What is the time complexity of Divide Two Integers?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Divide Two Integers?

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

What are the key insights for solving Divide Two Integers?

Insight 1: Bit manipulation can be used to simulate division. Specifically, left bit shifting ( >) is equivalent to dividing by powers of 2. Insight 2: Repeated subtraction can be done efficiently by subtracting increasingly larger multiples of the divisor (powers of 2) from the dividend until the dividend becomes smaller than the divisor.

Which companies ask LeetCode 29. Divide Two Integers in interviews?

This problem is frequently asked at Adobe, Amazon, Apple, Bloomberg, Meta and 6 more companies.

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Divide Two Integers - LeetCode 29 Solution

Divide Two Integers - Complete Solution Guide

Divide Two Integers is LeetCode problem 29, 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 dividend and divisor , divide two integers without using multiplication, division, and mod operator. The integer division should truncate toward zero, which means losing its fractional part. For example, 8.345 would be truncated to 8 , and -2.7335 would be truncated to -2 . Return the quotient after dividing dividend by divisor . Note: Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range: [−2 31 , 2 31 − 1]

Detailed Explanation

The problem asks us to implement integer division without using multiplication, division, or the modulo operator. Given two integers, `dividend` and `divisor`, we need to find the quotient of their division, truncating towards zero. The solution must also handle the 32-bit signed integer range constraint. If the quotient exceeds the maximum value (231 - 1) or falls below the minimum value (-231), we should return the respective limit.

Solution Approach

The provided solution employs a bit manipulation technique to simulate the division process. It iteratively subtracts multiples of the divisor (shifted by powers of 2) from the dividend, effectively determining how many times the divisor fits into the dividend. The algorithm first handles the edge case where dividing the minimum integer by -1 would result in integer overflow. It then determines the sign of the quotient. After that, both the dividend and divisor are converted to their absolute values to simplify the iterative subtraction. The core of the algorithm lies in iterating through the bits from the most significant bit (31st bit) down to the least significant bit. In each iteration, it checks if the divisor, shifted left by `i` bits (effectively multiplying it by 2i), is less than or equal to the dividend. If it is, it means that 2i multiples of the divisor can be subtracted from the dividend. The quotient is then updated by adding 2i, and the dividend is reduced by the shifted divisor. Finally, the solution returns the quotient multiplied by the correct sign, handling potential overflow issues at the end (in the C++ solution).

Step-by-Step Algorithm

Step 1: Handle the overflow edge case: If the dividend is INT_MIN and the divisor is -1, return INT_MAX.
Step 2: Determine the sign of the quotient. If the signs of the dividend and divisor are different, the sign is negative; otherwise, it's positive.
Step 3: Convert both dividend and divisor to their absolute values to simplify the subtraction process. This might involve casting to a `long` to prevent overflow of INT_MIN.
Step 4: Initialize the quotient to 0.
Step 5: Iterate from i = 31 down to 0. In each iteration:
Step 6: Check if the divisor shifted left by `i` bits (divisor << i) is less than or equal to the dividend.
Step 7: If it is, update the quotient by adding 2i (quotient |= (1 << i)) and reduce the dividend by divisor << i (dividend -= divisor << i).
Step 8: After the loop finishes, return the quotient multiplied by the sign.
Step 9: (In C++) perform a check to ensure that `sign * quotient` does not cause an overflow, and return INT_MAX or INT_MIN if it does.

Key Insights

Insight 1: Bit manipulation can be used to simulate division. Specifically, left bit shifting (<<) is equivalent to multiplying by powers of 2, and right bit shifting (>>) is equivalent to dividing by powers of 2.
Insight 2: Repeated subtraction can be done efficiently by subtracting increasingly larger multiples of the divisor (powers of 2) from the dividend until the dividend becomes smaller than the divisor.
Insight 3: Handling edge cases like division by zero (which is prohibited by the problem constraints, but good practice to acknowledge) and integer overflow is crucial for a robust solution. The most significant edge case is when dividend is INT_MIN and the divisor is -1, resulting in INT_MAX + 1.

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Math, Bit Manipulation.

Companies

Asked at: Adobe, Amazon, Apple, Bloomberg, Meta, Microsoft, Nextdoor, TikTok, Uber, Yahoo, Zepto.

Divide Two Integers | LeetCode 29

Medium

Math Bit Manipulation

Time: O(1)Space: O(1)

Solution Code

Available in:

class Solution {
    public int divide(int dividend, int divisor) {
        int INT_MAX = 2147483647;
        int INT_MIN = -2147483648;

        if (dividend == INT_MIN && divisor == -1) {
            return INT_MAX;
        }

        int sign = ((dividend < 0) != (divisor < 0)) ? -1 : 1;

        long dvd = Math.abs((long)dividend);
        long dvs = Math.abs((long)divisor);

        int quotient = 0;
        for (int i = 31; i >= 0; i--) {
            if ((dvd >> i) >= dvs) {
                quotient |= (1 << i);
                dvd -= (dvs << i);
            }
        }

        return sign * quotient;
    }
}

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Problem Info

Description

Problem 29: Divide Two Integers - Detailed explanation and solution approaches.

