What is the time complexity of Adding Two Negabinary Numbers?

The optimal solution has a time complexity of O(max(n, m)).

What is the space complexity of Adding Two Negabinary Numbers?

The space complexity is O(max(n, m)).

What data structures are used to solve Adding Two Negabinary Numbers?

This problem uses Array, Math. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Adding Two Negabinary Numbers?

Insight 1: Negabinary addition requires handling carries differently than standard binary addition. Instead of carrying over a '1' when the sum is 2, we need to adjust for the negative base, potentially requiring multiple carry adjustments. Insight 2: The carry can be negative. In each position, the sum can be 0, 1, or 2 from the input digits and the carry from the previous step. When the sum is greater than or equal to 2, or less than 0, we need to adjust the current digit to either 0 or 1 and propogate the 'carry' which can be negative.

Which companies ask LeetCode 1073. Adding Two Negabinary Numbers in interviews?

This problem is frequently asked at Grab.

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Adding Two Negabinary Numbers - LeetCode 1073 Solution

Adding Two Negabinary Numbers - Complete Solution Guide

Adding Two Negabinary Numbers is LeetCode problem 1073, 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 numbers arr1 and arr2 in base -2 , return the result of adding them together. Each number is given in array format : as an array of 0s and 1s, from most significant bit to least significant bit. For example, arr = [1,1,0,1] represents the number (-2)^3 + (-2)^2 + (-2)^0 = -3 . A number arr in array, format is also guaranteed to have no leading zeros: either arr == [0] or arr[0] == 1 . Return the result of adding arr1 and arr2 in the same format: as an array of 0s and 1s with no leading

Detailed Explanation

The problem asks us to add two numbers represented in base -2 (negabinary) and return the result in the same negabinary format (an array of 0s and 1s). The input arrays `arr1` and `arr2` represent the numbers, with the most significant bit at the beginning of the array. The output should also be an array of 0s and 1s, representing the sum in base -2, without any leading zeros (unless the result is 0 itself, represented as [0]).

Solution Approach

The provided solution performs addition digit by digit, from the least significant bit to the most significant bit, similar to how we add numbers manually. It maintains a `carry` variable that stores the carry from the previous addition. Since we're working in base -2, the carry can be negative. The core of the algorithm involves adding the corresponding digits from the input arrays and the carry, then determining the resulting digit (0 or 1) and the new carry for the next iteration. After the addition is complete, the result array is trimmed to remove any leading zeros, and then reversed to give the result in the correct order.

Step-by-Step Algorithm

Step 1: Initialize pointers `i` and `j` to the last indices of `arr1` and `arr2` respectively. Initialize `carry` to 0 and `result` to an empty list.
Step 2: Iterate while either `i` or `j` is within the bounds of their respective arrays, or `carry` is not 0.
Step 3: Inside the loop, add the digit at `arr1[i]` to `carry` if `i` is within bounds, and decrement `i`. Do the same for `arr2[j]` and `j`.
Step 4: Calculate the current digit to be added to the `result` list by taking the least significant bit of the `carry` (carry & 1). This essentially gets remainder when carry is divided by 2.
Step 5: Calculate the new carry by dividing the previous carry by -2 (carry = -(carry >> 1)). This is the carry to the next digit. Since carry can be negative it's important to carry out the result to the most significant digits.
Step 6: After the loop finishes, remove any leading zeros from the `result` list, ensuring that the list contains at least one element (0).
Step 7: Reverse the `result` list to obtain the final negabinary representation.

Key Insights

Insight 1: Negabinary addition requires handling carries differently than standard binary addition. Instead of carrying over a '1' when the sum is 2, we need to adjust for the negative base, potentially requiring multiple carry adjustments.
Insight 2: The carry can be negative. In each position, the sum can be 0, 1, or 2 from the input digits and the carry from the previous step. When the sum is greater than or equal to 2, or less than 0, we need to adjust the current digit to either 0 or 1 and propogate the 'carry' which can be negative.
Insight 3: After addition, leading zeros must be trimmed. Also, handling the carry bit after exhausting the input arrays is critical to the correct result.

Complexity Analysis

Time Complexity: O(max(n, m))

Space Complexity: O(max(n, m))

Topics

This problem involves: Array, Math.

Companies

Asked at: Grab.

