Given a matrix arr[][] of dimensions N * M where each cell consists of characters ‘R’ or ‘D’ except the cell arr[N][M] which contains ‘F’. ‘R’ and ‘D’ denotes that the player can move in the right and down direction respectively from the current cell. The task is to find the minimum number of characters required to be flipped from ‘R’ to ‘D’ or ‘D’ to ‘R’ such that it is possible to reach the finishing cell i.e., arr[N][M] from every cell.
Examples:
Input: N = 2, M = 3, arr[][] = {{D, D, R}, {R, R, F}}
Output: 1
Explanation: After changing the direction of (1, 3) to ‘D’, each cell can reach the finishing point.Input: N = 1, M = 3, arr[1][3] = {{D, D, F}}
Output: 2
Approach: The problem can be solved by observing that each cell can reach the finishing point after changing the following cells:
- Change all the ‘D’s to ‘R’ in the last row.
- Change all the ‘R’ to ‘D’ in the last column.
Follow the steps below to solve the problem:
- Initialize a variable, say ans, to store the minimum number of flips.
- Traverse from i = 0 to N – 1 and count the number of cells arr[i][M-1] which contains ‘R’.
- Traverse from i = 0 to M – 1 and count the number of cells arr[N – 1][i] which contains ‘D’.
- Print the sum of the two counts obtained in the above step as the required answer.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to calculate the minimum // number of flips required int countChanges(vector<vector< char > > mat)
{ // Dimensions of mat[][]
int n = mat.size();
int m = mat[0].size();
// Initialize answer
int ans = 0;
// Count all 'D's in the last row
for ( int j = 0; j < m - 1; j++) {
if (mat[n - 1][j] != 'R' )
ans++;
}
// Count all 'R's in the last column
for ( int i = 0; i < n - 1; i++) {
if (mat[i][m - 1] != 'D' )
ans++;
}
// Print answer
return ans;
} // Driver Code int main()
{ // Given matrix
vector<vector< char > > arr = { { 'R' , 'R' , 'R' , 'D' },
{ 'D' , 'D' , 'D' , 'R' },
{ 'R' , 'D' , 'R' , 'F' } };
// Function call
int cnt = countChanges(arr);
// Print answer
cout << cnt << endl;
return 0;
} |
// Java program for the above approach import java.io.*;
class GFG{
// Function to calculate the minimum // number of flips required public static int countChanges( char mat[][])
{ // Dimensions of mat[][]
int n = mat.length;
int m = mat[ 0 ].length;
// Initialize answer
int ans = 0 ;
// Count all 'D's in the last row
for ( int j = 0 ; j < m - 1 ; j++)
{
if (mat[n - 1 ][j] != 'R' )
ans++;
}
// Count all 'R's in the last column
for ( int i = 0 ; i < n - 1 ; i++)
{
if (mat[i][m - 1 ] != 'D' )
ans++;
}
// Print answer
return ans;
} // Driver Code public static void main(String[] args)
{ char arr[][] = { { 'R' , 'R' , 'R' , 'D' },
{ 'D' , 'D' , 'D' , 'R' },
{ 'R' , 'D' , 'R' , 'F' } };
// Function call
int cnt = countChanges(arr);
// Print answer
System.out.println(cnt);
} } // This code is contributed by Manu Pathria |
# Python3 program for the above approach # Function to calculate the minimum # number of flips required def countChanges(mat):
# Dimensions of mat[][]
n = len (mat)
m = len (mat[ 0 ])
# Initialize answer
ans = 0
# Count all 'D's in the last row
for j in range (m - 1 ):
if (mat[n - 1 ][j] ! = 'R' ):
ans + = 1
# Count all 'R's in the last column
for i in range (n - 1 ):
if (mat[i][m - 1 ] ! = 'D' ):
ans + = 1
# Print answer
return ans
# Driver Code # Given matrix arr = [ [ 'R' , 'R' , 'R' , 'D' ] ,
[ 'D' , 'D' , 'D' , 'R' ],
[ 'R' , 'D' , 'R' , 'F' ] ]
# Function call cnt = countChanges(arr)
# Print answer print (cnt)
# This code is contributed by susmitakundugoaldanga |
// C# program for the above approach using System;
class GFG{
// Function to calculate the minimum // number of flips required public static int countChanges( char [,]mat)
{ // Dimensions of [,]mat
int n = mat.GetLength(0);
int m = mat.GetLength(1);
// Initialize answer
int ans = 0;
// Count all 'D's in the last row
for ( int j = 0; j < m - 1; j++)
{
if (mat[n - 1,j] != 'R' )
ans++;
}
// Count all 'R's in the last column
for ( int i = 0; i < n - 1; i++)
{
if (mat[i,m - 1] != 'D' )
ans++;
}
// Print answer
return ans;
} // Driver Code public static void Main(String[] args)
{ char [,]arr = { { 'R' , 'R' , 'R' , 'D' },
{ 'D' , 'D' , 'D' , 'R' },
{ 'R' , 'D' , 'R' , 'F' } };
// Function call
int cnt = countChanges(arr);
// Print answer
Console.WriteLine(cnt);
} } // This code is contributed by Princi Singh |
<script> // Javascript program for the above approach // Function to calculate the minimum // number of flips required function countChanges(mat)
{ // Dimensions of mat[][]
let n = mat.length;
let m = mat[0].length;
// Initialize answer
let ans = 0;
// Count all 'D's in the last row
for (let j = 0; j < m - 1; j++)
{
if (mat[n - 1][j] != 'R' )
ans++;
}
// Count all 'R's in the last column
for (let i = 0; i < n - 1; i++)
{
if (mat[i][m - 1] != 'D' )
ans++;
}
// Print let answer
return ans;
} // Driver code let arr = [ [ 'R' , 'R' , 'R' , 'D' ],
[ 'D' , 'D' , 'D' , 'R' ],
[ 'R' , 'D' , 'R' , 'F' ] ];
// Function call let cnt = countChanges(arr); // Print let answer document.write(cnt); // This code is contributed by code_hunt </script> |
2
Time Complexity: O(N+M), where N and M are matrix dimensions.
Auxiliary Space: O(1) , since there is no extra space involved.