矩阵中距给定单元格最远的单元格

给定整数N 、 M 、 R和C ，其中N和M表示矩阵中的行数和列数， (R, C)表示该矩阵中的一个单元格，任务是找到离矩阵最远的单元格的距离细胞(R，C) 。
注意：矩阵一次只能水平或垂直遍历。

例子：

Input: N = 15, M = 12, R = 1, C = 6
Output: 20
Explanation: The maximum distance calculated from all the four corners are 20, 5, 19, 6. Therefore, 20 is the required answer.

Input: N = 15, M = 12, R = 2, C = 4
Output: 21

朴素的方法：最简单的方法是遍历矩阵并计算矩阵的每个单元格与给定单元格(R，C)的距离并保持所有距离的最大值。

时间复杂度： O(N * M)
辅助空间： O(1)

有效的方法：为了优化上述方法，观察到距矩阵中任何单元最远的单元将是四个最角落单元之一，即(1, 1), (1, M), (N, 1), (N, M) 。

请按照以下步骤解决问题：

• 初始化d1 , d2 , d3和d4分别等于N + M – R – C , R + C – 2 , N – R + C – 1和M – C + R – 1 。
• 打印d1 、 d2 、 d3和d4 中的最大值。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the farthest
// cell distance from the given cell
void farthestCellDistance(int N, int M,
                          int R, int C)
{
    // Distance from all the
    // cornermost cells
 
    // From cell(N, M)
    int d1 = N + M - R - C;
 
    // From Cell(1, 1)
    int d2 = R + C - 2;
 
    // From cell(N, 1)
    int d3 = N - R + C - 1;
 
    // From cell(1, M)
    int d4 = M - C + R - 1;
 
    // Finding out maximum
    int maxDistance = max(d1,
                          max(d2,
                              max(d3, d4)));
 
    // Print the answer
    cout << maxDistance;
}
 
// Driver Code
int main()
{
    int N = 15, M = 12, R = 1, C = 6;
    farthestCellDistance(N, M, R, C);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
 
// Function to find the farthest
// cell distance from the given cell
static void farthestCellDistance(int N, int M,
                                 int R, int C)
{
     
    // Distance from all the
    // cornermost cells
 
    // From cell(N, M)
    int d1 = N + M - R - C;
 
    // From Cell(1, 1)
    int d2 = R + C - 2;
 
    // From cell(N, 1)
    int d3 = N - R + C - 1;
 
    // From cell(1, M)
    int d4 = M - C + R - 1;
 
    // Finding out maximum
    int maxDistance = Math.max(d1, Math.max(
                  d2, Math.max(d3, d4)));
 
    // Print the answer
    System.out.println(maxDistance);
}
 
// Driver Code
public static void main(String[] args)
{
    int N = 15, M = 12, R = 1, C = 6;
     
    farthestCellDistance(N, M, R, C);
}
}
 
// This code is contributed by Dharanendra L V

Python3

# Python program for the above approach
 
# Function to find the farthest
# cell distance from the given cell
def farthestCellDistance(N, M, R, C):
   
    # Distance from all the
    # cornermost cells
 
    # From cell(N, M)
    d1 = N + M - R - C;
 
    # From Cell(1, 1)
    d2 = R + C - 2;
 
    # From cell(N, 1)
    d3 = N - R + C - 1;
 
    # From cell(1, M)
    d4 = M - C + R - 1;
 
    # Finding out maximum
    maxDistance = max(d1, max(d2, max(d3, d4)));
 
    # Prthe answer
    print(maxDistance);
 
# Driver Code
if __name__ == '__main__':
    N = 15;
    M = 12;
    R = 1;
    C = 6;
 
    farthestCellDistance(N, M, R, C);
 
# This code is contributed by shikhasingrajput

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the farthest
// cell distance from the given cell
static void farthestCellDistance(int N, int M,
                                 int R, int C)
{
     
    // Distance from all the
    // cornermost cells
 
    // From cell(N, M)
    int d1 = N + M - R - C;
 
    // From Cell(1, 1)
    int d2 = R + C - 2;
 
    // From cell(N, 1)
    int d3 = N - R + C - 1;
 
    // From cell(1, M)
    int d4 = M - C + R - 1;
 
    // Finding out maximum
    int maxDistance = Math.Max(d1, Math.Max(
                  d2, Math.Max(d3, d4)));
 
    // Print the answer
    Console.WriteLine(maxDistance);
}
 
// Driver Code
static public void Main()
{
    int N = 15, M = 12, R = 1, C = 6;
     
    farthestCellDistance(N, M, R, C);
}
}
 
// This code is contributed by Dharanendra L V

Javascript

// Java script program for the above approach
 
// Function to find the farthest
// cell distance from the given cell
function farthestCellDistance( N,  M,
                                 R,  C)
{
     
    // Distance from all the
    // cornermost cells
 
    // From cell(N, M)
    let d1 = N + M - R - C;
 
    // From Cell(1, 1)
    let d2 = R + C - 2;
 
    // From cell(N, 1)
    let d3 = N - R + C - 1;
 
    // From cell(1, M)
    let d4 = M - C + R - 1;
 
    // Finding out maximum
    let maxDistance = Math.max(d1, Math.max(
                d2, Math.max(d3, d4)));
 
    // Print the answer
    document.write(maxDistance);
}
 
// Driver Code
 
    let N = 15, M = 12, R = 1, C = 6;
     
    farthestCellDistance(N, M, R, C);
 
 
// This code is contributed by Bobby

输出：

20

时间复杂度： O(1)
辅助空间： O(1)

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