Zigzag (or diagonal) traversal of Matrix

Given a 2D matrix, print all elements of the given matrix in diagonal order. For example, consider the following 5 X 4 input matrix.  

Example:

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

Diagonal printing of the above matrix is

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

Another example: 

We strongly recommend that you click here and practice it, before moving on to the solution.

Following is the code for diagonal printing.
The diagonal printing of a given matrix “matrix[ROW][COL]” always has “ROW + COL – 1” lines in output. 

Implementation:

C++

// C++ program to print all elements
// of given matrix in diagonal order
#include <bits/stdc++.h>
using namespace std;

#define ROW 5
#define COL 4

// A utility function to find min
// of two integers
int minu(int a, int b)
{
    return (a < b) ? a : b;
}

// A utility function to find min
// of three integers
int min(int a, int b, int c)
{
    return minu(minu(a, b), c);
}

// A utility function to find
// max of two integers
int max(int a, int b)
{
    return (a > b) ? a : b;
}

// The main function that prints given
// matrix in diagonal order
void diagonalOrder(int matrix[][COL])
{

    // There will be ROW+COL-1 lines
    // in the output
    for(int line = 1;
            line <= (ROW + COL - 1);
            line++)
    {

        /* Get column index of the first element
           in this line of output.
           The index is 0 for first ROW lines and
           line - ROW for remaining lines  */
        int start_col =  max(0, line - ROW);

        /* Get count of elements in this line. The
           count of elements is equal to minimum of
           line number, COL-start_col and ROW */
         int count = min(line, (COL - start_col), ROW);

        /* Print elements of this line */
        for(int j = 0; j < count; j++)
            cout << setw(5) <<
            matrix[minu(ROW, line) - j - 1][start_col + j];

        /* Print elements of next
           diagonal on next line */
        cout << "\n";
    }
}

// Utility function to print a matrix
void printMatrix(int matrix[ROW][COL])
{
    for(int i = 0; i < ROW; i++)
    {
        for(int j = 0; j < COL; j++)
            cout << setw(5) << matrix[i][j];

        cout << "\n";
    }
}

// Driver code
int main()
{
    int M[ROW][COL] = { { 1, 2, 3, 4 },
                        { 5, 6, 7, 8 },
                        { 9, 10, 11, 12 },
                        { 13, 14, 15, 16 },
                        { 17, 18, 19, 20 },};
    cout << "Given matrix is " << endl;
    printMatrix(M);

    cout << "\nDiagonal printing of matrix is " << endl;
    diagonalOrder(M);
    return 0;
}

// This code is contributed by shubhamsingh10

C

// C program to print all elements
// of given matrix in diagonal order
#include <stdlib.h>

#define ROW 5
#define COL 4

// A utility function to find min of two integers
int minu(int a, int b)
{ return (a < b)? a: b; }

// A utility function to find min of three integers
int min(int a, int b, int c)
{ return minu(minu(a, b), c);}

// A utility function to find max of two integers
int max(int a, int b)
{ return (a > b)? a: b; }

// The main function that prints given matrix in
// diagonal order
void diagonalOrder(int matrix[][COL])
{
    // There will be ROW+COL-1 lines in the output
    for (int line=1; line<=(ROW + COL -1); line++)
    {
        /* Get column index of the first element
           in this line of output.
           The index is 0 for first ROW lines and
           line - ROW for remaining lines  */
        int start_col =  max(0, line-ROW);

        /* Get count of elements in this line. The
           count of elements is equal to minimum of
           line number, COL-start_col and ROW */
         int count = min(line, (COL-start_col), ROW);

        /* Print elements of this line */
        for (int j=0; j<count; j++)
            printf("%5d ",
                   matrix[minu(ROW, line)-j-1][start_col+j]);

        /* Print elements of next diagonal on next line */
        printf("\n");
    }
}

// Utility function to print a matrix
void printMatrix(int matrix[ROW][COL])
{
    for (int i=0; i< ROW; i++)
    {
        for (int j=0; j<COL; j++)
            printf("%5d ", matrix[i][j]);
        printf("\n");
    }
}

