Introduction

In this blog post, we will explore how to solve the LeetCode problem "397" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a positive integer n, you can apply one of the following operations: If n is even, replace n with n / 2. If n is odd, replace n with either n + 1 or n - 1. Return the minimum number of operations needed for n to become 1. Example 1: Input: n = 8 Output: 3 Explanation: 8 -> 4 -> 2 -> 1 Example 2: Input: n = 7 Output: 4 Explanation: 7 -> 8 -> 4 -> 2 -> 1 or 7 -> 6 -> 3 -> 2 -> 1 Example 3: Input: n = 4 Output: 2 Constraints: 1 <= n <= 231 - 1

Explanation

Here's a breakdown of the approach, complexity, and code:

• High-Level Approach:

  • Greedy Strategy: Prioritize division by 2 whenever possible. This leads to faster convergence towards 1.
  • Odd Number Handling: For odd numbers, decide whether to increment or decrement based on the following bitwise insight: If the two least significant bits are '11', incrementing is usually better (leading to more trailing zeros). If the two least significant bits are '01', decrementing is better (again, leading to faster division). Edge case: when n is 3, decrementing is the only path.
  • Bitwise Operations: Leverage bitwise operations for efficiency (right shifts for division, bitwise AND for checking parity).

• Complexity:

  • Runtime Complexity: O(log n) - Each operation roughly halves the number in the best case.
  • Storage Complexity: O(1) - Constant extra space.

• Python Code:

Code

    def integerReplacement(n: int) -> int:
    """
    Calculates the minimum number of operations to transform n to 1 using the given rules.
    """
    operations = 0
    while n > 1:
        if n % 2 == 0:
            n //= 2
        else:
            if n == 3:
                n -= 1
            elif (n & 2) == 0:  # Equivalent to (n % 4 == 1)
                n -= 1
            else:
                n += 1
        operations += 1
    return operations