What is the time complexity of Minimum One Bit Operations to Make Integers Zero?

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

What data structures are used to solve Minimum One Bit Operations to Make Integers Zero?

This problem uses Dynamic Programming, Bit Manipulation, Memoization. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum One Bit Operations to Make Integers Zero?

Insight 1: The operations are related to Gray code. Converting n to 0 using the minimum operations is equivalent to finding the Gray code representation of n. Insight 2: The number of operations to convert n to 0 is given by the XOR of n with right-shifted n until n becomes 0. This elegantly utilizes bitwise operations to achieve the goal.

Which companies ask LeetCode 1611. Minimum One Bit Operations to Make Integers Zero in interviews?

This problem is frequently asked at Expedia, McKinsey, ServiceNow.

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Minimum One Bit Operations to Make Integers Zero - LeetCode 1611 Solution

Minimum One Bit Operations to Make Integers Zero - Complete Solution Guide

Minimum One Bit Operations to Make Integers Zero is LeetCode problem 1611, a Hard 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 n , you must transform it into 0 using the following operations any number of times: Change the rightmost ( 0 th ) bit in the binary representation of n . Change the i th bit in the binary representation of n if the (i-1) th bit is set to 1 and the (i-2) th through 0 th bits are set to 0 . Return the minimum number of operations to transform n into 0 . Example 1: Input: n = 3 Output: 2 Explanation: The binary representation of 3 is "11". " 1 1" -> " 0 1" with the 2 nd operation

Detailed Explanation

The problem asks us to find the minimum number of operations required to transform a given integer `n` into 0. The allowed operations are: 1. Change the rightmost bit (0th bit). 2. Change the ith bit if the (i-1)th bit is 1 and all bits from (i-2)th down to the 0th bit are 0. The input is a non-negative integer `n`, and the output is the minimum number of operations to convert `n` to 0, adhering to the specified bit manipulation rules.

Solution Approach

The solution leverages the property that the minimum number of operations to transform 'n' to 0 is equivalent to converting 'n' to its Gray code representation. Gray code can be calculated efficiently using the formula `n ^ (n >> 1)`. The provided code iteratively applies this concept. In each iteration, we XOR the current number `n` with the `res` variable that stores the total number of operations. Simultaneously, we right-shift `n` by 1 to process each bit. The loop continues until `n` becomes 0. Effectively, this code calculates the Gray code equivalent, and this value represents the minimum number of operations required.

Step-by-Step Algorithm

Step 1: Initialize `res` to 0. This variable will accumulate the minimum number of operations.
Step 2: Start a `while` loop that continues as long as `n` is greater than 0.
Step 3: Inside the loop, perform a bitwise XOR operation between `res` and `n`, and store the result back in `res`. `res ^= n`
Step 4: Right-shift `n` by 1 bit using the `>>=` operator. `n >>= 1`.
Step 5: Repeat steps 3 and 4 until `n` becomes 0.
Step 6: Return the value of `res`, which represents the minimum number of operations.

Key Insights

Insight 1: The operations are related to Gray code. Converting n to 0 using the minimum operations is equivalent to finding the Gray code representation of n.
Insight 2: The number of operations to convert n to 0 is given by the XOR of n with right-shifted n until n becomes 0. This elegantly utilizes bitwise operations to achieve the goal.
Insight 3: The Gray code sequence's property is that consecutive Gray codes differ by only one bit, which directly corresponds to the allowed operations in the problem statement.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

This problem involves: Dynamic Programming, Bit Manipulation, Memoization.

Companies

Asked at: Expedia, McKinsey, ServiceNow.

Minimum One Bit Operations to Make Integers Zero | LeetCode 1611

Hard

Dynamic Programming Bit Manipulation Memoization

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

Solution Code

Available in:

class Solution {
    public int minimumOneBitOperations(int n) {
        int res = 0;
        while (n > 0) {
            res ^= n;
            n >>= 1;
        }
        return res;
    }
}

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Problem Info

Description

Problem 1611: Minimum One Bit Operations to Make Integers Zero - Detailed explanation and solution approaches.

Topics

Dynamic Programming Bit Manipulation Memoization

Companies

ExpediaMcKinseyServiceNow

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:1611

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 `res` to 0. This variable will accumulate the minimum number of operations.
2
Step 2: Start a `while` loop that continues as long as `n` is greater than 0.
3
Step 3: Inside the loop, perform a bitwise XOR operation between `res` and `n`, and store the result back in `res`. `res ^= n`
4
Step 4: Right-shift `n` by 1 bit using the `>>=` operator. `n >>= 1`.
5
Step 5: Repeat steps 3 and 4 until `n` becomes 0.
6
Step 6: Return the value of `res`, which represents the minimum number of operations.

💡

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 O(log n) because the `while` loop iterates a number of times proportional to the number of bits in `n`, which is logarithmic with respect to `n`. Each iteration involves constant-time bitwise operations. The space complexity is O(1) because the algorithm uses only a fixed number of variables (`res` and `n`) regardless of the size of the input `n`.

📖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

Recursion with Memoization

A recursive approach can be used to solve the problem by breaking it down into smaller subproblems. Memoization can be employed to store the results of previously computed subproblems and avoid redundant calculations. The base case is when n = 0, where the result is 0. For n > 0, consider the two allowed operations and recursively calculate the minimum operations required after applying each operation.

Time: O(log n)Space: O(log n) - due to the recursion depth and memoization table.

Trade-offs: The recursive approach with memoization has the same time complexity as the iterative approach but incurs additional space overhead due to the recursion stack and memoization table. While conceptually straightforward, it might be less efficient in practice compared to the bit manipulation solution.

✨

Key Insights

1Insight 1: The operations are related to Gray code. Converting n to 0 using the minimum operations is equivalent to finding the Gray code representation of n.
2Insight 2: The number of operations to convert n to 0 is given by the XOR of n with right-shifted n until n becomes 0. This elegantly utilizes bitwise operations to achieve the goal.
3Insight 3: The Gray code sequence's property is that consecutive Gray codes differ by only one bit, which directly corresponds to the allowed operations in the problem statement.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 0. The `while` loop condition `n > 0` is immediately false, so the loop doesn't execute, and the initial value of `res`, which is 0, is returned. This correctly handles the case when the input is already 0.
Edge case 2: n = 1. The loop executes once. res becomes 1 (0 ^ 1), and n becomes 0 (1 >> 1). The loop terminates, and 1 is returned, which is the correct answer.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly applying the bit manipulation operations. It's crucial to understand the XOR and right-shift operations to correctly implement the Gray code conversion.

✗Common mistake 2: Trying to simulate the operations directly instead of recognizing the underlying Gray code pattern. Simulating each operation step-by-step would lead to a much more complex and less efficient 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: Gray codes are used in rotary encoders to minimize errors. This problem shows how Gray codes can be efficiently generated using bitwise operations.

🏦Real-world scenario 2: Gray codes are also used in communication systems to reduce the probability of errors due to single-bit changes during transmission. The algorithm can be applied when dealing with data requiring efficient transition between states.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

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: November 19, 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 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

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 "Minimum One Bit Operations to Make Integers Zero"?

For this hard-level Dynamic Programming 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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