// Driver code
int main()
{
    int M[ROW][COL] = {{1, 2, 3, 4},
                       {5, 6, 7, 8},
                       {9, 10, 11, 12},
                       {13, 14, 15, 16},
                       {17, 18, 19, 20},
                      };
    printf ("Given matrix is \n");
    printMatrix(M);

    printf ("\nDiagonal printing of matrix is \n");
    diagonalOrder(M);
    return 0;
}

Java

// Java program to print all elements
// of given matrix in diagonal order
class GFG {
    static final int ROW = 5;
    static final int COL = 4;

    // A utility function to find min
    // of two integers
    static int min(int a, int b)
    {
        return (a < b) ? a : b;
    }

    // A utility function to find min
    // of three integers
    static int min(int a, int b, int c)
    {
        return min(min(a, b), c);
    }

    // A utility function to find max
    // of two integers
    static int max(int a, int b)
    {
         return (a > b) ? a : b;
    }

    // The main function that prints given
    // matrix in diagonal order
    static void diagonalOrder(int matrix[][])
    {

        // There will be ROW+COL-1 lines in the output
        for (int line = 1;
             line <= (ROW + COL - 1);
             line++) {

            // Get column index of the first
            // element in this line of output.
            // The index is 0 for first ROW
            // lines and line - ROW for remaining lines
            int start_col = max(0, line - ROW);

            // Get count of elements in this line.
            // The count of elements is equal to
            // minimum of line number, COL-start_col and ROW
            int count = min(line, (COL - start_col),
                            ROW);

            // Print elements of this line
            for (int j = 0; j < count; j++)
                System.out.print(matrix[min(ROW, line)
                                        - j- 1][start_col + j]
                                 + " ");

            // Print elements of next diagonal on next line
            System.out.println();
        }
    }

    // Utility function to print a matrix
    static void printMatrix(int matrix[][])
    {
        for (int i = 0; i < ROW; i++)
        {
            for (int j = 0; j < COL; j++)
                System.out.print(matrix[i][j] + " ");
            System.out.print("\n");
        }
    }

    // Driver code
    public static void main(String[] args)
    {
        int M[][] = {
            { 1, 2, 3, 4 },     { 5, 6, 7, 8 },
            { 9, 10, 11, 12 },  { 13, 14, 15, 16 },
            { 17, 18, 19, 20 },
        };
        System.out.print("Given matrix is \n");
        printMatrix(M);

        System.out.print(
            "\nDiagonal printing of matrix is \n");
        diagonalOrder(M);
    }
}
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to print all elements
# of given matrix in diagonal order
ROW = 5
COL = 4

# Main function that prints given
# matrix in diagonal order


def diagonalOrder(matrix):

    # There will be ROW+COL-1 lines in the output
    for line in range(1, (ROW + COL)):
        # Get column index of the first element
        # in this line of output. The index is 0
        # for first ROW lines and line - ROW for
        # remaining lines
        start_col = max(0, line - ROW)

        # Get count of elements in this line.
        # The count of elements is equal to
        # minimum of line number, COL-start_col and ROW
        count = min(line, (COL - start_col), ROW)

        # Print elements of this line
        for j in range(0, count):
            print(matrix[min(ROW, line) - j - 1]
                        [start_col + j], end="\t")

        print()


# Utility function to print a matrix
def printMatrix(matrix):
    for i in range(0, ROW):
        for j in range(0, COL):
            print(matrix[i][j], end="\t")

        print()


# Driver Code
M = [[1, 2, 3, 4],
     [5, 6, 7, 8],
     [9, 10, 11, 12],
     [13, 14, 15, 16],
     [17, 18, 19, 20]]
print("Given matrix is ")
printMatrix(M)

print("\nDiagonal printing of matrix is ")
diagonalOrder(M)

# This code is contributed by Nikita Tiwari.

C#

// C# program to print all elements
// of given matrix in diagonal order
using System;
using static System.Math;

class GFG {

    static int ROW = 5;
    static int COL = 4;

