What is the time complexity of Bitwise AND of Numbers Range?

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

What data structures are used to solve Bitwise AND of Numbers Range?

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

What are the key insights for solving Bitwise AND of Numbers Range?

Insight 1: The bitwise AND of a range of numbers will have common prefix bits. Once the bits start to differ between the left and right bounds, all bits to the right will become zero due to the inclusion of numbers with differing bits. Insight 2: Repeatedly right-shifting both 'left' and 'right' until they become equal effectively identifies the common prefix. The number of shifts represents the power of 2 that was factored out.

Which companies ask LeetCode 201. Bitwise AND of Numbers Range in interviews?

This problem is frequently asked at Google.

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Bitwise AND of Numbers Range - LeetCode 201 Solution

Q: How to solve Bitwise AND of Numbers Range?

The solution leverages the observation that the bitwise AND of a range of numbers will only have '1's in the most significant bits that are common to all numbers in the range. The core idea is to repeatedly right-shift both 'left' and 'right' until they become equal. This effectively removes the differing least significant bits, isolating the common prefix. The number of right shifts is then used to shift the common prefix back to its original position, yielding the final result.

Bitwise AND of Numbers Range - Complete Solution Guide

Bitwise AND of Numbers Range is LeetCode problem 201, 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 two integers left and right that represent the range [left, right] , return the bitwise AND of all numbers in this range, inclusive . Example 1: Input: left = 5, right = 7 Output: 4 Example 2: Input: left = 0, right = 0 Output: 0 Example 3: Input: left = 1, right = 2147483647 Output: 0 Constraints: 0 <= left <= right <= 2 31 - 1

Detailed Explanation

The problem asks us to find the bitwise AND of all numbers within a given range [left, right] (inclusive). Bitwise AND operation compares each bit of the two numbers. If the bit is 1 in both numbers, the corresponding result bit is set to 1. Otherwise, the result bit is set to 0.

Solution Approach

The solution leverages the observation that the bitwise AND of a range of numbers will only have '1's in the most significant bits that are common to all numbers in the range. The core idea is to repeatedly right-shift both 'left' and 'right' until they become equal. This effectively removes the differing least significant bits, isolating the common prefix. The number of right shifts is then used to shift the common prefix back to its original position, yielding the final result.

Step-by-Step Algorithm

Step 1: Initialize a 'shift' counter to 0. This counter tracks how many times 'left' and 'right' are right-shifted.
Step 2: Enter a 'while' loop that continues as long as 'left' is strictly less than 'right'.
Step 3: Inside the loop, right-shift both 'left' and 'right' by 1 bit (i.e., divide by 2 and discard the remainder). This removes the least significant bit.
Step 4: Increment the 'shift' counter by 1 for each right-shift operation.
Step 5: After the loop terminates (when 'left' equals 'right'), left-shift 'left' by 'shift' bits (i.e., multiply 'left' by 2 raised to the power of 'shift'). This restores the common prefix to its original place value.
Step 6: Return the final value of 'left'.

Key Insights

Insight 1: The bitwise AND of a range of numbers will have common prefix bits. Once the bits start to differ between the left and right bounds, all bits to the right will become zero due to the inclusion of numbers with differing bits.
Insight 2: Repeatedly right-shifting both 'left' and 'right' until they become equal effectively identifies the common prefix. The number of shifts represents the power of 2 that was factored out.
Insight 3: The result is the common prefix, shifted back by the same number of positions to restore the original scale.

Complexity Analysis

Time Complexity: O(log(n))

Space Complexity: O(1)

Topics

This problem involves: Bit Manipulation.

Companies

Asked at: Google.

Bitwise AND of Numbers Range | LeetCode 201

Medium

Bit Manipulation

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

Solution Code

Available in:

class Solution {
    public int rangeBitwiseAnd(int left, int right) {
        int shift = 0;
        while (left < right) {
            left >>= 1;
            right >>= 1;
            shift += 1;
        }
        return left << shift;
    }
}

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Problem Info

Description

Problem 201: Bitwise AND of Numbers Range - Detailed explanation and solution approaches.

Topics

Bit Manipulation

Companies

Google

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:201

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: Initialize a 'shift' counter to 0. This counter tracks how many times 'left' and 'right' are right-shifted.
2
Step 2: Enter a 'while' loop that continues as long as 'left' is strictly less than 'right'.
3
Step 3: Inside the loop, right-shift both 'left' and 'right' by 1 bit (i.e., divide by 2 and discard the remainder). This removes the least significant bit.
4
Step 4: Increment the 'shift' counter by 1 for each right-shift operation.
5
Step 5: After the loop terminates (when 'left' equals 'right'), left-shift 'left' by 'shift' bits (i.e., multiply 'left' by 2 raised to the power of 'shift'). This restores the common prefix to its original place value.
6
Step 6: Return the final value of 'left'.

💡

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(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is logarithmic because we are repeatedly dividing the range by 2 in each step. The space complexity is constant because we are only using a few integer variables, regardless of the input size.

📖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

Iterative AND

Iterate through the range [left, right] and perform bitwise AND on all numbers. This is a brute-force approach.

Time: O(n), where n is the range between left and right.Space: O(1)

Trade-offs: This approach is simple to understand but inefficient for large ranges. The provided solution is significantly more efficient because it avoids iterating through the entire range.

✨

Key Insights

1Insight 1: The bitwise AND of a range of numbers will have common prefix bits. Once the bits start to differ between the left and right bounds, all bits to the right will become zero due to the inclusion of numbers with differing bits.
2Insight 2: Repeatedly right-shifting both 'left' and 'right' until they become equal effectively identifies the common prefix. The number of shifts represents the power of 2 that was factored out.
3Insight 3: The result is the common prefix, shifted back by the same number of positions to restore the original scale.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: left = 0, right = 0. The while loop is skipped because the condition (left < right) is false, and the function returns left << 0 = 0.
Edge case 2: left = right. The while loop is skipped because the condition (left < right) is false, and the function returns left << 0 = left (which equals right).
Edge case 3: left = 1, right = 2147483647. The loop continues shifting until left and right converge to the same value. The large difference between the two values requires a significant number of shifts. This is handled correctly because the `shift` variable and the `left` and `right` variables are large enough to contain the intermediate and final results.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Performing bitwise AND without considering the common prefix. A naive implementation might calculate the AND of just left and right, which would not produce the correct answer.

✗Common mistake 2: Not understanding bitwise operations. A lack of understanding of right shift and left shift can lead to incorrect implementations.

✗Common mistake 3: Integer overflow. Although the solution handles large values, one must be careful about integer overflows if implementing a brute force solution.

✅

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: Determining network address from a range of IP addresses. The bitwise AND can be used to find the common network address that all IPs in a subnet share.

🏦Real-world scenario 2: In embedded systems, this algorithm might be used to determine the common configuration bits from a range of possible configurations, where the configuration is represented by an integer.

💡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

XOR is your friend for "find the single number" problems. Remember: n & (n-1) clears the lowest set bit.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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: June 22, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

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

@CodeNinjaJun 29

❓

Frequently Asked Questions

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 "Bitwise AND of Numbers Range"?

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

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