What data structures are used to solve XOR Queries of a Subarray?

This problem uses Array, Bit Manipulation, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving XOR Queries of a Subarray?

Insight 1: Calculating the XOR of a subarray naively for each query would result in a high time complexity. The key is to precompute XOR prefixes. Insight 2: The XOR operation is its own inverse, i.e., `a ^ a = 0`. This property is crucial for efficiently calculating the XOR of subarrays using prefix XORs.

Which companies ask LeetCode 1310. XOR Queries of a Subarray in interviews?

This problem is frequently asked at Airtel.

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XOR Queries of a Subarray - LeetCode 1310 Solution

Q: What is the time complexity of XOR Queries of a Subarray?

The optimal solution has a time complexity of O(N + Q), where N is the length of the array `arr` and Q is the number of queries. Building the prefix XOR array takes O(N) time. Calculating the XOR for each query takes O(1) time, and since we have Q queries, the total time for processing queries is O(Q). Therefore, the overall time complexity is O(N + Q)..

Q: What is the space complexity of XOR Queries of a Subarray?

The space complexity is O(N), where N is the length of the array `arr`. The prefix XOR array `prefix` requires O(N) space. The `answer` array requires O(Q) space which is dominated by O(N) given the constraints, so it's still O(N)..

XOR Queries of a Subarray - Complete Solution Guide

XOR Queries of a Subarray is LeetCode problem 1310, 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

You are given an array arr of positive integers. You are also given the array queries where queries[i] = [left i, right i ] . For each query i compute the XOR of elements from left i to right i (that is, arr[left i ] XOR arr[left i + 1] XOR ... XOR arr[right i ] ). Return an array answer where answer[i] is the answer to the i th query. Example 1: Input: arr = [1,3,4,8], queries = [[0,1],[1,2],[0,3],[3,3]] Output: [2,7,14,8] Explanation: The binary representation of the elements in the array are:

Detailed Explanation

The problem asks us to calculate the XOR of elements within specified subarrays of a given array `arr`. We are provided with a list of queries, where each query `queries[i]` is a pair `[left_i, right_i]` representing the start and end indices of a subarray. For each query, we need to compute the XOR of all elements in `arr` from index `left_i` to `right_i` (inclusive) and return an array containing the XOR results for each query.

Solution Approach

The solution employs the prefix XOR technique. First, it computes the prefix XOR array, where each element `prefix[i]` stores the XOR of all elements from `arr[0]` to `arr[i-1]`. Then, for each query `[left, right]`, it calculates the XOR of the subarray `arr[left...right]` using the formula `prefix[right + 1] ^ prefix[left]`. This approach efficiently computes the XOR for each query in O(1) time after the initial prefix XOR array is built.

Step-by-Step Algorithm

Step 1: Create a prefix XOR array `prefix` of size `len(arr) + 1`. Initialize `prefix[0] = 0`.
Step 2: Iterate through the input array `arr` from index 0 to `len(arr) - 1`.
Step 3: For each element `arr[i]`, calculate `prefix[i+1] = prefix[i] ^ arr[i]`. This builds the prefix XOR array.
Step 4: Create an empty array `answer` to store the XOR results for each query.
Step 5: Iterate through the `queries` array.
Step 6: For each query `[left, right]`, calculate the XOR of the subarray using the formula `prefix[right + 1] ^ prefix[left]`.
Step 7: Append the calculated XOR value to the `answer` array.
Step 8: Return the `answer` array.

Key Insights

Insight 1: Calculating the XOR of a subarray naively for each query would result in a high time complexity. The key is to precompute XOR prefixes.
Insight 2: The XOR operation is its own inverse, i.e., `a ^ a = 0`. This property is crucial for efficiently calculating the XOR of subarrays using prefix XORs.
Insight 3: Using prefix XORs allows us to calculate the XOR of a subarray `arr[left...right]` as `prefix[right+1] ^ prefix[left]`. This is because `prefix[right+1]` represents the XOR of `arr[0...right]`, and `prefix[left]` represents the XOR of `arr[0...left-1]`. XORing these two cancels out the common prefix, leaving only the XOR of `arr[left...right]`.

Complexity Analysis

Time Complexity: O(N + Q), where N is the length of the array `arr` and Q is the number of queries. Building the prefix XOR array takes O(N) time. Calculating the XOR for each query takes O(1) time, and since we have Q queries, the total time for processing queries is O(Q). Therefore, the overall time complexity is O(N + Q).

Space Complexity: O(N), where N is the length of the array `arr`. The prefix XOR array `prefix` requires O(N) space. The `answer` array requires O(Q) space which is dominated by O(N) given the constraints, so it's still O(N).

Topics

This problem involves: Array, Bit Manipulation, Prefix Sum.

Companies

Asked at: Airtel.

