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):

package main

import (
	"fmt"
)

func zigZagMatrix(mat *[][]int) {    //use pointer to avoid unnecessary function recopying of Array
	// Determine dimensions automatically
	var n = len(*mat)
	var m = len((*mat)[0])

	var i = 0; var j = 0;

	for (i < n && j < m) {
	    if i == 0 && j == 0 {    //top-left corner
		fmt.Print((*mat)[i][j], ", ")
		j++ //go right once
		fmt.Print((*mat)[i][j], ", ")
		for j > 0 && i < n-1 {    //end at left or bottom edge
		    i++ ; j-- //go down and left
		    fmt.Print((*mat)[i][j], ", ")
		}
	    }

	    if j == 0 && i > 0 && i < n-1 {    //left edge and not top or bottom ends
	        i++ //go down once
		fmt.Print((*mat)[i][j], ", ")
		for i > 0 && j < m-1 {    //end at top or right edge
		    i-- ; j++    //go right and up
		    fmt.Print((*mat)[i][j], ", ")
		}
	    }

	    if i == 0 && j > 0 && j < m-1 {    //top edge and not left or right ends
	        j++ //go right once
		fmt.Print((*mat)[i][j], ", ")
		for j > 0 && i < n-1 {     //end at left or bottom edge
		    i++ ; j--    //go down and left
		    fmt.Print((*mat)[i][j], ", ")
		}
	    }

	    if i == n-1 && j == 0 {    //left-bottom corner
	        j++ //go right once
		fmt.Print((*mat)[i][j], ", ")
		for i > 0 && j < m-1 {    //end at top or right edge
		    i-- ; j++    //go right and up
		    fmt.Print((*mat)[i][j], ", ")
		}
	    }

	    if i == 0 && j == m-1 {     //top-right corner
	        i++    //go down once
		fmt.Print((*mat)[i][j], ", ")
		for j > 0 && i < n-1 {    //end at left or bottom edge
		    i++ ; j--    //go down and left
		    fmt.Print((*mat)[i][j], ", ")
		}
	    }

	    if i > 0 && i < n-1 && j == m-1 {     //right edge and not top or bottom ends
	        i++ //go down once
		fmt.Print((*mat)[i][j], ", ")
		for j > 0 && i < n-1 {    //end at left or bottom edge
		    i++ ; j--    //go down and left
		    fmt.Print((*mat)[i][j], ", ")
		}
		//bottom-right end of matrix
		if i == n-1 && j == m-1 {
		    i++ ; j++     //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++ //go right once
		fmt.Print((*mat)[i][j], ", ")
		for i > 0 && j < m-1 {     //end at top or right edge
		    i-- ; j++ //go right and up
		    fmt.Print((*mat)[i][j], ", ")
		}
		//bottom-right end of matrix
		if i == n-1 && j == m-1 {
		    i++; j++    //increase i and j to stop the big while-loop
		}
	    }
	}
	fmt.Println()
}

func main() {
  /*
    var mat = [][]int{{1, 2, 3},
		      {4, 5, 6},
		      {7, 8, 9}}
    zigZagMatrix(&mat)    //use pointer to avoid unnecessary function recopying of Array
  */
  /*
    var mat2 = [][]int{{1, 2, 3, 4, 5},
		       {6, 7, 8, 9, 10},
		       {11, 12, 13, 14, 15},
		       {16, 17, 18, 19, 20}}
    zigZagMatrix(&mat2)    //use pointer to avoid unnecessary function recopying of Array
  */
}

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.