What is the time complexity of Complement of Base 10 Integer?

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

What data structures are used to solve Complement of Base 10 Integer?

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

What are the key insights for solving Complement of Base 10 Integer?

The core problem lies in *selectively* flipping bits. We are not performing a standard bitwise NOT (`~`) operation that would flip all bits in an integer's fixed-size storage (e.g., 32-bit), but rather only the bits that constitute the number itself in its minimal positive binary form. The solution cleverly generates a 'mask' of all ones that has the exact same number of bits as the input number `n`. This mask is then used with the XOR operator to flip precisely those bits without affecting any leading (insignificant) zeros.

Which companies ask LeetCode 1009. Complement of Base 10 Integer in interviews?

This problem is frequently asked at Cloudera, Snap.

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Complement of Base 10 Integer - LeetCode 1009 Solution

Complement of Base 10 Integer - Complete Solution Guide

Complement of Base 10 Integer is LeetCode problem 1009, 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

The complement of an integer is the integer you get when you flip all the 0 's to 1 's and all the 1 's to 0 's in its binary representation. For example, The integer 5 is "101" in binary and its complement is "010" which is the integer 2 . Given an integer n , return its complement . Example 1: Input: n = 5 Output: 2 Explanation: 5 is "101" in binary, with complement "010" in binary, which is 2 in base-10. Example 2: Input: n = 7 Output: 0 Explanation: 7 is "111" in binary, with complement "000

Detailed Explanation

This problem asks us to find the "complement" of a given base 10 integer `n`. The key here is understanding what "complement" means in this specific context. It's not the mathematical complement or the two's complement used for negative numbers. Instead, we are instructed to take the binary representation of `n` and flip all its `0`s to `1`s and `1`s to `0`s. The crucial detail, subtly conveyed by the example, is that we only flip the *significant* bits. For instance, `n = 5` is `"101"` in binary. Its complement is `"010"`, which translates back to `2` in base 10. If we were to consider a fixed-width 32-bit representation (e.g., `00...0101`), flipping all bits would yield `11...1010`, which is a large negative number, not `2`. This distinction between fixed-width bitwise NOT and flipping *only* the relevant bits is where the problem gets interesting. The challenge, therefore, is to determine the precise length of the binary string we need to consider. We can't just use a language's built-in bitwise NOT operator (`~`) because that typically operates on the underlying fixed-size integer type and would flip leading zeros as well. For example, in Python, `~5` yields `-6`, which is not `2`. We need a method to confine the bit-flipping operation to just the bits that constitute the number itself, ignoring any leading zeros that are merely artifacts of the integer's storage size. The problem statement also presents a simple edge case: what is the complement of `0`? Its binary is `0`, so its complement would be `1`.

Solution Approach

The provided solution elegantly solves this by first identifying the number of bits required to represent `n`, and then constructing a 'mask' of all ones with that exact length. Let's trace it: first, it explicitly handles the `n=0` case, returning `1`. For any `n > 0`, it determines `num_bits`, which is the count of significant bits in `n`'s binary representation. This is achieved by repeatedly right-shifting `n` (`temp >>= 1`) until it becomes zero, incrementing `num_bits` each time. For `n=5` (binary `101`), `num_bits` would become `3`. Once `num_bits` is known, a mask is created using `mask = (1 << num_bits) - 1`. This is a powerful bit manipulation trick: `1 << num_bits` gives a `1` followed by `num_bits` zeros (e.g., `1 << 3` is `1000` binary, or `8` decimal). Subtracting `1` from this yields a number consisting of `num_bits` ones (e.g., `1000 - 1 = 111` binary, or `7` decimal). Finally, `n` is XORed (`^`) with this mask. The XOR operation has the property that `X ^ 1 = ~X` (flips the bit) and `X ^ 0 = X` (leaves the bit unchanged). Since our mask consists solely of ones in the positions corresponding to `n`'s significant bits, XORing `n` with this mask effectively flips exactly those bits, producing the desired complement. Any bits beyond `num_bits` are implicitly `0` in `n` and `0` in our mask (assuming sufficient integer width for `mask`), so `0 ^ 0 = 0`, preserving the effect of only flipping relevant bits.

