Naive Approach

The most straightforward approach is to simply increment the starting number (1) until it reaches the target. However, this would only use the increment operation and ignore the double operation. This results in target - 1 moves. This is inefficient, especially when maxDoubles is significant and target is large.

Optimal Approach

The optimal approach is to work backward from the target. If the target is even, we can halve it if we have doubles available. If the target is odd, we must decrement it by 1. We repeat this process until we reach 1. This minimizes the number of moves.

Algorithm

1. Initialize moves = 0.
2. While target > 1:
   • If target is even and maxDoubles > 0:
     • target = target / 2
     • maxDoubles--
     • moves++
   • Else:
     • If maxDoubles == 0: moves += target - 1; break;
     • target--
     • moves++
3. Return moves.

Edge Cases

• If target is 1, the number of moves is 0.
• If maxDoubles is 0, the number of moves is target - 1.

Code (Java)

class Solution {
    public int minMoves(int target, int maxDoubles) {
        int moves = 0;
        while (target > 1) {
            if (target % 2 == 0 && maxDoubles > 0) {
                target /= 2;
                maxDoubles--;
                moves++;
            } else {
                if (maxDoubles == 0) {
                   moves += target - 1; 
                   break;
                }
                target--;
                moves++;
            }
        }
        return moves;
    }
}

Time and Space Complexity

• Time Complexity: O(log(target)) in the best case (many doubles), O(target) in the worst case (no doubles).
• Space Complexity: O(1) - constant extra space.