    // The main function that prints given
    // matrix in diagonal order
    static void diagonalOrder(int[, ] matrix)
    {

        // There will be ROW+COL-1 lines in the output
        for (int line = 1;
             line <= (ROW + COL - 1);
             line++) {

            // Get column index of the first element
            // in this line of output.The index is 0
            // for first ROW lines and line - ROW for
            // remaining lines
            int start_col = Max(0, line - ROW);

            // Get count of elements in this line. The
            // count of elements is equal to minimum of
            // line number, COL-start_col and ROW
            int count = Min(line, Math.Min((COL - start_col), ROW));

            // Print elements of this line
            for (int j = 0; j < count; j++)
                Console.Write(matrix[Min(ROW, line) - j - 1, start_col + j] + " ");

            // Print elements of next diagonal
            // on next line
            Console.WriteLine();
        }
    }

    // Utility function to print a matrix
    static void printMatrix(int[, ] matrix)
    {
        for (int i = 0; i < ROW; i++)
        {
            for (int j = 0; j < COL; j++)
                Console.Write(matrix[i, j] + " ");

            Console.WriteLine("\n");
        }
    }

    // Driver code
    public static void Main()
    {
        int[, ] M = { { 1, 2, 3, 4 },
                      { 5, 6, 7, 8 },
                      { 9, 10, 11, 12 },
                      { 13, 14, 15, 16 },
                      { 17, 18, 19, 20 } };

        Console.Write("Given matrix is \n");

        printMatrix(M);

        Console.Write("\nDiagonal printing" + " of matrix is \n");

        diagonalOrder(M);
    }
}

// This code is contributed by Sam007.

PHP

<?php
// PHP Code for Zigzag (or diagonal)
// traversal of Matrix
$ROW = 5;
$COL = 4;

// The main function that prints
// given matrix in diagonal order
function diagonalOrder(&$matrix)
{
    global $ROW, $COL;

    // There will be ROW+COL-1
    // lines in the output
    for ($line = 1;
         $line <= ($ROW + $COL - 1); $line++)
    {
        /* Get column index of the first
           element in this line of output.
           The index is 0 for first ROW lines
           and line - ROW for remaining lines */
        $start_col = max(0, $line - $ROW);

        /* Get count of elements in this line.
           The count of elements is equal to
           minimum of line number, COL-start_col
           and ROW */
        $count = min($line, ($COL -
                     $start_col), $ROW);

        /* Print elements of this line */
        for ($j = 0; $j < $count; $j++)
        {
            echo $matrix[min($ROW, $line) -
                 $j - 1][$start_col + $j];
            echo "\t";
        }

        /* Print elements of next
           diagonal on next line */
        print("\n");
    }
}

// Utility function
// to print a matrix
function printMatrix(&$matrix)
{
    global $ROW, $COL;
    for ($i = 0; $i < $ROW; $i++)
    {
        for ($j = 0; $j < $COL; $j++)
        {
            echo $matrix[$i][$j] ;
            echo "\t";
        }
        print("\n");
    }
}

// Driver Code
$M = array(array(1, 2, 3, 4),
           array(5, 6, 7, 8),
           array(9, 10, 11, 12),
           array(13, 14, 15, 16),
           array(17, 18, 19, 20));
echo "Given matrix is \n";
printMatrix($M);

printf ("\nDiagonal printing " .
        "of matrix is \n");
diagonalOrder($M);

// This code is contributed
// by ChitraNayal
?>

Javascript

<script>
// Javascript program to print all elements
// of given matrix in diagonal order

let ROW = 5;
let COL = 4;

// A utility function to find min
    // of two integers
function min(a, b)
{
    return (a < b) ? a : b;
}

// A utility function to find min
    // of three integer
function _min(a, b, c)
{
    return min(min(a, b), c);
}

// A utility function to find max
    // of two integers
function max(a,b)
{
    return (a > b) ? a : b;
}

// The main function that prints given
    // matrix in diagonal order
function diagonalOrder(matrix)
{
    // There will be ROW+COL-1 lines in the output
        for (let line = 1;
             line <= (ROW + COL - 1);
             line++) {

