How do you solve Range Sum Query - Mutable on LeetCode?

Range Sum Query - Mutable (LeetCode #307) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(n) space complexity. Key insight: Understanding the difference between direct iteration and more complex data structures like Segment Trees is crucial for optimizing performance.

What is the brute force solution for Range Sum Query - Mutable?

The brute force approach for Range Sum Query - Mutable is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves directly calculating the sum for every query by iterating through the specified range. This is straightforward but inefficient for large arrays or many queries. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Range Sum Query - Mutable?

Range Sum Query - Mutable uses the following concepts: Array, Divide and Conquer, Design, Binary Indexed Tree, Segment Tree. The recommended patterns to study are: Segment Trees, Binary Indexed Trees.

What are the interview tips for Range Sum Query - Mutable?

Always clarify the constraints and expected input sizes before choosing an approach. Discuss the trade-offs of different data structures during the interview to show your understanding. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Range Sum Query - Mutable?

Forgetting to update the Segment Tree correctly after an update operation. Not considering edge cases, such as querying a range that includes the entire array.

#307

Range Sum Query - Mutable

Medium

ArrayDivide and ConquerDesignBinary Indexed TreeSegment TreeSegment TreesBinary Indexed Trees

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(log n)Space O(n)

Using a Segment Tree allows us to efficiently update elements and calculate range sums in logarithmic time. This is crucial for handling many queries quickly.

⚙️

Algorithm

3 steps

1Step 1: Build a Segment Tree from the input array, where each node represents the sum of a segment of the array.
2Step 2: For the update operation, update the specific index in the Segment Tree, which takes O(log n) time.
3Step 3: For the sumRange operation, query the Segment Tree to get the sum of the specified range, also in O(log n) time.

solution.py40 lines

1class SegmentTree:
2    def __init__(self, nums):
3        self.n = len(nums)
4        self.tree = [0] * (2 * self.n)
5        for i in range(self.n):
6            self.tree[self.n + i] = nums[i]
7        for i in range(self.n - 1, 0, -1):
8            self.tree[i] = self.tree[i * 2] + self.tree[i * 2 + 1]
9
10    def update(self, index, val):
11        index += self.n
12        self.tree[index] = val
13        while index > 1:
14            index //= 2
15            self.tree[index] = self.tree[index * 2] + self.tree[index * 2 + 1]
16
17    def sumRange(self, left, right):
18        left += self.n
19        right += self.n + 1
20        sum_ = 0
21        while left < right:
22            if left % 2:
23                sum_ += self.tree[left]
24                left += 1
25            if right % 2:
26                right -= 1
27                sum_ += self.tree[right]
28            left //= 2
29            right //= 2
30        return sum_
31
32class NumArray:
33    def __init__(self, nums):
34        self.segment_tree = SegmentTree(nums)
35
36    def update(self, index, val):
37        self.segment_tree.update(index, val)
38
39    def sumRange(self, left, right):
40        return self.segment_tree.sumRange(left, right)

ℹ

Complexity note: The time complexity is O(log n) for both update and sumRange operations due to the properties of the Segment Tree, which allows us to efficiently manage updates and queries on the array.

1Understanding the difference between direct iteration and more complex data structures like Segment Trees is crucial for optimizing performance.
2Recognizing when to use a data structure based on the types of operations required (updates vs. queries) can greatly enhance efficiency.

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