What is the time complexity of Range Sum Query - Mutable?

The optimal solution has a time complexity of O(log n).

What data structures are used to solve Range Sum Query - Mutable?

This problem uses Array, Divide and Conquer, Design, Binary Indexed Tree, Segment Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Range Sum Query - Mutable?

Insight 1: A naive approach of iterating through the range for `sumRange` and direct assignment for `update` would result in O(n) time complexity for `sumRange`, which is inefficient given the number of queries. Therefore, using a more efficient data structure or algorithm is necessary. Insight 2: Binary Indexed Tree (BIT), also known as Fenwick Tree, is an efficient data structure for solving range query problems where both point updates and prefix sum queries are required. BIT can perform both operations in O(log n) time.

Which companies ask LeetCode 307. Range Sum Query - Mutable in interviews?

This problem is frequently asked at Google.

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Range Sum Query - Mutable - LeetCode 307 Solution

Q: How to solve Range Sum Query - Mutable?

The provided code implements the `NumArray` class using a Binary Indexed Tree (BIT). The BIT is used to store prefix sums in a way that allows for both efficient updates and range sum queries. During initialization, the BIT is constructed based on the input array `nums`. The `update` operation updates the BIT to reflect the change in value at a given index. The `sumRange` operation calculates the sum of elements within the specified range by utilizing the prefix sum property of the BIT. Specifically, it finds the sum from index 0 up to `right` and subtracts the sum from index 0 up to `left - 1` to get the desired range sum.

Range Sum Query - Mutable - Complete Solution Guide

Range Sum Query - Mutable is LeetCode problem 307, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 . Implement the NumArray class: NumArray(int[] nums) Initializes the object with the integer array nums . void update(int index, int val) Updates the value of nums[index] to be val . int sumRange(int left, int right) Returns the sum of the elements of nums between indices left

Detailed Explanation

The problem asks us to implement a `NumArray` class that can handle two types of queries on an integer array `nums`: updating the value of an element at a given index and calculating the sum of elements within a given range (inclusive). The `NumArray` class should be initialized with the initial array `nums`, and we need to efficiently implement both `update` and `sumRange` operations, adhering to the given constraints. Essentially, we need a data structure that allows for both point updates and range queries on an array.

Solution Approach

The provided code implements the `NumArray` class using a Binary Indexed Tree (BIT). The BIT is used to store prefix sums in a way that allows for both efficient updates and range sum queries. During initialization, the BIT is constructed based on the input array `nums`. The `update` operation updates the BIT to reflect the change in value at a given index. The `sumRange` operation calculates the sum of elements within the specified range by utilizing the prefix sum property of the BIT. Specifically, it finds the sum from index 0 up to `right` and subtracts the sum from index 0 up to `left - 1` to get the desired range sum.

Step-by-Step Algorithm

Step 1: **Initialization:** The `NumArray` class is initialized with the input array `nums`. A BIT (Binary Indexed Tree) is created with a size of `n + 1`, where `n` is the length of `nums`.
Step 2: **BIT Construction:** The BIT is constructed by iterating through the `nums` array and updating the BIT for each element. The `_update_bit` helper function is used for this, which iteratively adds the element's value to the appropriate nodes in the BIT.
Step 3: **Update Operation:** The `update(index, val)` function updates the value at the given `index` in the `nums` array and propagates this change to the BIT. The `delta` is computed as the difference between the `val` and the old value at `index`. This delta is then used to update the BIT using the `_update_bit` function.
Step 4: **Sum Range Operation:** The `sumRange(left, right)` function calculates the sum of the elements within the range `[left, right]` (inclusive). It achieves this by calculating the prefix sum up to `right` using the `_query_bit` function and subtracting the prefix sum up to `left - 1` using the `_query_bit` function. The result is the sum of the desired range.
Step 5: **Query BIT ( _query_bit):** This helper function calculates the prefix sum up to a given `index`. It iteratively traverses the BIT, adding the values of the nodes that contribute to the prefix sum.
Step 6: **Update BIT ( _update_bit):** This helper function updates the BIT when a value at a specific `index` in the original array changes. It iteratively traverses the BIT, adding the `delta` to the appropriate nodes to reflect the change.

Key Insights

Insight 1: A naive approach of iterating through the range for `sumRange` and direct assignment for `update` would result in O(n) time complexity for `sumRange`, which is inefficient given the number of queries. Therefore, using a more efficient data structure or algorithm is necessary.
Insight 2: Binary Indexed Tree (BIT), also known as Fenwick Tree, is an efficient data structure for solving range query problems where both point updates and prefix sum queries are required. BIT can perform both operations in O(log n) time.
Insight 3: The BIT data structure requires a one-based indexing scheme, meaning we will need to shift the indices by one when accessing it from the original zero-based `nums` array.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Divide and Conquer, Design, Binary Indexed Tree, Segment Tree.

Companies

Asked at: Google.