XOR Queries of a Subarray | LeetCode 1310

Medium

Array Bit Manipulation Prefix Sum

Time: O(N + Q), where N is the length of the array `arr` and Q is the number of queries. Building the prefix XOR array takes O(N) time. Calculating the XOR for each query takes O(1) time, and since we have Q queries, the total time for processing queries is O(Q). Therefore, the overall time complexity is O(N + Q).Space: O(N), where N is the length of the array `arr`. The prefix XOR array `prefix` requires O(N) space. The `answer` array requires O(Q) space which is dominated by O(N) given the constraints, so it's still O(N).

Solution Code

Available in:

class Solution {
    public int[] xorQueries(int[] arr, int[][] queries) {
        int[] prefix = new int[arr.length + 1];
        prefix[0] = 0;
        for (int i = 0; i < arr.length; i++) {
            prefix[i + 1] = prefix[i] ^ arr[i];
        }

        int[] ans = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int left = queries[i][0];
            int right = queries[i][1];
            ans[i] = prefix[right + 1] ^ prefix[left];
        }

        return ans;
    }
}

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Problem Info

Description

Problem 1310: XOR Queries of a Subarray - Detailed explanation and solution approaches.

Topics

Array Bit Manipulation Prefix Sum

Companies

Airtel

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1310

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

8 steps

1
Step 1: Create a prefix XOR array `prefix` of size `len(arr) + 1`. Initialize `prefix[0] = 0`.
2
Step 2: Iterate through the input array `arr` from index 0 to `len(arr) - 1`.
3
Step 3: For each element `arr[i]`, calculate `prefix[i+1] = prefix[i] ^ arr[i]`. This builds the prefix XOR array.
4
Step 4: Create an empty array `answer` to store the XOR results for each query.
5
Step 5: Iterate through the `queries` array.
6
Step 6: For each query `[left, right]`, calculate the XOR of the subarray using the formula `prefix[right + 1] ^ prefix[left]`.
7
Step 7: Append the calculated XOR value to the `answer` array.
8
Step 8: Return the `answer` array.

💡

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(N + Q), where N is the length of the array `arr` and Q is the number of queries. Building the prefix XOR array takes O(N) time. Calculating the XOR for each query takes O(1) time, and since we have Q queries, the total time for processing queries is O(Q). Therefore, the overall time complexity is O(N + Q).

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

💾

Space Complexity

O(N), where N is the length of the array `arr`. The prefix XOR array `prefix` requires O(N) space. The `answer` array requires O(Q) space which is dominated by O(N) given the constraints, so it's still O(N).

📝

Why?

The prefix XOR array is of size N+1. Each query is processed in O(1) time after the prefix array is built. The constraints specify the maximum size of array arr and queries, so the space complexity is linear to the size of arr.

📖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 Approach

Calculate the XOR for each query by iterating through the subarray specified by the query.

Time: O(N*Q), where N is the average size of subarrays and Q is the number of queries.Space: O(1)

Trade-offs: This approach has a much higher time complexity, making it impractical for large inputs. It's only suitable when the number of queries or the size of the subarrays is very small. It saves space but at a huge cost of time.

✨

Key Insights

1Insight 1: Calculating the XOR of a subarray naively for each query would result in a high time complexity. The key is to precompute XOR prefixes.
2Insight 2: The XOR operation is its own inverse, i.e., `a ^ a = 0`. This property is crucial for efficiently calculating the XOR of subarrays using prefix XORs.
3Insight 3: Using prefix XORs allows us to calculate the XOR of a subarray `arr[left...right]` as `prefix[right+1] ^ prefix[left]`. This is because `prefix[right+1]` represents the XOR of `arr[0...right]`, and `prefix[left]` represents the XOR of `arr[0...left-1]`. XORing these two cancels out the common prefix, leaving only the XOR of `arr[left...right]`.

📚

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 == right`. The solution correctly handles this case because `prefix[right+1] ^ prefix[left]` becomes `prefix[left+1] ^ prefix[left]` which equals `arr[left]` (since `prefix[left+1] = prefix[left] ^ arr[left]`).
Edge case 2: `left == 0`. The solution handles this correctly because `prefix[right+1] ^ prefix[0]` becomes `prefix[right+1] ^ 0`, which equals `prefix[right+1]`, the XOR of the elements from index 0 up to `right`.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not understanding the properties of XOR and trying to iterate through the subarray for each query, leading to a very slow solution.

✗Common mistake 2: Incorrectly calculating the prefix XOR. It's crucial to remember that `prefix[i]` stores the XOR of elements from index 0 up to `i-1` in `arr`.

✅

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: Error detection and correction in data transmission. XOR can be used to generate parity bits, which can detect errors in transmitted data.

🏦Real-world scenario 2: Cryptography. XOR is a fundamental operation used in many encryption algorithms due to its reversibility.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

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

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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: July 21, 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

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

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "XOR Queries of a Subarray"?

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 bit manipulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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