Integer Replacement

Problem Description

Given a positive integer n, you can apply one of the following operations:

1. If n is even, replace n with n / 2.
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.

Key Insights

• If n is even, the fastest way to reduce n is to divide by 2.
• If n is odd, we have two choices: either incrementing or decrementing, which leads to different paths.
• For odd numbers, incrementing n when the next number (n + 1) is more favorable in terms of subsequent divisions can reduce the number of operations.
• A recursive approach with memoization can optimize the solution by storing already computed results for specific n values.

Space and Time Complexity

Time Complexity: O(log n) on average due to the division operation, but can be higher due to odd number choices.
Space Complexity: O(log n) for the recursion stack in the worst case, plus O(n) for the memoization storage.

Solution

The solution can be approached using either a recursive method with memoization or an iterative method. The key is to reduce the number of operations by choosing the optimal path depending on whether n is odd or even.

1. If n is even, simply divide it by 2.
2. If n is odd, check if (n - 1) or (n + 1) leads to fewer operations, and choose the best path.

This can be implemented using a dictionary for memoization to avoid recalculating results for the same n.

Code Solutions