Range Sum Query - Mutable | LeetCode 307

Medium

Array Divide and Conquer Design Binary Indexed Tree Segment Tree

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

Solution Code

Available in:

class NumArray {

    private int[] nums;
    private int[] bit;
    private int n;

    public NumArray(int[] nums) {
        this.n = nums.length;
        this.nums = nums;
        this.bit = new int[this.n + 1];
        for (int i = 0; i < this.n; i++) {
            _update_bit(i, this.nums[i]);
        }
    }

    private void _update_bit(int index, int delta) {
        int i = index + 1;
        while (i <= this.n) {
            this.bit[i] += delta;
            i += i & (-i);
        }
    }

    public void update(int index, int val) {
        int delta = val - this.nums[index];
        this.nums[index] = val;
        _update_bit(index, delta);
    }

    private int _query_bit(int index) {
        int i = index + 1;
        int prefix_sum = 0;
        while (i > 0) {
            prefix_sum += this.bit[i];
            i -= i & (-i);
        }
        return prefix_sum;
    }

    public int sumRange(int left, int right) {
        return _query_bit(right) - _query_bit(left - 1);
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * obj.update(index,val);
 * int param_2 = obj.sumRange(left,right);
 */

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Problem Info

Description

Problem 307: Range Sum Query - Mutable - Detailed explanation and solution approaches.

Topics

Array Divide and Conquer Design Binary Indexed Tree Segment Tree

Companies

Google

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:307

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: **Initialization:** The `NumArray` class is initialized with the input array `nums`. A BIT (Binary Indexed Tree) is created with a size of `n + 1`, where `n` is the length of `nums`.
2
Step 2: **BIT Construction:** The BIT is constructed by iterating through the `nums` array and updating the BIT for each element. The `_update_bit` helper function is used for this, which iteratively adds the element's value to the appropriate nodes in the BIT.
3
Step 3: **Update Operation:** The `update(index, val)` function updates the value at the given `index` in the `nums` array and propagates this change to the BIT. The `delta` is computed as the difference between the `val` and the old value at `index`. This delta is then used to update the BIT using the `_update_bit` function.
4
Step 4: **Sum Range Operation:** The `sumRange(left, right)` function calculates the sum of the elements within the range `[left, right]` (inclusive). It achieves this by calculating the prefix sum up to `right` using the `_query_bit` function and subtracting the prefix sum up to `left - 1` using the `_query_bit` function. The result is the sum of the desired range.
5
Step 5: **Query BIT ( _query_bit):** This helper function calculates the prefix sum up to a given `index`. It iteratively traverses the BIT, adding the values of the nodes that contribute to the prefix sum.
6
Step 6: **Update BIT ( _update_bit):** This helper function updates the BIT when a value at a specific `index` in the original array changes. It iteratively traverses the BIT, adding the `delta` to the appropriate nodes to reflect the change.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log n)

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The BIT allows for logarithmic time complexity for both updates and range queries. Each update or query requires traversing at most log n nodes in the BIT. The space complexity is O(n) because the BIT array has the same size as the input array.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Segment Tree

A segment tree can also be used to solve this problem. A segment tree represents the array in a tree structure where each node represents a segment of the array. Updates and range queries can be performed in logarithmic time.

Time: O(log n) for both update and query.Space: O(n)

Trade-offs: Segment trees are more versatile than BITs and can handle a wider range of queries, such as finding the minimum or maximum in a range. However, they are generally more complex to implement and may have a slightly larger constant factor in their time and space complexity compared to BIT.

Square Root Decomposition

Divide the array into blocks of size sqrt(n). For each block, store the sum of the elements in the block. Update operations can be performed by updating the individual element and the block sum it belongs to. Range queries can be performed by iterating through the blocks that fully lie within the range and summing their block sums, plus iterating through the elements in the partially covered blocks.

Time: O(sqrt(n)) for both update and query.Space: O(n)

Trade-offs: Square root decomposition is simpler to implement than BIT or segment tree, but its time complexity is higher (O(sqrt(n))). It is a good option when the number of updates and queries is relatively small, or when simplicity is more important than performance.

✨

Key Insights

1Insight 1: A naive approach of iterating through the range for `sumRange` and direct assignment for `update` would result in O(n) time complexity for `sumRange`, which is inefficient given the number of queries. Therefore, using a more efficient data structure or algorithm is necessary.
2Insight 2: Binary Indexed Tree (BIT), also known as Fenwick Tree, is an efficient data structure for solving range query problems where both point updates and prefix sum queries are required. BIT can perform both operations in O(log n) time.
3Insight 3: The BIT data structure requires a one-based indexing scheme, meaning we will need to shift the indices by one when accessing it from the original zero-based `nums` array.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Empty input array. The constructor should handle this case gracefully. The given solutions handle it implicitly; if the input `nums` is empty, the loops in the constructor won't execute.
Edge case 2: `left` and `right` being equal. The `sumRange` function correctly calculates the sum for a single element range in this scenario.
Edge case 3: `index` out of bounds during `update`. The problem statement states that 0 <= index < nums.length so we can assume that index is always within bounds. In practice, input validation should be performed to handle invalid index values.
Edge case 4: All values in input array being zero. The BIT will be all zeros after initialization and the `sumRange` method will return zero.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Off-by-one errors when calculating indices for updating or querying the BIT. Remember that the BIT uses 1-based indexing, so you need to add 1 to the original index when accessing the BIT.

✗Common mistake 2: Incorrectly calculating the next index to update or query in the BIT. The formula `i += i & (-i)` for updating and `i -= i & (-i)` for querying must be used correctly.

✗Common mistake 3: Forgetting to update the original `nums` array during the `update` operation. The `nums` array needs to be kept consistent with the BIT.

✗Common mistake 4: Not understanding how the binary indexed tree stores cumulative information.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Financial data analysis where you need to track and update stock prices and calculate cumulative returns over a period.

🏦Real-world scenario 2: Tracking website traffic and calculating the number of visits within a specific time range.

🎮Real-world scenario 3: Game development, where you might need to update the state of objects in a game world and query the total health or score within a region.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 6, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Range Sum Query - Mutable"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether divide and conquer techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?