Minimum Moves to Reach Target Score

• State Representation: Model the problem as a shortest path on a graph. Each state is defined by a pair (current_value, doubles_used).
• Initialization: Start a Breadth-First Search (BFS) from the initial state (1, 0) with moves = 0.
• Data Structures: Use a queue to manage states to visit and a Set to keep track of visited states to avoid redundant computations.
• Traversal:
  1. Dequeue a state (curr, doubles).
  2. If curr equals target, return the current number of moves.
  3. Generate next possible states:
     • Increment: A new state (curr + 1, doubles).
     • Double: A new state (curr * 2, doubles + 1), only if doubles < maxDoubles.
  4. For each new state, if it's within the target boundary and has not been visited, add it to the queue and the visited set.
• Level-by-Level: Process the BFS level by level, incrementing the moves count after each level is fully processed. This ensures finding the shortest path.

We can solve this problem by exploring all possible sequences of operations in a breadth-first manner. This guarantees that we find the path with the minimum number of moves first. The state in our search needs to track not only the current number but also the number of double operations used, as this is a limited resource.

• We initialize a queue with the starting state (value=1, doubles_used=0).
• We also use a visited set to store states we have already processed to avoid getting into cycles or doing redundant work. A state is uniquely identified by both its value and the number of doubles used to reach it.
• The BFS proceeds in levels. Each level corresponds to one additional move. In each step, we explore all possible next states from the states in the current level.
• From a state (curr, doubles), we can transition to (curr + 1, doubles) (increment) and (curr * 2, doubles + 1) (double), provided the new value does not exceed target and we have not exhausted maxDoubles.
• The first time we reach the target value, we are guaranteed to have done so in the minimum number of moves. However, due to the large constraints on target, this approach is not practical.

1import java.util.Queue;2import java.util.LinkedList;3import java.util.Set;4import java.util.HashSet;5import java.util.Objects;6 7class Solution {8    // This approach is illustrative but will time out for large inputs.9    class State {10        long value;11        int doubles;12 13        State(long value, int doubles) {14            this.value = value;15            this.doubles = doubles;16        }17 18        @Override19        public boolean equals(Object o) {20            if (this == o) return true;21            if (o == null || getClass() != o.getClass()) return false;22            State state = (State) o;23            return value == state.value && doubles == state.doubles;24        }25 26        @Override27        public int hashCode() {28            return Objects.hash(value, doubles);29        }30    }31 32    public int minMoves(int target, int maxDoubles) {33        if (target == 1) {34            return 0;35        }36 37        Queue<State> queue = new LinkedList<>();38        Set<State> visited = new HashSet<>();39 40        State startState = new State(1, 0);41        queue.add(startState);42        visited.add(startState);43 44        int moves = 0;45 46        while (!queue.isEmpty()) {47            int levelSize = queue.size();48            moves++;49            for (int i = 0; i < levelSize; i++) {50                State current = queue.poll();51 52                // Try incrementing53                long nextValInc = current.value + 1;54                if (nextValInc == target) {55                    return moves;56                }57                State nextStateInc = new State(nextValInc, current.doubles);58                if (nextValInc < target && !visited.contains(nextStateInc)) {59                    visited.add(nextStateInc);60                    queue.add(nextStateInc);61                }62 63                // Try doubling64                if (current.doubles < maxDoubles) {65                    long nextValDouble = current.value * 2;66                    if (nextValDouble == target) {67                        return moves;68                    }69                    State nextStateDouble = new State(nextValDouble, current.doubles + 1);70                    if (nextValDouble < target && !visited.contains(nextStateDouble)) {71                        visited.add(nextStateDouble);72                        queue.add(nextStateDouble);73                    }74                }75            }76        }77        return -1; // Should not be reached78    }79}

• Strategy: Work backward from target to 1.
• Inverse Operations: The forward operations increment and double correspond to backward operations decrement and halve.
• Initialization: Start with moves = 0 and current_value = target.
• Greedy Loop: While target > 1 and maxDoubles > 0:
  • If target is even, it's always optimal to have reached it from target / 2. Perform the reverse operation: target /= 2, decrement maxDoubles, and increment moves.
  • If target is odd, the only possible previous step was an increment from target - 1. Perform the reverse: target--, and increment moves.
• Final Steps: After the loop, if target > 1, it means we have exhausted all maxDoubles. The remaining distance to 1 must be covered by increments. Add target - 1 to the total moves.

The logic for this greedy strategy is based on the observation that to minimize moves, we should prioritize the double operation, which has a much larger effect than an increment. When working backward from target:

• If the current target is odd, it could not have been the result of a double operation. Therefore, the last move must have been an increment. So, we reverse this by decrementing target by 1, which costs one move.
• If the current target is even, it could have been reached from target - 1 (by increment) or target / 2 (by double). Reaching target from target / 2 is always more efficient than reaching it from target - 1, as it gets us closer to our goal of 1 in a single step. Thus, if we have maxDoubles left, we should always choose to reverse the double operation by halving the target.

This process is repeated until target becomes 1 or we run out of maxDoubles. If we run out of maxDoubles before target reaches 1, the remaining steps must all be increments, so we simply add target - 1 to our move count.

1class Solution {2    public int minMoves(int target, int maxDoubles) {3        int moves = 0;4        5        // We can't use doubles once target is 1 or we run out of maxDoubles6        while (target > 1 && maxDoubles > 0) {7            if (target % 2 == 0) {8                // If target is even, it's optimal to divide by 2.9                // This corresponds to a "double" operation in the forward direction.10                target /= 2;11                maxDoubles--;12            } else {13                // If target is odd, we must have reached it by an increment.14                // So we reverse it by decrementing.15                target--;16            }17            moves++;18        }19        20        // If target is still greater than 1, it means we've used up all maxDoubles.21        // The rest of the moves must be increments from 1 to the current target.22        // The number of such moves is target - 1.23        if (target > 1) {24            moves += (target - 1);25        }26        27        return moves;28    }29}