.query(left)
# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# obj.update(index,val)
# param_2 = obj.sumRange(left,right)
Complexity
The time complexity is O(n log n) (typical). The space complexity is O(log n).
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.