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Minimum Knight Moves

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In an infinite chess board with coordinates from -infinity to +infinity, you have a knight at square [0, 0].

A knight has 8 possible moves it can make, as illustrated below. Each move is two squares in a cardinal direction, then one square in an orthogonal direction.

Return the minimum number of steps to move the knight to the square [x, y]. It is guaranteed the answer exists.

Example 1:

Input: x = 2, y = 1
Output: 1
Explanation: [0, 0] -> [2, 1]

Example 2:

Input: x = 5, y = 5
Output: 4
Explanation: [0, 0] -> [2, 1] -> [4, 2] -> [3, 4] -> [5, 5]

Constraints:

-300 <= x, y <= 300
The answer is guaranteed to exist.

Minimum Knight Moves

Medium

1 view

Topics:

Graphs

In an infinite chess board with coordinates from -infinity to +infinity, you have a knight at square [0, 0].

A knight has 8 possible moves it can make, as illustrated below. Each move is two squares in a cardinal direction, then one square in an orthogonal direction.

Return the minimum number of steps to move the knight to the square [x, y]. It is guaranteed the answer exists.

Example 1:

Input: x = 2, y = 1
Output: 1
Explanation: [0, 0] -> [2, 1]

Example 2:

Input: x = 5, y = 5
Output: 4
Explanation: [0, 0] -> [2, 1] -> [4, 2] -> [3, 4] -> [5, 5]

Constraints:

-300 <= x, y <= 300
The answer is guaranteed to exist.

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:

Are the target coordinates guaranteed to be reachable by the knight?
Can the target coordinates be negative, zero, or non-integer values?
What is the maximum magnitude of the target coordinates (e.g., is there a limit to how far the knight can be asked to move)?
If the target coordinates are the same as the starting coordinates (0, 0), should I return 0?
Is there a specific return type expected (e.g., integer representing the minimum number of moves)?

Brute Force Solution

Approach

To find the fewest moves for a knight to reach a destination, the brute force method explores every possible path. It starts at the beginning and tries all possible knight moves, one by one, without any prior knowledge of the best route. This process continues until the target is reached.

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

Start the knight at its initial position.
Consider all eight possible moves a knight can make from its current spot.
For each of those possible moves, imagine the knight takes that move and lands on a new spot.
From each of those new spots, again consider all eight possible knight moves.
Keep repeating this process, branching out in all directions, until the knight lands on the destination.
As you explore each path, count how many moves it took to get to the destination.
Once you have found the destination through different paths, compare the number of moves each path required.
The path with the fewest moves is the solution.

Code Implementation

def minimum_knight_moves_brute_force(target_x, target_y):
    queue = [(0, 0, 0)] # Initial position and moves
    visited = set()
    visited.add((0, 0))

    while queue:
        current_x, current_y, moves = queue.pop(0)

        if current_x == target_x and current_y == target_y:
            return moves

        knight_moves = [
            (1, 2), (1, -2), (-1, 2), (-1, -2),
            (2, 1), (2, -1), (-2, 1), (-2, -1)
        ]

        for move_x, move_y in knight_moves:
            next_x = current_x + move_x
            next_y = current_y + move_y

            # Avoid revisiting already explored positions
            if (next_x, next_y) not in visited:
                queue.append((next_x, next_y, moves + 1))
                visited.add((next_x, next_y))

    return -1 # If target is unreachable

Big(O) Analysis

Time Complexity

O(8^d) – The brute force approach explores all possible knight moves until the destination is reached. In the worst case, the algorithm might explore a vast number of paths before finding the optimal one. At each position, the knight has up to 8 possible moves. If 'd' represents the minimum distance (number of moves) to the target, in the worst-case scenario, the algorithm explores all possible paths up to length 'd', branching out exponentially with each move. This results in approximately 8^d operations, where 'd' is the minimum number of moves. Therefore, the time complexity is O(8^d).

Space Complexity

O(N^2) – The brute force method explores all possible paths on a grid. It implicitly maintains a queue (or similar data structure) to keep track of the locations to visit, which grows as the search expands outwards from the starting position. In the worst case, the algorithm might visit every cell in a grid defined by the possible knight moves needed to reach the destination. Thus, the space complexity is determined by the size of this queue, which can grow up to N^2, where N represents the maximum coordinate value of the destination.

Optimal Solution

Approach

This problem asks for the fewest moves a knight makes on a chessboard to reach a target square. We'll use a technique that explores outwards from the starting position, always checking the closest squares first, to find the target in the fewest steps.

