Farthest cell from a given cell in a Matrix
Given the integers N, M, R and C where N and M denotes the number of rows and columns in a matrix and (R, C) denotes a cell in that matrix, the task is to find the distance of the farthest cell from the cell (R, C).
Note: The matrix can only be traversed either horizontally or vertically at a time.
Examples:
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
Naive Approach: The simplest approach is to traverse the matrix and calculate the distance of each cell of the matrix from the given cell (R, C) and maintain the maximum of all distances.
Time Complexity: O(N*M), as we are using nested loops to traverse N*M times.
Auxiliary Space: O(1), as we are not using any extra space.
Efficient Approach: To optimize the above the approach, the observation is that the farthest distant cell from any cell in a matrix will be one of the four corner-most cells i.e. (1, 1), (1, M), (N, 1), (N, M).
Follow the steps below to solve the problem:
- Initialize d1, d2, d3 and d4 be equal to N + M – R – C, R + C – 2, N – R + C – 1 and M – C + R – 1 respectively.
- Print the maximum among d1, d2, d3 and d4.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> 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))); # Print the 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
<script> // 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 </script> |
20
Time Complexity: O(1), as we are not using any loops or recursion to traverse.
Auxiliary Space: O(1), as we are not using any extra space.
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