Minimum Moves to Reach Target Score

Medium

Asked by:

13 views

Topics:

Greedy Algorithms

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

Try coding it in LeetCode

Minimum Moves to Reach Target Score

Medium

Asked by:

13 views

Topics:

Greedy Algorithms

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible ranges for the target score and the starting number?
Can the target score or starting number be zero or negative?
If it is impossible to reach the target score, what should be returned?
Does multiplying by 2 count as a move, even if it doesn't get me closer to the target?
Is the starting number always less than or equal to the target score?

Brute Force Solution

Approach

The brute force method for finding the minimum moves involves exploring every possible sequence of actions. We simulate performing different operations (either doubling or decrementing) step-by-step, until the target is reached. We then pick the sequence that required the fewest moves.

Here's how the algorithm would work step-by-step:

Start with the initial score and explore both possible moves: doubling it, or subtracting one.
For each of those new scores, again explore both possible moves: doubling it, or subtracting one.
Keep repeating this process of branching out, exploring every move from every possible score reached so far.
Continue doing this until you either reach the target score or go past it in a way that makes it clear you can't reach it efficiently (e.g., you subtract one to get to 0, and now need to add one many times).
Keep track of the number of moves it took to reach the target score for each different path you explored.
Once you've explored enough paths, compare the number of moves it took for each one to reach the target score.
Choose the path that took the fewest moves. That's your answer.

Code Implementation

def minimum_moves_to_reach_target_score_brute_force(
    current_score_start,
    target_score,
    max_doubles
):
    queue = [(current_score_start, 0)]
    minimum_moves = float('inf')

    while queue:
        current_score, number_of_moves = queue.pop(0)

        if current_score == target_score:
            minimum_moves = min(minimum_moves, number_of_moves)
            continue

        if current_score > target_score:
            number_of_moves_needed =
            number_of_moves + (current_score - target_score)
            minimum_moves = min(minimum_moves, number_of_moves_needed)
            continue

        # Explore doubling, but only if we have doubles left.
        if max_doubles > 0:

            queue.append((current_score * 2, number_of_moves + 1))

        # Always explore decrementing.

        queue.append((current_score - 1, number_of_moves + 1))

    return minimum_moves

Big(O) Analysis

Time Complexity

O(2^target) – The brute force approach explores all possible sequences of doubling and decrementing the score until the target is reached. In the worst-case scenario, particularly when the target value is large and the starting value is small, the algorithm essentially builds a binary tree where each node represents a possible score and each branch represents either doubling or decrementing. The depth of this tree can grow proportionally to the target value. Therefore, the number of nodes explored, and hence the time complexity, grows exponentially with the target, resulting in a time complexity of O(2^target).

Space Complexity

O(2^target) – The brute force method explores every possible sequence of moves using a branching approach. Each score can potentially lead to two new scores (doubling or decrementing), creating a binary tree-like structure. In the worst case, where we explore all possible paths until reaching or exceeding the target, the number of nodes in this tree can grow exponentially with the target value. Therefore, to store the intermediate scores and the moves taken to reach them for all paths, the algorithm uses auxiliary space proportional to the number of nodes, which is O(2^target). This space is used implicitly to store the call stack during the recursive exploration of different paths.

Optimal Solution

Approach

To minimize the moves, we want to work backward from the target. The optimal strategy involves prioritizing division by two whenever possible because it reduces the score much faster than adding one.

Here's how the algorithm would work step-by-step:

Start with the target score and the number of allowed moves.
Check if the target is divisible by two. If it is, divide the target by two. This reduces the target faster than adding one.
If the target is not divisible by two, add one to the score until it is. Count the number of 'add one' moves it takes to make the target divisible by two.
Each time we divide the target by two, we count it as one move.
Repeat steps two and three until the target score becomes one.
The total number of division and addition moves is the minimum number of moves needed.

Code Implementation

def minimum_moves_to_reach_target_score(target_score, max_doubles):
    moves_count = 0

    while target_score > 1:
        if max_doubles > 0 and target_score % 2 == 0:
            # Prioritize dividing by two when possible.
            target_score //= 2
            max_doubles -= 1
            moves_count += 1

        elif max_doubles == 0:
            # When doubles are exhausted, only addition remains.
            moves_count += target_score - 1
            target_score = 1

        else:
            # If not divisible by 2, increment to make it so.
            target_score -= 1
            moves_count += 1

    return moves_count

Big(O) Analysis

Time Complexity

O(log target) – The algorithm's runtime is primarily determined by how many times the target score can be divided by two. Each division effectively halves the target value. The number of additions needed to make the target divisible by two is at most one per iteration, as we only add one at a time to ensure divisibility. Since the target is repeatedly halved, the number of iterations is proportional to the logarithm base 2 of the initial target value. Thus, the time complexity is O(log target).

Space Complexity

O(1) – The provided algorithm operates iteratively and does not utilize any auxiliary data structures like lists, hash maps, or trees. The algorithm only uses a few variables to store the current target score and the move count. Therefore, the space required remains constant regardless of the initial target value. The space complexity is O(1).

Edge Cases

Case	How to Handle
target is 1	Return 0 immediately as no moves are needed to reach 1.
maxDoubles is 0 and target is greater than 1	Only increment operations are possible; return target - 1.
target is negative	Return -1 or throw an error, as the problem statement implies positive values; clarify assumptions beforehand.
maxDoubles is negative	Return -1 or throw an error, as maxDoubles should be non-negative; clarify assumptions beforehand.
target is close to the maximum integer value and maxDoubles is high, causing possible overflow during doubling.	Use long or consider that integer overflow may cause calculation errors, and appropriate measures such as modular arithmetic if possible, need to be implemented.
target is much larger than maxDoubles	Optimize for the case where decrementing is more efficient by checking `target % 2 == 0` before doubling
maxDoubles is extremely large compared to target	Even if doubling is allowed, optimize by decrementing until target becomes closer to a power of 2 if appropriate.
target equals Integer.MAX_VALUE and maxDoubles is also very large	Handle Integer.MAX_VALUE and avoid unnecessary calculations, possibly optimizing decrements instead of doubling where it's more efficient.