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

Imagine the chessboard. The knight starts at its initial square and wants to get to the target square.
Think of spreading out from the starting square like ripples in a pond. The first ripple are all the squares the knight can reach in one move.
Then, the next ripple consists of the squares reachable in two moves, and so on.
We explore these ripples systematically, keeping track of the number of moves it takes to reach each square we visit.
As we explore, we mark each square with the minimum number of moves needed to get there from the starting square.
We stop when we finally reach the target square. The number of moves marked on that square is the answer.
This guarantees the fewest moves because we always explore the closest squares first. If there were a shorter route to the target, we would have found it earlier.

Code Implementation

from collections import deque

def minimum_knight_moves(start_row, start_column):
    queue = deque([(0, 0, 0)])
    visited = set([(0, 0)])
    
    possible_moves = [(1, 2), (1, -2), (-1, 2), (-1, -2),
                      (2, 1), (2, -1), (-2, 1), (-2, -1)]

    target_row = start_row
    target_column = start_column

    while queue:
        current_row, current_column, moves = queue.popleft()

        if current_row == target_row and current_column == target_column:
            return moves

        for move_row, move_column in possible_moves:
            new_row = current_row + move_row
            new_column = current_column + move_column

            # Only explore valid and unvisited squares.
            if (new_row, new_column) not in visited:
                visited.add((new_row, new_column))
                queue.append((new_row, new_column, moves + 1))

def minKnightMoves(target_row, target_column):
    # Ensure target coordinates are positive.
    target_row = abs(target_row)
    target_column = abs(target_column)

    # BFS to find the shortest path.
    queue = deque([(0, 0, 0)])
    visited = set([(0, 0)])
    
    possible_moves = [(1, 2), (1, -2), (-1, 2), (-1, -2),
                      (2, 1), (2, -1), (-2, 1), (-2, -1)]

    while queue:
        current_row, current_column, moves = queue.popleft()

        if current_row == target_row and current_column == target_column:
            return moves

        for move_row, move_column in possible_moves:
            new_row = current_row + move_row
            new_column = current_column + move_column

            # Only explore valid and unvisited squares.
            if new_row >= -1 and new_column >= -1 and (new_row, new_column) not in visited:
                
                # Mark as visited to avoid re-exploration.
                visited.add((new_row, new_column))

                # Add to queue for further exploration.
                queue.append((new_row, new_column, moves + 1))

Big(O) Analysis

Time Complexity

O(n²) – The algorithm explores the chessboard using a breadth-first search approach, potentially visiting each square. In the worst-case scenario, the knight might need to traverse a significant portion of the board to reach the target. Since the board's dimensions can be conceptually considered as n x n, where n represents the maximum coordinate value, the algorithm could explore on the order of n² squares. The queue operations (enqueue and dequeue) performed during the BFS contribute linearly to the number of visited squares. Thus, the overall time complexity is O(n²).

Space Complexity

O(N^2) – The algorithm explores the chessboard using a breadth-first search approach. In the worst-case scenario, the algorithm might need to visit all the cells on the board to reach the target square. This requires storing the visited cells in a queue or similar data structure (implicit in step 4 and 5 to keep track of visited locations and minimum moves). If we consider N to be the maximum of the target row and target column value, then we can say the maximum grid size would be N * N. Therefore, the space complexity is proportional to the size of the board which is O(N^2).

Edge Cases

Case	How to Handle
Target destination is the same as the starting position (0, 0).	Return 0 immediately since no moves are needed.
Destination coordinates are extremely large, potentially leading to integer overflow or excessive computation.	Use a data structure like a hash set to prune the search space and explore only necessary coordinates within a reasonable boundary.
Knight can only reach the target coordinate from one direction.	The BFS explores all possible paths, so this scenario is naturally covered and doesn't require special handling.
Coordinates involve negative numbers.	Shift the coordinate system to the positive quadrant by adding an offset to both x and y coordinates.
Very large target coordinates that might cause memory issues when storing visited coordinates.	Use a more memory-efficient data structure or a different search algorithm if memory becomes a constraint.
The shortest path involves visiting the same coordinate multiple times.	The visited set correctly tracks the coordinates the algorithm has already visited to prevent infinite loops.
The shortest path length exceeds a certain threshold, indicating an infinite or extremely long path.	Implement a maximum move limit to prevent the algorithm from running indefinitely.
The target coordinate is on an 'edge' such that knight movements frequently take the knight out of bounds and then back.	Expand the boundaries of the visited set to consider coordinates slightly beyond the target coordinate and prevent out of bounds issues.

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:

Are the target coordinates guaranteed to be reachable by the knight?
Can the target coordinates be negative, zero, or non-integer values?
What is the maximum magnitude of the target coordinates (e.g., is there a limit to how far the knight can be asked to move)?
If the target coordinates are the same as the starting coordinates (0, 0), should I return 0?
Is there a specific return type expected (e.g., integer representing the minimum number of moves)?

