How do you solve Divide Two Integers on LeetCode?

Divide Two Integers (LeetCode #29) can be solved using Brute Force or Optimal Solution (Bit Manipulation). The optimal approach is Optimal Solution (Bit Manipulation) with O(log n) time complexity and O(1) space complexity. Key insight: Recognizing that division can be simulated through repeated subtraction or bit manipulation can help in optimizing the solution.

What is the brute force solution for Divide Two Integers?

The brute force approach for Divide Two Integers is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach involves repeatedly subtracting the divisor from the dividend until the dividend is less than the divisor. This simulates the division process but is inefficient for large numbers. The optimal Optimal Solution (Bit Manipulation) approach improves this to O(log n).

What algorithm or data structure is used to solve Divide Two Integers?

Divide Two Integers uses the following concepts: Math, Bit Manipulation. The recommended patterns to study are: Bit Manipulation, Math, Greedy.

What are the interview tips for Divide Two Integers?

When you first see this problem, clarify the constraints and edge cases with the interviewer. Communicate your thought process clearly, explaining why you chose a particular approach and how it works. Mention edge cases like overflow, and consider discussing how you would test your solution.

What are common mistakes when solving Divide Two Integers?

Students often forget to handle the edge case of overflow when dividend is -2^31 and divisor is -1. Another common mistake is not correctly adjusting the sign of the quotient based on the input signs.

#29

Divide Two Integers

Medium

MathBit ManipulationBit ManipulationMathGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution (Bit Manipulation)★
Time	O(n)	O(log n)
Space	O(1)	O(1)

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Intuition

Time O(log n)Space O(1)

The optimal approach uses bit manipulation to perform division more efficiently. By leveraging the properties of binary numbers, we can double the divisor and count how many times it fits into the dividend, which reduces the number of iterations significantly.

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Algorithm

6 steps

1Step 1: Handle edge cases for overflow.
2Step 2: Determine the sign of the quotient based on the signs of dividend and divisor.
3Step 3: Work with positive values of dividend and divisor.
4Step 4: Use a loop to double the divisor and a counter to track how many times it fits into the dividend.
5Step 5: Subtract the largest doubled divisor from the dividend and add to the quotient accordingly.
6Step 6: Return the quotient, ensuring it is within the 32-bit signed integer range.

solution.py19 lines

1# Full working Python code with comments
2
3def divide(dividend: int, divisor: int) -> int:
4    # Edge case for overflow
5    if dividend == -2**31 and divisor == -1:
6        return 2**31 - 1
7    # Determine the sign of the quotient
8    sign = -1 if (dividend < 0) ^ (divisor < 0) else 1
9    # Work with positive values
10    dividend, divisor = abs(dividend), abs(divisor)
11    quotient = 0
12    while dividend >= divisor:
13        temp, multiple = divisor, 1
14        while dividend >= temp:
15            dividend -= temp
16            quotient += multiple
17            temp <<= 1  # Double the temp
18            multiple <<= 1  # Double the multiple
19    return sign * quotient

ℹ

Complexity note: The time complexity is O(log n) because we are effectively halving the dividend in each iteration by doubling the divisor, which reduces the number of iterations significantly compared to the brute force approach.

1Recognizing that division can be simulated through repeated subtraction or bit manipulation can help in optimizing the solution.
2Understanding how to handle edge cases, especially related to integer overflow, is crucial for robust solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.