Traverse the matrix in Diagonally Bottum-Up fashion using Recursion

Given a matrix mat[][] of size N x N, the task is to traverse the matrix Diagonally in Bottom-up fashion using recursion.

Diagonally Bottum-up Traversal:

Traverse the major-diagonal of the matrix.
Traverse the bottom diagonal to the major-diagonal of the matrix.
Traverse the up diagonal to the major-diagonal of the matrix.
Similary, Traverse the matrix for every diagonal.

Below image shows the Bottom-up Traversal of the matrix.

Examples:

Input: 
M[][] = {{11, 42, 25, 51}, 
         {14, 17, 61, 23},
         {22, 38, 19, 12},
         {27, 81, 29, 71}} 
Output: 
11 17 19 71 
14 38 29 
42 61 12 
22 81 
25 23 
27 
51 

Input: 
M[][] = {{3, 2, 5}, 
         {4, 7, 6},
         {2, 8, 9}}
Output: 
3 7 9 
4 8 
2 6 
2 
5

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to traverse the major-diagonal elements of the matrix and then recursively call the for the bottom diagonal of the matrix and the diagonal above to the major-diagonal of the matrix. Recursive Definition of the approach is described as follows:

Function Definition: For this problem, there will be following arguments as follows:
- mat[][] // Matrix to be Traversed
- Current Row (say i) // Current Row to be Traversed
- Current Column (say j) // Current Column to be Traversed
- Number of rows (say row)
- Number of columns (say col)
Base Case: The base case for this problem can be the when the current row or the current column is out of bounds. In this case traverse the other bottom diagonal or if the bottom diagonal is choosen last time then traverse the major-diagonal just above it.
```
if (i >= row or j >= col)
    if (flag)
        // Change the Current index
        // to the bottom diagonal
    else
        // Change the current index
        // to the up diagonal of matrix
```
Recursive Case: There will be two cases of the recursive traversal of the matrix which is defined as follows:
- Traversal of the Current Diagonal: To traverse the current diagonal increment the current row and column by 1 at the same time and recursively call the function.
- Travesal of Bottom / Up Diagonal: To traverse the bottom / up diagonal call the recusive function with the static variables storing the next traversal start point of the matrix.

Below is the implementation of the above approach:

C++

// C++ implementation to traverse the 
// matrix in the bottom-up fashion 
// using Recursion 
  
#include <iostream> 
  
using namespace std; 
  
// Recursive function to traverse the 
// matrix Diagonally Bottom-up 
bool traverseMatrixDiagonally(int m[][5],  
          int i, int j, int row, int col) 
{ 
      
    // Static variable for changing 
    // Row and coloumn 
    static int k1 = 0, k2 = 0; 
      
    // Flag variable for handling 
    // Bottum up diagonal traversing 
    static bool flag = true; 
      
    // Base Condition 
    if (i >= row || j >= col) { 
          
        // Condition when to traverse 
        // Bottom Diagonal of the matrix 
        if (flag) { 
            int a = k1; 
            k1 = k2; 
            k2 = a; 
            flag = !flag; 
            k1++; 
        } 
        else { 
  
            int a = k1; 
            k1 = k2; 
            k2 = a; 
            flag = !flag; 
        } 
        cout << endl; 
        return false; 
    } 
      
    // Print matrix cell value 
    cout << m[i][j] << " "; 
      
    // Recursive function to traverse 
    // The matrix diagonally 
    if (traverseMatrixDiagonally( 
           m, i + 1, j + 1, row, col)) { 
        return true; 
    } 
    // Recursive function  
    // to change diagonal 
    if (traverseMatrixDiagonally( 
            m, k1, k2, row, col)) { 
        return true; 
    } 
      
    return true; 
} 
  
// Driver Code 
int main() 
{ 
    // Intialize the 5 x 5 matrix 
    int mtrx[5][5] = { 
        { 10, 11, 12, 13, 14 }, 
        { 15, 16, 17, 18, 19 }, 
        { 20, 21, 22, 23, 24 }, 
        { 25, 26, 27, 28, 29 }, 
        { 30, 31, 32, 33, 34 } 
    }; 
  
    // Function call  
    // for traversing matrix 
    traverseMatrixDiagonally( 
            mtrx, 0, 0, 5, 5); 
} 

Java

// Java implementation to traverse  
// the matrix in the bottom-up  
// fashion using recursion 
class GFG{ 
      