Brute Force Solution

Approach

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

Start the knight at its initial position.
Consider all eight possible moves a knight can make from its current spot.
For each of those possible moves, imagine the knight takes that move and lands on a new spot.
From each of those new spots, again consider all eight possible knight moves.
Keep repeating this process, branching out in all directions, until the knight lands on the destination.
As you explore each path, count how many moves it took to get to the destination.
Once you have found the destination through different paths, compare the number of moves each path required.
The path with the fewest moves is the solution.

Code Implementation

def minimum_knight_moves_brute_force(target_x, target_y):
    queue = [(0, 0, 0)] # Initial position and moves
    visited = set()
    visited.add((0, 0))

    while queue:
        current_x, current_y, moves = queue.pop(0)

        if current_x == target_x and current_y == target_y:
            return moves

        knight_moves = [
            (1, 2), (1, -2), (-1, 2), (-1, -2),
            (2, 1), (2, -1), (-2, 1), (-2, -1)
        ]

        for move_x, move_y in knight_moves:
            next_x = current_x + move_x
            next_y = current_y + move_y

            # Avoid revisiting already explored positions
            if (next_x, next_y) not in visited:
                queue.append((next_x, next_y, moves + 1))
                visited.add((next_x, next_y))

    return -1 # If target is unreachable

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine the chessboard. The knight starts at its initial square and wants to get to the target square.
Think of spreading out from the starting square like ripples in a pond. The first ripple are all the squares the knight can reach in one move.
Then, the next ripple consists of the squares reachable in two moves, and so on.
We explore these ripples systematically, keeping track of the number of moves it takes to reach each square we visit.
As we explore, we mark each square with the minimum number of moves needed to get there from the starting square.
We stop when we finally reach the target square. The number of moves marked on that square is the answer.
This guarantees the fewest moves because we always explore the closest squares first. If there were a shorter route to the target, we would have found it earlier.

Code Implementation

from collections import deque

def minimum_knight_moves(start_row, start_column):
    queue = deque([(0, 0, 0)])
    visited = set([(0, 0)])
    
    possible_moves = [(1, 2), (1, -2), (-1, 2), (-1, -2),
                      (2, 1), (2, -1), (-2, 1), (-2, -1)]

    target_row = start_row
    target_column = start_column

    while queue:
        current_row, current_column, moves = queue.popleft()

        if current_row == target_row and current_column == target_column:
            return moves

        for move_row, move_column in possible_moves:
            new_row = current_row + move_row
            new_column = current_column + move_column

            # Only explore valid and unvisited squares.
            if (new_row, new_column) not in visited:
                visited.add((new_row, new_column))
                queue.append((new_row, new_column, moves + 1))

def minKnightMoves(target_row, target_column):
    # Ensure target coordinates are positive.
    target_row = abs(target_row)
    target_column = abs(target_column)

    # BFS to find the shortest path.
    queue = deque([(0, 0, 0)])
    visited = set([(0, 0)])
    
    possible_moves = [(1, 2), (1, -2), (-1, 2), (-1, -2),
                      (2, 1), (2, -1), (-2, 1), (-2, -1)]

    while queue:
        current_row, current_column, moves = queue.popleft()

        if current_row == target_row and current_column == target_column:
            return moves

        for move_row, move_column in possible_moves:
            new_row = current_row + move_row
            new_column = current_column + move_column

            # Only explore valid and unvisited squares.
            if new_row >= -1 and new_column >= -1 and (new_row, new_column) not in visited:
                
                # Mark as visited to avoid re-exploration.
                visited.add((new_row, new_column))

                # Add to queue for further exploration.
                queue.append((new_row, new_column, moves + 1))

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Target destination is the same as the starting position (0, 0).	Return 0 immediately since no moves are needed.
Destination coordinates are extremely large, potentially leading to integer overflow or excessive computation.	Use a data structure like a hash set to prune the search space and explore only necessary coordinates within a reasonable boundary.
Knight can only reach the target coordinate from one direction.	The BFS explores all possible paths, so this scenario is naturally covered and doesn't require special handling.
Coordinates involve negative numbers.	Shift the coordinate system to the positive quadrant by adding an offset to both x and y coordinates.
Very large target coordinates that might cause memory issues when storing visited coordinates.	Use a more memory-efficient data structure or a different search algorithm if memory becomes a constraint.
The shortest path involves visiting the same coordinate multiple times.	The visited set correctly tracks the coordinates the algorithm has already visited to prevent infinite loops.
The shortest path length exceeds a certain threshold, indicating an infinite or extremely long path.	Implement a maximum move limit to prevent the algorithm from running indefinitely.
The target coordinate is on an 'edge' such that knight movements frequently take the knight out of bounds and then back.	Expand the boundaries of the visited set to consider coordinates slightly beyond the target coordinate and prevent out of bounds issues.