Topics

Math Bit Manipulation

Companies

AdobeAmazonAppleBloombergMetaMicrosoftNextdoorTikTok

Stats

Difficulty:Medium

Languages:4

Companies:11

Problem #:29

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

9 steps

1
Step 1: Handle the overflow edge case: If the dividend is INT_MIN and the divisor is -1, return INT_MAX.
2
Step 2: Determine the sign of the quotient. If the signs of the dividend and divisor are different, the sign is negative; otherwise, it's positive.
3
Step 3: Convert both dividend and divisor to their absolute values to simplify the subtraction process. This might involve casting to a `long` to prevent overflow of INT_MIN.
4
Step 4: Initialize the quotient to 0.
5
Step 5: Iterate from i = 31 down to 0. In each iteration:
6
Step 6: Check if the divisor shifted left by `i` bits (divisor << i) is less than or equal to the dividend.
7
Step 7: If it is, update the quotient by adding 2i (quotient |= (1 << i)) and reduce the dividend by divisor << i (dividend -= divisor << i).
8
Step 8: After the loop finishes, return the quotient multiplied by the sign.
9
Step 9: (In C++) perform a check to ensure that `sign * quotient` does not cause an overflow, and return INT_MAX or INT_MIN if it does.

💡

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(1)

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

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The loop iterates a fixed number of times (32), regardless of the input values. Therefore, the time complexity is constant. The algorithm uses only a few constant space variables to store the sign, absolute values, and the quotient. Hence, the space complexity is also constant.

📖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

Repeated Subtraction

Repeatedly subtract the divisor from the dividend until the dividend becomes smaller than the divisor. Count the number of subtractions to get the quotient.

Time: O(N), where N is the quotient. In the worst case, N could be close to 2^31.Space: O(1)

Trade-offs: This approach is simple to understand, but very inefficient for large dividends and small divisors. The bit manipulation approach is significantly faster.

✨

Key Insights

1Insight 1: Bit manipulation can be used to simulate division. Specifically, left bit shifting (<<) is equivalent to multiplying by powers of 2, and right bit shifting (>>) is equivalent to dividing by powers of 2.
2Insight 2: Repeated subtraction can be done efficiently by subtracting increasingly larger multiples of the divisor (powers of 2) from the dividend until the dividend becomes smaller than the divisor.
3Insight 3: Handling edge cases like division by zero (which is prohibited by the problem constraints, but good practice to acknowledge) and integer overflow is crucial for a robust solution. The most significant edge case is when dividend is INT_MIN and the divisor is -1, resulting in INT_MAX + 1.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: `dividend = INT_MIN, divisor = -1`. The result would be INT_MAX + 1, which overflows. The solution explicitly handles this case and returns INT_MAX.
Edge case 2: `divisor = 0`. The problem statement specifies that the divisor will not be zero, so no handling is added to specifically address this possibility.
Edge case 3: `dividend = 0`. The algorithm will correctly return 0 in this case.
Edge case 4: `dividend = INT_MIN, divisor = 1`. The dividend remains INT_MIN and the sign is negative; this results in the correct answer of INT_MIN being returned.
Edge case 5: `dividend = INT_MAX, divisor = -1`. The dividend becomes INT_MAX, the sign is negative, and the function will return -INT_MAX which is a valid return value.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to handle the edge case of `dividend = INT_MIN` and `divisor = -1`. This can lead to integer overflow and incorrect results.

✗Common mistake 2: Not correctly handling the sign of the quotient. Ensure the sign is correctly calculated and applied to the final result.

✗Common mistake 3: Incorrectly shifting bits, leading to wrong multiples of the divisor being subtracted.

✗Common mistake 4: Assuming time complexity is O(32) and incorrectly labeling it as O(log n). The loop always executes 32 times, which is a constant. Therefore time complexity is O(1).

✅

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 system programming where division hardware is unavailable or restricted. Bit manipulation can be used to implement division algorithms in embedded systems or device drivers.

🏦Real-world scenario 2: Cryptography algorithms often rely on modular arithmetic and bitwise operations. The principles used in this division algorithm can be applied to implement modular division efficiently.

💡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

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 30, 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 13

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

@DevExpert92Jul 6

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

@AlgoMasterJun 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 "Divide Two Integers"?

For this medium-level Math 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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