// Static variable for changing 
// row and coloumn 
static int k1 = 0, k2 = 0; 
  
// Flag variable for handling 
// bottum up diagonal traversing 
static boolean flag = true;  
  
// Recursive function to traverse the 
// matrix diagonally bottom-up 
static boolean traverseMatrixDiagonally(int m[][], int i, 
                                        int j, int row, 
                                        int col) 
{ 
      
    // Base Condition 
    if (i >= row || j >= col) 
    { 
          
        // Condition when to traverse 
        // Bottom Diagonal of the matrix 
        if (flag) 
        { 
            int a = k1; 
            k1 = k2; 
            k2 = a; 
            flag = !flag; 
            k1++; 
        } 
        else
        { 
            int a = k1; 
            k1 = k2; 
            k2 = a; 
            flag = !flag; 
        } 
          
        System.out.println(); 
        return false; 
    } 
      
    // Print matrix cell value 
    System.out.print(m[i][j] + " "); 
      
    // Recursive function to traverse 
    // The matrix diagonally 
    if (traverseMatrixDiagonally(m, i + 1, 
                                    j + 1, row, col)) 
    { 
        return true; 
    } 
      
    // Recursive function  
    // to change diagonal 
    if (traverseMatrixDiagonally(m, k1, k2, row, col)) 
    { 
        return true; 
    } 
      
    return true; 
} 
  
// Driver Code 
public static void main(String[] args) 
{ 
    // Intialize the 5 x 5 matrix 
    int mtrx[][] = { { 10, 11, 12, 13, 14 }, 
                     { 15, 16, 17, 18, 19 }, 
                     { 20, 21, 22, 23, 24 }, 
                     { 25, 26, 27, 28, 29 }, 
                     { 30, 31, 32, 33, 34 } }; 
  
    // Function call  
    // for traversing matrix 
    traverseMatrixDiagonally(mtrx, 0, 0, 5, 5); 
} 
} 
  
// This code is contributed by sapnasingh4991 

C#

// C# implementation to traverse  
// the matrix in the bottom-up  
// fashion using recursion 
using System; 
  
class GFG{ 
      
// Static variable for changing 
// row and coloumn 
static int k1 = 0, k2 = 0; 
  
// Flag variable for handling 
// bottum up diagonal traversing 
static bool flag = true;  
  
// Recursive function to traverse the 
// matrix diagonally bottom-up 
static bool traverseMatrixDiagonally(int [,]m, int i, 
                                     int j, int row, 
                                     int col) 
{ 
      
    // Base Condition 
    if (i >= row || j >= col) 
    { 
          
        // Condition when to traverse 
        // Bottom Diagonal of the matrix 
        if (flag) 
        { 
            int a = k1; 
            k1 = k2; 
            k2 = a; 
            flag = !flag; 
            k1++; 
        } 
        else
        { 
            int a = k1; 
            k1 = k2; 
            k2 = a; 
            flag = !flag; 
        } 
          
        Console.WriteLine(); 
        return false; 
    } 
      
    // Print matrix cell value 
    Console.Write(m[i, j] + " "); 
      
    // Recursive function to traverse 
    // The matrix diagonally 
    if (traverseMatrixDiagonally(m, i + 1, 
                                    j + 1, row, col)) 
    { 
        return true; 
    } 
      
    // Recursive function  
    // to change diagonal 
    if (traverseMatrixDiagonally(m, k1, k2, row, col)) 
    { 
        return true; 
    } 
    return true; 
} 
  
// Driver Code 
public static void Main(String[] args) 
{ 
      
    // Intialize the 5 x 5 matrix 
    int [,]mtrx = { { 10, 11, 12, 13, 14 }, 
                    { 15, 16, 17, 18, 19 }, 
                    { 20, 21, 22, 23, 24 }, 
                    { 25, 26, 27, 28, 29 }, 
                    { 30, 31, 32, 33, 34 } }; 
  
    // Function call for   
    // traversing matrix 
    traverseMatrixDiagonally(mtrx, 0, 0, 5, 5); 
} 
} 
  
// This code is contributed by Amit Katiyar 

Output:

10 16 22 28 34 
15 21 27 33 
11 17 23 29 
20 26 32 
12 18 24 
25 31 
13 19 
30 
14

Time Complexity: O(N²)

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

C#

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