            // Get column index of the first
            // element in this line of output.
            // The index is 0 for first ROW
            // lines and line - ROW for remaining lines
            let start_col = max(0, line - ROW);

            // Get count of elements in this line.
            // The count of elements is equal to
            // minimum of line number, COL-start_col and ROW
            let count = min(line, (COL - start_col),
                            ROW);

            // Print elements of this line
            for (let j = 0; j < count; j++)
                document.write(matrix[min(ROW, line)
                                        - j- 1][start_col + j]
                                 + " ");

            // Print elements of next diagonal on next line
            document.write("<br>");
        }
}

// Utility function to print a matrix
function printMatrix(matrix)
{
    for (let i = 0; i < ROW; i++)
        {
            for (let j = 0; j < COL; j++)
                document.write(matrix[i][j] + " ");
            document.write("<br>");
        }
}

// Driver code
let M = [[ 1, 2, 3, 4 ], [ 5, 6, 7, 8 ],
            [ 9, 10, 11, 12 ],  [ 13, 14, 15, 16 ],
            [ 17, 18, 19, 20 ]];
document.write("Given matrix is <br>");
printMatrix(M);

document.write(
"<br>Diagonal printing of matrix is <br>");
diagonalOrder(M);

// This code is contributed by ab2127
</script>

Output

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

Diagonal printing of matrix is 
    1 
    5     2 
    9     6     3 
   13    10     7     4 
   17    14    11     8 
   18    15    12 
   19    16 
   20 

Time Complexity: O(row x col) 
Auxiliary Space: O(1)

Below is an Alternate Method to solve the above problem.

Matrix =>       1     2     3     4
                5     6     7     8
                9     10    11   12
                13    14    15   16
                17    18    19   20 
   
Observe the sequence
          1 /  2 /  3 /  4
           / 5  /  6 /  7 /  8
               /  9 / 10 / 11 / 12
                   / 13 / 14 / 15 / 16
                       / 17 / 18 / 19 / 20

Implementation:

C++

#include <bits/stdc++.h>
#define R 5
#define C 4
using namespace std;

bool isValid(int i, int j)
{
    if (i < 0 || i >= R
        || j >= C || j < 0)
        return false;
    return true;
}

void diagonalOrder(int arr[][C])
{
    /* through this for loop we choose
       each element of first column as
       starting point and print diagonal
       starting at it.
    arr[0][0], arr[1][0]....arr[R-1][0]
    are all starting points */
    for (int k = 0; k < R; k++)
    {
        cout << arr[k][0] << " ";

        // set row index for next point in
        // diagonal
        int i = k - 1;

        // set column index for next point in
        //    diagonal
        int j = 1;

        /* Print Diagonally upward */
        while (isValid(i, j)) {
            cout << arr[i][j] << " ";
            i--;

            // move in upright direction
            j++;
        }
        cout << endl;
    }

    /* through this for loop we choose
       each element of last row as starting
       point (except the [0][c-1] it has
       already been processed in previous
       for loop) and print diagonal starting
       at it. arr[R-1][0], arr[R-1][1]....arr[R-1][c-1]
       are all starting points
     */

    // Note : we start from k = 1 to C-1;
    for (int k = 1; k < C; k++)
    {
        cout << arr[R - 1][k] << " ";

        // set row index for next point in
        // diagonal
        int i = R - 2;

        // set column index for next point in
        // diagonal
        int j = k + 1;

        /* Print Diagonally upward */
        while (isValid(i, j))
        {
            cout << arr[i][j] << " ";
            i--;

            // move in upright direction
            j++;
        }
        cout << endl;
    }
}

// Driver Code
int main()
{

    int arr[][C] = {
        { 1, 2, 3, 4 },
        { 5, 6, 7, 8 },
        { 9, 10, 11, 12 },
        { 13, 14, 15, 16 },
        { 17, 18, 19, 20 },
    };
    diagonalOrder(arr);
    return 0;
}

