Introduction

In this blog post, we will explore how to solve the LeetCode problem "2139" 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

You are playing a game with integers. You start with the integer 1 and you want to reach the integer target. In one move, you can either: Increment the current integer by one (i.e., x = x + 1). Double the current integer (i.e., x = 2 * x). You can use the increment operation any number of times, however, you can only use the double operation at most maxDoubles times. Given the two integers target and maxDoubles, return the minimum number of moves needed to reach target starting with 1. Example 1: Input: target = 5, maxDoubles = 0 Output: 4 Explanation: Keep incrementing by 1 until you reach target. Example 2: Input: target = 19, maxDoubles = 2 Output: 7 Explanation: Initially, x = 1 Increment 3 times so x = 4 Double once so x = 8 Increment once so x = 9 Double again so x = 18 Increment once so x = 19 Example 3: Input: target = 10, maxDoubles = 4 Output: 4 Explanation: Initially, x = 1 Increment once so x = 2 Double once so x = 4 Increment once so x = 5 Double again so x = 10 Constraints: 1 <= target <= 109 0 <= maxDoubles <= 100

Explanation

Here's the solution to the problem:

• Reverse Thinking: Instead of incrementing and doubling from 1 to the target, work backward from the target to 1.
• Prioritize Halving: If the target is even and maxDoubles is available, halve the target (reverse of doubling). This is generally more efficient than incrementing repeatedly.

• Handle Odd Numbers: If the target is odd, decrement it by 1 (reverse of incrementing), as it must have been reached by incrementing the previous even number.

• Runtime Complexity: O(log(target)) - due to the halving operations.

• Storage Complexity: O(1) - constant extra space.

Code

    def minMoves(target: int, maxDoubles: int) -> int:
    moves = 0
    while target > 1:
        if maxDoubles > 0 and target % 2 == 0:
            target //= 2
            maxDoubles -= 1
            moves += 1
        elif maxDoubles == 0:
            moves += target - 1
            target = 1
        else:
            target -= 1
            moves += 1
    return moves