所需的最少翻转次数，以便可以从任何其他单元格到达矩阵的最后一个单元格

给定一个维度为N * M的矩阵arr[][] ，其中每个单元格由字符‘R’或‘D’ 组成，除了包含‘F’的单元格arr[N][M] 。 ‘R’和‘D’表示玩家可以从当前单元格分别向右和向下移动。任务是找到从‘R’到‘D’或从‘D’到‘R’翻转所需的最少字符数，以便可以到达完成单元，即arr[N][M]从每个细胞。

例子：

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

处理方法：通过观察每个单元格在更改以下单元格后可以到达终点来解决问题：

• 将最后一行中的所有“D”更改为“R” 。
• 将最后一列中的所有“R”更改为“D” 。

请按照以下步骤解决问题：

1. 初始化一个变量，比如ans ，以存储最小翻转次数。
2. 从i = 0遍历到N – 1并计算包含‘R’的单元格arr[i][M-1]的数量。
3. 从i = 0遍历到 M – 1并计算包含‘D’的单元格arr[N – 1][i]的数量。
4. 打印在上述步骤中获得的两个计数的总和作为所需答案。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to calculate the minimum
// number of flips required
int countChanges(vector > 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 > 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

// 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

# 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#

// 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

Javascript

 
// 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
  

输出：

2

时间复杂度： O(N*M)
辅助空间：O(N*M)

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