The reader is advised to read all the lessons (tutorials) in this full course, in the order presented.

Problem

Given a matrix of 2D array of n rows and m columns, print this matrix in ZIG-ZAG fashion as shown in the following figure:

Example 1: 
Input: mat[][] = {{1, 2, 3}
                  {4, 5, 6}
                  {7, 8, 9}}

Output : 1, 2, 4, 7, 5, 3, 6, 8, 9

Example 2: 
Input: mat[][] = {{1,  2,  3,  4,  5}, 
                  {6,  7,  8,  9,  10},
                  {11, 12, 13, 14, 15},
                  {16, 17, 18, 19, 20}};

Output : 1, 2, 6, 11, 7, 3, 4, 8, 12, 16, 17, 13, 9, 5, 10, 14, 18, 19, 15, 20

Solution in O(n x m) Time

The algorithm is as follows:

- Visit the top-left element first. Visit the next element on the right.
- Continue visiting the current minor diagonal elements in the left-downwards direction until meeting the left edge or bottom edge of the matrix.
- If the left edge of the matrix is met and not the bottom edge, visit the next element below.
- Continue visiting the current minor diagonal elements in the right-upwards direction until meeting the top edge or right edge of the matrix.
- If the bottom edge of the matrix was met, and not the left edge, then visit the next element on the right at the bottom edge.
- Continue visiting the current minor diagonal elements in the right-upwards direction until meeting the top edge or right edge of the matrix.
- If the top edge of the matrix is met and not the right edge, visit the next element on the right.
- Continue visiting the current minor diagonal elements in the left-downwards direction until meeting the left edge or bottom edge of the matrix.
- If the right edge of the matrix was met, and not the top edge, then visit the next element below.
- Continue visiting the current minor diagonal elements in the left-downwards direction until meeting the left edge or bottom edge of the matrix.
- Continue repeating the above steps accordingly, until the element at the bottom-right corner of the matrix is met, which will always be met, either from the left or from above.
- Do not forget to consider the cases, where the visit is at the four corners.

There are eight cases to consider: when the visit is at the four corners and when the visit is at the four edges. The case of what happens after the visit of the bottom-right element, does not have to be addressed. This gives rise to seven code segments in a major while-loop.

Indexes

The top most row of the matrix is identified by i=0.
The leftmost column of the matrix is identified by j=0.
The bottommost row of the matrix is identified by i=n-1.
The rightmost column of the matrix is identified by j=m-1.

The program

The program is (read through the code and comments):

fn zig_zag_matrix(mat: &[Vec<i32>]) {
    // Determine dimensions automatically
    let n = mat.len() as isize;
    let m = mat[0].len() as isize;
    
    let mut i: isize = 0;
    let mut j: isize = 0;
    
    while i < n && j < m {
        if i == 0 && j == 0 {
            // top-left corner
            print!("{}, ", mat[i as usize][j as usize]);
            j += 1; // go right once
            print!("{}, ", mat[i as usize][j as usize]);
            while j > 0 && i < n - 1 {
                // end at left or bottom edge
                i += 1;
                j -= 1; // go down and left
                print!("{}, ", mat[i as usize][j as usize]);
            }
        }
        if j == 0 && i > 0 && i < n - 1 {
            // left edge and not top or bottom ends
            i += 1; // go down once
            print!("{}, ", mat[i as usize][j as usize]);
            while i > 0 && j < m - 1 {
                // end at top or right edge
                i -= 1;
                j += 1; // go right and up
                print!("{}, ", mat[i as usize][j as usize]);
            }
        }
        if i == 0 && j > 0 && j < m - 1 {
            // top edge and not left or right ends
            j += 1; // go right once
            print!("{}, ", mat[i as usize][j as usize]);
            while j > 0 && i < n - 1 {
                // end at left or bottom edge
                i += 1;
                j -= 1; // go down and left
                print!("{}, ", mat[i as usize][j as usize]);
            }
        }
        if i == n - 1 && j == 0 {
            // left-bottom corner
            j += 1; // go right once
            print!("{}, ", mat[i as usize][j as usize]);
            while i > 0 && j < m - 1 {
                // end at top or right edge
                i -= 1;
                j += 1; // go right and up
                print!("{}, ", mat[i as usize][j as usize]);
            }
        }
        if i == 0 && j == m - 1 {
            // top-right corner
            i += 1; // go down once
            print!("{}, ", mat[i as usize][j as usize]);
            while j > 0 && i < n - 1 {
                // end at left or bottom edge
                i += 1;
                j -= 1; // go down and left
                print!("{}, ", mat[i as usize][j as usize]);
            }
        }
        if i > 0 && i < n - 1 && j == m - 1 {
            // right edge and not top or bottom ends
            i += 1; // go down once
            print!("{}, ", mat[i as usize][j as usize]);
            while j > 0 && i < n - 1 {
                // end at left or bottom edge
                i += 1;
                j -= 1; // go down and left
                print!("{}, ", mat[i as usize][j as usize]);
            }
            // bottom-right end of matrix
            if i == n - 1 && j == m - 1 {
                i += 1;
                j += 1; // increase i and j to stop the big while-loop
            }
        }
        if j > 0 && j < m - 1 && i == n - 1 {
            // bottom edge and not left or right ends
            j += 1; // go right once
            print!("{}, ", mat[i as usize][j as usize]);
            while i > 0 && j < m - 1 {
                // end at top or right edge
                i -= 1;
                j += 1; // go right and up
                print!("{}, ", mat[i as usize][j as usize]);
            }
            // bottom-right end of matrix
            if i == n - 1 && j == m - 1 {
                i += 1;
                j += 1; // increase i and j to stop the big while-loop
            }
        }
    }
    println!();
}

fn main() {
/*
    let mat = vec![
        vec![1, 2, 3], 
        vec![4, 5, 6], 
        vec![7, 8, 9]
    ];
    zig_zag_matrix(&mat);
*/
/*
    let mat2 = vec![
        vec![1, 2, 3, 4, 5],
        vec![6, 7, 8, 9, 10],
        vec![11, 12, 13, 14, 15],
        vec![16, 17, 18, 19, 20],
    ];
    zig_zag_matrix(&mat2);
*/
}

Remove the first pair of the block comment markers (/* */). Make the first argument of the called function, "int mat[][3]". Run the program and the output should be:

    1, 2, 4, 7, 5, 3, 6, 8, 9, 

Put back the first pair of the block comment markers (/* */). Remove the second pair of the block comment markers (/* */). Make the first argument of the called function, "int mat[][5]". Run the program and the output should be:

1, 2, 6, 11, 7, 3, 4, 8, 12, 16, 17, 13, 9, 5, 10, 14, 18, 19, 15, 20,

The time complexity is actually O(2 n x m), with the first O(n x m) for populating the input matrix with all the elements, and the second O(n x m) for iterating over all the elements in the matrix in the called function. Since the coefficient (multiplicand of 2) is usually omitted, the time complexity is quoted as O(n x m).

The space complexity is O(n x m), for the input matrix, which is the same as the matrix in the called function (same reference).

Thanks for reading.