Determine if a Cell Is Reachable at a Given Time

class Solution {
public:
    bool canReachDestination(int startX, int startY, int finalX, int finalY, int time) {
        int deltaX = abs(startX - finalX);
        int deltaY = abs(startY - finalY);
        if (deltaX == 0 && deltaY == 0 && time == 1) {
            return false;
        }
        return time >= max(deltaX, deltaY);
    }
};

This code snippet defines a method in C++ to determine whether it is possible to reach a target cell in a matrix from a given starting position within a specified time. The function canReachDestination() receives five parameters: the starting coordinates (startX, startY), the destination coordinates (finalX, finalY), and the available time to reach the destination. The approach primarily revolves around calculating the absolute differences between the corresponding coordinates of the start and final positions (deltaX and deltaY). The condition checks:

• If the starting position coincides with the destination and only one unit of time is allowed, the function returns false.
• In all other scenarios, the function checks whether the given time is greater than or equal to the maximum of deltaX and deltaY.

This ensures flexibility in movement pattern, prioritizing the maximum required steps in any direction to conclude if the destination is reachable within the specified duration.

class Solution {
    public boolean canReachDestination(int startX, int startY, int endX, int endY, int time) {
        int deltaX = Math.abs(startX - endX);
        int deltaY = Math.abs(startY - endY);
        if (deltaX == 0 && deltaY == 0 && time == 1) {
            return false;
        }
        return time >= Math.max(deltaX, deltaY);
    }
}

The problem "Determine if a Cell Is Reachable at a Given Time" involves finding whether one can reach a destination (cell) from a starting point within a certain number of moves or time units in a grid. The solution is implemented in Java.

In this Java solution, the canReachDestination method calculates whether the destination is reachable from the start position within the given time. Utilize the Math.abs() function to compute the absolute differences in the x and y coordinates between the start and destination points, labeled as deltaX and deltaY, respectively.

The method first checks if the starting point is the same as the destination and the time is exactly 1. This specific condition returns false, which is a special case where the move isn't possible due to constraints probably intended to handle cases where staying in place costs a move.

For other general cases, compare the maximum of deltaX and deltaY to the available time. Return true if the time is greater than or equal to the maximum of these deltas, indicating that moving horizontally or vertically or both in any sequence could cover the distance in the given time.

Therefore:

• Determine the absolute differences between start and destination coordinates with Math.abs().
• Verify if starting at the destination with time equals 1 falls under no movement allowed.
• Return true if time supports at least the longest single-axis movement derived from the maximum of deltaX and deltaY.

bool can_arrive(int start_x, int start_y, int finish_x, int finish_y, int time_given){
    int delta_x = abs(start_x - finish_x);
    int delta_y = abs(start_y - finish_y);
    if (delta_x == 0 && delta_y == 0 && time_given == 1) {
        return false;
    }
    return time_given >= (delta_x > delta_y ? delta_x : delta_y);
}

This solution in C determines if it's possible to reach a cell (defined by its coordinates) from a starting point within a given time frame. The function can_arrive accepts coordinates for the starting (start_x, start_y) and finishing positions (finish_x, finish_y), along with the time_given.

The approach utilizes the Manhattan distance between the start and end points by calculating the absolute differences between their respective coordinates (delta_x and delta_y). The conditional check returns:

• false if both starting and ending points are the same and the time given is 1 - which under default conditions might incorrectly return true.
• true if the time_given is greater than or equal to the maximum of delta_x and delta_y. This ensures checking the minimum time required to either shift entirely horizontally or vertically to reach the target cell.

Overall, this function effectively calculates if a direct move from the start to the finish within the stipulated time is feasible, assuming a unit step takes equivalent time horizontally or vertically.

var canReachDestination = function(startX, startY, finishX, finishY, time) {
    var deltaX = Math.abs(startX - finishX);
    var deltaY = Math.abs(startY - finishY);
    if (deltaX === 0 && deltaY === 0 && time === 1) {
        return false;
    }
    return time >= Math.max(deltaX, deltaY);
};

This JavaScript function, canReachDestination, checks whether a destination cell in a grid can be reached from a starting position within a given time. The function takes as inputs the coordinates of the start (startX, startY) and finish (finishX, finishY) positions, along with the time expected to reach the destination.

• Calculate the absolute differences between the start and finish coordinates for both X and Y dimensions using Math.abs. Store these differences in deltaX and deltaY.
• An initial condition checks if the start position is the same as the finish position and the provided time is exactly 1, in which case it returns false.
• For all other cases, compare the provided time with the maximum of deltaX and deltaY values. If the time is greater than or equal to this value, return true, indicating that reaching the destination is possible within the given time. Otherwise, return false.

This approach ensures that distance constraints are respected relative to the time allowed, effectively solving the problem of reachability in a grid environment based on Manhattan distance constraints.

class Solution:
    def canReachInTime(self, start_x: int, start_y: int, end_x: int, end_y: int, time: int) -> bool:
        dx = abs(start_x - end_x)
        dy = abs(start_y - end_y)
        if dx == 0 and dy == 0 and time == 1:
            return False
        return time >= max(dx, dy)

This Python solution addresses the problem of determining if a cell in a grid can be reached from a starting position within a given time. The canReachInTime method takes starting coordinates (start_x, start_y), ending coordinates (end_x, end_y), and the maximum time allowed (time). The method calculates the Manhattan distance between the start and end positions using the absolute differences dx and dy. If both dx and dy are zero and the time is only a single step, then reaching the cell isn't possible and it returns False. Otherwise, the method checks if the time provided is greater than or equal to the maximum of dx and dy. If so, the cell is reachable, and the function returns True; otherwise, it returns False. This approach ensures that the shortest path in terms of grid steps is considered.