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

The optimal solution has a time complexity of O(1).

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

This problem uses Array, Design, Prefix Sum. Understanding these concepts is essential for solving this Easy problem efficiently.

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

Insight 1: Using a prefix sum array drastically reduces the time complexity of each range sum query. Instead of iterating through the array for each query, we can calculate the sum in constant time. Insight 2: The prefix sum array `prefix_sums[i]` stores the cumulative sum of elements from `nums[0]` to `nums[i-1]`. This allows us to calculate the sum of any range [left, right] as `prefix_sums[right + 1] - prefix_sums[left]`.

Which companies ask LeetCode 303. Range Sum Query - Immutable in interviews?

This problem is frequently asked at Palantir Technologies.

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Range Sum Query - Immutable - LeetCode 303 Solution

Q: How to solve Range Sum Query - Immutable?

The provided solutions utilize the prefix sum technique. A prefix sum array is created during initialization. This array stores the cumulative sum of the input array at each index. When a `sumRange` query is made, the difference between the prefix sums at `right + 1` and `left` gives the sum of the elements in the specified range. This approach leverages the mathematical property that the sum of elements from index `left` to `right` is equal to the cumulative sum up to `right` minus the cumulative sum up to `left - 1`.

Range Sum Query - Immutable - Complete Solution Guide

Range Sum Query - Immutable is LeetCode problem 303, a Easy 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 type: 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 . int sumRange(int left, int right) Returns the sum of the elements of nums between indices left and right inclusive (i.e. nums[left] + nums[left + 1] + ... + nums[right] ). Example 1: Input ["NumArray", "sumRange",

Detailed Explanation

The problem asks you to design a data structure that efficiently handles range sum queries on an integer array. Given an array `nums`, you need to create a structure that allows you to quickly calculate the sum of elements within any given range [left, right] (inclusive) of the array. The input consists of the initial array `nums` and multiple queries specifying the `left` and `right` indices for the sum calculation. The output is a series of sums, one for each query.

Solution Approach

The provided solutions utilize the prefix sum technique. A prefix sum array is created during initialization. This array stores the cumulative sum of the input array at each index. When a `sumRange` query is made, the difference between the prefix sums at `right + 1` and `left` gives the sum of the elements in the specified range. This approach leverages the mathematical property that the sum of elements from index `left` to `right` is equal to the cumulative sum up to `right` minus the cumulative sum up to `left - 1`.

Step-by-Step Algorithm

Step 1: Initialize the `NumArray` object with the input array `nums`. Create a prefix sum array `prefix_sums` of size `len(nums) + 1` and initialize the first element to 0.
Step 2: Iterate through the input array `nums`. For each element `nums[i]`, calculate the cumulative sum by adding `nums[i]` to the previous cumulative sum in `prefix_sums`, and store the result in `prefix_sums[i + 1]`.
Step 3: For each `sumRange` query with indices `left` and `right`, return `prefix_sums[right + 1] - prefix_sums[left]`. This efficiently calculates the sum of the specified range using the pre-computed prefix sums.

Key Insights

Insight 1: Using a prefix sum array drastically reduces the time complexity of each range sum query. Instead of iterating through the array for each query, we can calculate the sum in constant time.
Insight 2: The prefix sum array `prefix_sums[i]` stores the cumulative sum of elements from `nums[0]` to `nums[i-1]`. This allows us to calculate the sum of any range [left, right] as `prefix_sums[right + 1] - prefix_sums[left]`.
Insight 3: The space complexity is increased by using the prefix sum array, but this trade-off significantly improves query time. This is a classic space-time trade-off optimization.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(n)

Topics

This problem involves: Array, Design, Prefix Sum.

Companies

Asked at: Palantir Technologies.

Range Sum Query - Immutable | LeetCode 303

Easy

Array Design Prefix Sum

Time: O(1)Space: O(n)

Solution Code

Available in:

class NumArray {
    private int[] prefixSum;

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

    public int sumRange(int left, int right) {
        return prefixSum[right + 1] - prefixSum[left];
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(left,right);
 */

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Problem Info

Description

Problem 303: Range Sum Query - Immutable - Detailed explanation and solution approaches.

Topics

Array Design Prefix Sum

Companies

Palantir Technologies

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:303

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

3 steps

1
Step 1: Initialize the `NumArray` object with the input array `nums`. Create a prefix sum array `prefix_sums` of size `len(nums) + 1` and initialize the first element to 0.
2
Step 2: Iterate through the input array `nums`. For each element `nums[i]`, calculate the cumulative sum by adding `nums[i]` to the previous cumulative sum in `prefix_sums`, and store the result in `prefix_sums[i + 1]`.
3
Step 3: For each `sumRange` query with indices `left` and `right`, return `prefix_sums[right + 1] - prefix_sums[left]`. This efficiently calculates the sum of the specified range using the pre-computed prefix sums.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The `__init__` function iterates through the input array once to create the prefix sum array, resulting in O(n) time complexity. The `sumRange` function performs a constant-time subtraction operation, resulting in O(1) time complexity. The space complexity is dominated by the prefix sum array, which has a size proportional to the input array size, leading to O(n) space complexity.

📖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

Naive Iteration

Iterate through the input array `nums` from `left` to `right` and sum the elements directly.

Time: O(n) for `sumRange`Space: O(1)

Trade-offs: This approach is simpler to implement, but significantly slower for multiple queries because it re-calculates the sum for each range query. The prefix sum approach is superior when there are many `sumRange` queries.

✨

Key Insights

1Insight 1: Using a prefix sum array drastically reduces the time complexity of each range sum query. Instead of iterating through the array for each query, we can calculate the sum in constant time.
2Insight 2: The prefix sum array `prefix_sums[i]` stores the cumulative sum of elements from `nums[0]` to `nums[i-1]`. This allows us to calculate the sum of any range [left, right] as `prefix_sums[right + 1] - prefix_sums[left]`.
3Insight 3: The space complexity is increased by using the prefix sum array, but this trade-off significantly improves query time. This is a classic space-time trade-off optimization.

📚

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: `left` equals `right`. The solution correctly handles this by returning `prefix_sums[right + 1] - prefix_sums[left]`, which is equivalent to `nums[left]`.
Edge case 2: `left` equals 0. The solution correctly calculates the sum from the beginning of the array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the prefix sum, often off by one index in the array.

✗Common mistake 2: Not handling edge cases correctly, such as when `left` or `right` is at the beginning or end of the array, or when `left` equals `right`.

✅

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: Data analytics – quickly calculating sums of data points within specific time ranges or other dimensions. For example, calculating total sales within a specific month from a sales dataset.

🏦Real-world scenario 2: Financial modeling – efficiently calculating cumulative returns or profits over different periods. Imagine a stock price dataset; you could use this method to quickly find the total gain/loss for specific time frames.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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 2, 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

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

@CodeNinjaJul 13

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

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 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.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

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 - Immutable"?

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

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