Java

// JAVA Code for Zigzag (or diagonal)
// traversal of Matrix

class GFG {

    public static int R, C;

    private static void diagonalOrder(int[][] arr)
    {

        /* through this for loop we choose each
        element of first column as starting point
        and print diagonal starting at it. arr[0][0],
        arr[1][0]....arr[R-1][0] are all starting points */
        for (int k = 0; k < R; k++) {
            System.out.print(arr[k][0] + " ");

            // set row index for next
            // point in diagonal
            int i = k - 1;

            //  set column index for
            // next point in diagonal
            int j = 1;

            /* Print Diagonally upward */
            while (isValid(i, j))
            {
                System.out.print(arr[i][j] + " ");

                i--;

                // move in upright direction
                j++;
            }

            System.out.println("");
        }

        /* through this for loop we choose each element
           of last row as starting point (except the
           [0][c-1] it has already been processed in
           previous for loop) and print diagonal
           starting at it. arr[R-1][0], arr[R-1][1]....
           arr[R-1][c-1] are all starting points */

        // Note : we start from k = 1 to C-1;
        for (int k = 1; k < C; k++) {
            System.out.print(arr[R - 1][k] + " ");

            // set row index for next
            // point in diagonal
            int i = R - 2;

            // set column index for
            // next point in diagonal
            int j = k + 1;

            /* Print Diagonally upward */
            while (isValid(i, j))
            {
                System.out.print(arr[i][j] + " ");

                // move in upright direction
                i--;
                j++;
            }

            System.out.println("");
        }
    }

    public static boolean isValid(int i, int j)
    {
        if (i < 0 || i >= R
            || j >= C || j < 0)
            return false;
        return true;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[][] = {
            { 1, 2, 3, 4 },
            { 5, 6, 7, 8 },
            { 9, 10, 11, 12 },
            { 13, 14, 15, 16 },
            { 17, 18, 19, 20 },
        };

        R = arr.length;
        C = arr[0].length;

        // Function call
        diagonalOrder(arr);
    }
}

// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 program to print all elements
# of given matrix in diagonal order
R = 5
C = 4


def isValid(i, j):
    if (i < 0 or i >= R or j >= C or j < 0):
        return False
    return True


def diagonalOrder(arr):

    # through this for loop we choose each element
    # of first column as starting point and print
    # diagonal starting at it.
    # arr[0][0], arr[1][0]....arr[R-1][0]
    # are all starting points
    for k in range(0, R):
        print(arr[k][0], end="  ")

        # set row index for next point in diagonal
        i = k - 1

        # set column index for next point in diagonal
        j = 1

        # Print Diagonally upward
        while (isValid(i, j)):
            print(arr[i][j], end="  ")
            i -= 1
            j += 1  # move in upright direction

        print()

    # Through this for loop we choose each
    # element of last row as starting point
    # (except the [0][c-1] it has already been
    # processed in previous for loop) and print
    # diagonal starting at it.
    # arr[R-1][0], arr[R-1][1]....arr[R-1][c-1]
    # are all starting points

    # Note : we start from k = 1 to C-1;
    for k in range(1, C):
        print(arr[R-1][k], end="  ")

        # set row index for next point in diagonal
        i = R - 2

        # set column index for next point in diagonal
        j = k + 1

        # Print Diagonally upward
        while (isValid(i, j)):
            print(arr[i][j], end="  ")
            i -= 1
            j += 1  # move in upright direction

        print()


# Driver Code
arr = [[1, 2, 3, 4],
       [5, 6, 7, 8],
       [9, 10, 11, 12],
       [13, 14, 15, 16],
       [17, 18, 19, 20]]

# Function call
diagonalOrder(arr)

# This code is contributed by Nikita Tiwari.

C#

// C# Code for Zigzag (or diagonal)
// traversal of Matrix
using System;

class GFG {
    public static int R, C;

    private static void diagonalOrder(int[, ] arr)
    {

        /* through this for loop we
        choose each element of first
        column as starting point and
        print diagonal starting at it.
        arr[0,0], arr[1,0]....arr[R-1,0]
        are all starting points */
        for (int k = 0; k < R; k++)
        {
            Console.Write(arr[k, 0] + " ");