Adding Two Negabinary Numbers | LeetCode 1073

Medium

Array Math

Time: O(max(n, m))Space: O(max(n, m))

Solution Code

Available in:

class Solution {
    public int[] addNegabinary(int[] arr1, int[] arr2) {
        int i = arr1.length - 1;
        int j = arr2.length - 1;
        int carry = 0;
        java.util.List<Integer> result = new java.util.ArrayList<>();

        while (i >= 0 || j >= 0 || carry != 0) {
            if (i >= 0) {
                carry += arr1[i];
                i -= 1;
            }
            if (j >= 0) {
                carry += arr2[j];
                j -= 1;
            }
            
            result.add(carry & 1);
            carry = -(carry >> 1);
        }

        while (result.size() > 1 && result.get(result.size() - 1) == 0) {
            result.remove(result.size() - 1);
        }
        
        java.util.Collections.reverse(result);
        
        int[] resultArray = new int[result.size()];
        for (int k = 0; k < result.size(); k++) {
            resultArray[k] = result.get(k);
        }
        
        return resultArray;
    }
}

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Problem Info

Description

Problem 1073: Adding Two Negabinary Numbers - Detailed explanation and solution approaches.

Topics

Array Math

Companies

Grab

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1073

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

7 steps

1
Step 1: Initialize pointers `i` and `j` to the last indices of `arr1` and `arr2` respectively. Initialize `carry` to 0 and `result` to an empty list.
2
Step 2: Iterate while either `i` or `j` is within the bounds of their respective arrays, or `carry` is not 0.
3
Step 3: Inside the loop, add the digit at `arr1[i]` to `carry` if `i` is within bounds, and decrement `i`. Do the same for `arr2[j]` and `j`.
4
Step 4: Calculate the current digit to be added to the `result` list by taking the least significant bit of the `carry` (carry & 1). This essentially gets remainder when carry is divided by 2.
5
Step 5: Calculate the new carry by dividing the previous carry by -2 (carry = -(carry >> 1)). This is the carry to the next digit. Since carry can be negative it's important to carry out the result to the most significant digits.
6
Step 6: After the loop finishes, remove any leading zeros from the `result` list, ensuring that the list contains at least one element (0).
7
Step 7: Reverse the `result` list to obtain the final negabinary representation.

💡

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(max(n, m))

💾

Space Complexity

O(max(n, m))

📝

Why?

The time complexity is determined by the while loop which runs at most max(n, m) + k times, where k is the number of carry bits. Since k <= max(n, m) , the time complexity is O(max(n, m)). The space complexity is determined by the size of the result array which is at most max(n, m) + k, giving O(max(n, m)).

📖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

Conversion to Decimal, Addition, and Conversion Back

Convert both negabinary numbers to their decimal equivalents, add the decimal numbers, and then convert the decimal sum back to negabinary.

Time: O(n) for conversion to decimal, O(m) for conversion back to negabinary, where n is the combined length of input arrays, m is length of result. Typically O(n) where n=max(arr1.length, arr2.length)Space: O(1) in terms of input size, but can be O(log(N)) where N is the decimal value of the result, depending on implementation.

Trade-offs: Potentially easier to understand conceptually, but less efficient than the direct addition approach due to the overhead of conversions. Might be susceptible to integer overflow issues if the input arrays represent large numbers.

✨

Key Insights

1Insight 1: Negabinary addition requires handling carries differently than standard binary addition. Instead of carrying over a '1' when the sum is 2, we need to adjust for the negative base, potentially requiring multiple carry adjustments.
2Insight 2: The carry can be negative. In each position, the sum can be 0, 1, or 2 from the input digits and the carry from the previous step. When the sum is greater than or equal to 2, or less than 0, we need to adjust the current digit to either 0 or 1 and propogate the 'carry' which can be negative.
3Insight 3: After addition, leading zeros must be trimmed. Also, handling the carry bit after exhausting the input arrays is critical to the correct result.

📚

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: Both input arrays are [0]. The algorithm correctly returns [0]. The trailing zero removal handles this case.
Edge case 2: One input array is [0] and the other is non-zero. The algorithm correctly handles this case by effectively copying the non-zero array and adjusting for any carries.
Edge case 3: Large input arrays with significant carries. The algorithm is designed to handle carries correctly, even when they propagate through multiple digits.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling the carry in negabinary. Remembering that the carry can be negative and propagates differently than in standard binary addition is crucial.

✗Common mistake 2: Forgetting to remove leading zeros from the result array. The problem specifies that the output should not have leading zeros (unless the result is 0).

✗Common mistake 3: Incorrectly reversing the result array. The addition is performed from least significant bit to most significant bit, so the result needs to be reversed to match the problem's specified output format.

✅

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: Negabinary numbers are a specific case of a more general concept of mixed-radix numeral systems. While not directly used in typical computing, understanding negabinary arithmetic can be helpful in designing custom numeral systems for specialized applications, such as cryptography or error detection.

🏦Real-world scenario 2: The algorithm's core idea of digit-by-digit addition with carry propagation is applicable to various other number systems, like balanced ternary or other exotic radices. It demonstrates a general principle in computer arithmetic.

💡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.

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 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

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.

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 "Adding Two Negabinary Numbers"?

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

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