Step-by-Step Algorithm

**Step 1: Handle the base case:** If the input `n` is 0, return 1.
**Step 2: Determine the number of bits:** Find the number of bits required to represent `n` in binary. This can be done efficiently using a loop (as shown) or using `log2(n) + 1` (logarithm method).
**Step 3: Create a bitmask:** Generate a mask with all bits set to 1, having the same number of bits as `n`. This is done by left-shifting 1 by the number of bits and subtracting 1. This creates a number where all bits are 1s (e.g., if `num_bits` is 3, the mask will be 111 in binary, which is 7 in decimal).
**Step 4: Calculate the complement:** Perform a bitwise XOR operation between the input `n` and the created mask. The XOR operation will flip each bit, thus generating the complement.

Key Insights

The core problem lies in *selectively* flipping bits. We are not performing a standard bitwise NOT (`~`) operation that would flip all bits in an integer's fixed-size storage (e.g., 32-bit), but rather only the bits that constitute the number itself in its minimal positive binary form.
The solution cleverly generates a 'mask' of all ones that has the exact same number of bits as the input number `n`. This mask is then used with the XOR operator to flip precisely those bits without affecting any leading (insignificant) zeros.
An efficient way to construct a mask of `k` ones is `(1 << k) - 1`. This technique is valuable in bit manipulation problems for creating specific bit patterns quickly. The `while` loop simultaneously identifies `k` (the number of significant bits) and prepares for this mask generation.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

This problem involves: Bit Manipulation.

Companies

Asked at: Cloudera, Snap.

Complement of Base 10 Integer | LeetCode 1009

Easy

Bit Manipulation

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

Solution Code

Available in:

class Solution {
    public int bitwiseComplement(int n) {
        if (n == 0) return 1;
        int numBits = (int)(Math.log(n) / Math.log(2)) + 1;
        int mask = (1 << numBits) - 1;
        return n ^ mask;
    }
}

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Problem Info

Description

Problem 1009: Complement of Base 10 Integer - Detailed explanation and solution approaches.

Topics

Bit Manipulation

Companies

ClouderaSnap

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:1009

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

4 steps

1
**Step 1: Handle the base case:** If the input `n` is 0, return 1.
2
**Step 2: Determine the number of bits:** Find the number of bits required to represent `n` in binary. This can be done efficiently using a loop (as shown) or using `log2(n) + 1` (logarithm method).
3
**Step 3: Create a bitmask:** Generate a mask with all bits set to 1, having the same number of bits as `n`. This is done by left-shifting 1 by the number of bits and subtracting 1. This creates a number where all bits are 1s (e.g., if `num_bits` is 3, the mask will be 111 in binary, which is 7 in decimal).
4
**Step 4: Calculate the complement:** Perform a bitwise XOR operation between the input `n` and the created mask. The XOR operation will flip each bit, thus generating the complement.

💡

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 determined by the loop to find the number of bits in the input number, which is logarithmic with respect to the input `n` (the number of iterations is proportional to the number of bits). The space complexity is constant because we are using only a few 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

Using bit manipulation with a loop directly

Instead of creating a mask, you can iteratively flip each bit of the number using bitwise operations within a loop. This method might be slightly less efficient for larger numbers.

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

Trade-offs: This approach might be more intuitive for beginners but slightly less efficient than the mask approach.

✨

Key Insights

1The core problem lies in *selectively* flipping bits. We are not performing a standard bitwise NOT (`~`) operation that would flip all bits in an integer's fixed-size storage (e.g., 32-bit), but rather only the bits that constitute the number itself in its minimal positive binary form.
2The solution cleverly generates a 'mask' of all ones that has the exact same number of bits as the input number `n`. This mask is then used with the XOR operator to flip precisely those bits without affecting any leading (insignificant) zeros.
3An efficient way to construct a mask of `k` ones is `(1 << k) - 1`. This technique is valuable in bit manipulation problems for creating specific bit patterns quickly. The `while` loop simultaneously identifies `k` (the number of significant bits) and prepares for this mask generation.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: n = 0. The solution correctly handles this by returning 1, which is the complement of 0.
Edge case 2: n = a power of 2 (e.g., 4, 8, 16). The algorithm works correctly for these cases because the mask creation will correctly generate the appropriate number of 1s in the mask to flip all the bits of the input.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the number of bits needed to represent the integer. Off-by-one errors are common here.

✗Common mistake 2: Not handling the edge case where n=0. Failing to account for this will lead to incorrect results.

✅

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: Low-level programming and embedded systems, where bit manipulation is frequently used for efficient data processing.

🏦Real-world scenario 2: Network programming, where bit manipulation is useful for manipulating network packets and managing flags.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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: September 18, 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

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

@TechLeadJul 14

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

@DevExpert92Jul 7

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

@AlgoMasterJun 30

❓

Frequently Asked Questions

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 "Complement of Base 10 Integer"?

For this easy-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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