            // set row index for next
            // point in diagonal
            int i = k - 1;

            // set column index for
            // next point in diagonal
            int j = 1;

            /* Print Diagonally upward */
            while (isValid(i, j))
            {
                Console.Write(arr[i, j] + " ");

                i--;

                // move in upright direction
                j++;
            }

            Console.Write("\n");
        }

        /*  through this for loop we
            choose each element of last
            row as starting point (except
            the [0][c-1] it has already
            been processed in previous for
            loop) and print diagonal starting
            at it. arr[R-1,0], arr[R-1,1]....
            arr[R-1,c-1] are all starting points */

        // Note : we start from k = 1 to C-1;
        for (int k = 1; k < C; k++)
        {
            Console.Write(arr[R - 1, k] + " ");

            // set row index for next
            // point in diagonal
            int i = R - 2;

            // set column index for
            // next point in diagonal
            int j = k + 1;

            /* Print Diagonally upward */
            while (isValid(i, j))
            {
                Console.Write(arr[i, j] + " ");

                i--;
                j++; // move in upright direction
            }

            Console.Write("\n");
        }
    }

    public static bool isValid(int i, int j)
    {
        if (i < 0 || i >= R || j >= C || j < 0)
            return false;
        return true;
    }

    // Driver code
    public static void Main()
    {
        int[, ] arr = { { 1, 2, 3, 4 },
                        { 5, 6, 7, 8 },
                        { 9, 10, 11, 12 },
                        { 13, 14, 15, 16 },
                        { 17, 18, 19, 20 } };

        R = arr.GetLength(0);
        C = arr.GetLength(1);

        // Function call
        diagonalOrder(arr);
    }
}

// This code is contributed
// by ChitraNayal

PHP

<?php
// PHP code for Zigzag (or diagonal)
// traversal of Matrix
define("R", 5);
define("C", 4);

function isValid($i, $j)
{
    if ($i < 0 || $i >= R ||
        $j >= C || $j < 0) return false;
    return true;
}

function diagonalOrder(&$arr)
{
    /* through this for loop we choose
    each element of first column as
    starting point and print diagonal
    starting at it.
    arr[0][0], arr[1][0]....arr[R-1][0]
    are all starting points */
    for ($k = 0; $k < R; $k++)
    {
        echo $arr[$k][0] . " ";
        $i = $k - 1; // set row index for next
                     // point in diagonal
        $j = 1; // set column index for next
                // point in diagonal

        /* Print Diagonally upward */
        while (isValid($i,$j))
        {
            echo $arr[$i][$j] . " ";
            $i--;
            $j++; // move in upright direction
        }
        echo "\n";
    }

    /* through this for loop we choose each
    element of last row as starting point
    (except the [0][c-1] it has already been
    processed in previous for loop) and print
    diagonal starting at it. arr[R-1][0],
    arr[R-1][1]....arr[R-1][c-1] are all
    starting points */

    //Note : we start from k = 1 to C-1;
    for ($k = 1; $k < C; $k++)
    {
        echo $arr[R - 1][$k] . " ";
        $i = R - 2; // set row index for next
                    // point in diagonal
        $j = $k + 1; // set column index for next
                     // point in diagonal

        /* Print Diagonally upward */
        while (isValid($i, $j))
        {
            echo $arr[$i][$j] . " ";
            $i--;
            $j++; // move in upright direction
        }
        echo "\n";
    }
}

// Driver Code
$arr = array(array(1, 2, 3, 4),
             array(5, 6, 7, 8),
             array(9, 10, 11, 12),
             array(13, 14, 15, 16),
             array(17, 18, 19, 20));

// Function call
diagonalOrder($arr);

// This code is contributed
// by rathbhupendra
?>

Javascript

<script>
// JAVA Code for Zigzag (or diagonal)
// traversal of Matrix

var R, C;

    function diagonalOrder( arr)
    {

        /* through this for loop we choose each
        element of first column as starting point
        and print diagonal starting at it. arr[0][0],
        arr[1][0]....arr[R-1][0] are all starting points */
        for (var k = 0; k < R; k++) {
            document.write(arr[k][0] + " ");

            // set row index for next
            // point in diagonal
            var i = k - 1;

            //  set column index for
            // next point in diagonal
            var j = 1;

            /* Print Diagonally upward */
            while (isValid(i, j))
            {
                document.write(arr[i][j] + " ");

                i--;

                // move in upright direction
                j++;
            }

            document.writeln("<br>");
        }

        /* through this for loop we choose each element
           of last row as starting point (except the
           [0][c-1] it has already been processed in
           previous for loop) and print diagonal
           starting at it. arr[R-1][0], arr[R-1][1]....
           arr[R-1][c-1] are all starting points */

        // Note : we start from k = 1 to C-1;
        for (var k = 1; k < C; k++) {
            document.write(arr[R - 1][k] + " ");

            // set row index for next
            // point in diagonal
            var i = R - 2;

            // set column index for
            // next point in diagonal
            var j = k + 1;

            /* Print Diagonally upward */
            while (isValid(i, j))
            {
                document.write(arr[i][j] + " ");

                // move in upright direction
                i--;
                j++;
            }

            document.writeln("<br>");
        }
    }

    function isValid( i,  j)
    {
        if (i < 0 || i >= R
            || j >= C || j < 0)
            return false;
        return true;
    }

    // Driver code
            var arr = [
            [ 1, 2, 3, 4 ],
            [ 5, 6, 7, 8 ],
            [ 9, 10, 11, 12 ],
            [ 13, 14, 15, 16 ],
            [ 17, 18, 19, 20 ],
        ];

        R = arr.length;
        C = arr[0].length;

        // Function call
        diagonalOrder(arr);

// This code is contributed by shivanisinghss2110
</script>

Output

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

Time Complexity: O(row x col) 
Auxiliary Space: O(1)

Thanks to Gaurav Ahirwar for suggesting this method. 

• Print matrix in antispiral form
• Print matrix in spiral form
• Print a given matrix in zigzag form

This article is compiled by Ashish Anand and reviewed by GeeksforGeeks team. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Another Approach:

It’s a keen observation that the sum of [i+j] that is the indexes of the array remains the same throughout the diagonal. So we will exploit this property of the matrix to make our code short and simple.

Below is the implementation of the above idea:

C++

#include <bits/stdc++.h>
#define R 5
#define C 4
using namespace std;

void diagonalOrder(int arr[][C],
                   int n, int m)
{
    // we will use a 2D vector to
    // store the diagonals of our array
    // the 2D vector will have (n+m-1)
    // rows that is equal to the number of
    // diagonals
    vector<vector<int> > ans(n + m - 1);

    for (int i = 0; i < m; i++)
    {
        for (int j = 0; j < n; j++)
        {
            ans[i + j].push_back(arr[j][i]);
        }
    }

    for (int i = 0; i < ans.size(); i++)
    {
        for (int j = 0; j < ans[i].size(); j++)
            cout << ans[i][j] << " ";

        cout << endl;
    }
}

// Driver Code
int main()
{
    // we have a matrix of n rows
    // and m columns
    int n = 5, m = 4;
    int arr[][C] = {
        { 1, 2, 3, 4 },
        { 5, 6, 7, 8 },
        { 9, 10, 11, 12 },
        { 13, 14, 15, 16 },
        { 17, 18, 19, 20 },
    };

    // Function call
    diagonalOrder(arr, n, m);
    return 0;
}

Java

import java.util.*;
import java.io.*;

class GFG
{
  public static int R = 5, C = 4;
  public static void diagonalOrder(int[][] arr, int n, int m)
  {

    // we will use a 2D vector to
    // store the diagonals of our array
    // the 2D vector will have (n+m-1)
    // rows that is equal to the number of
    // diagonals
    ArrayList<ArrayList<Integer>> ans = new ArrayList<ArrayList<Integer>>(n+m-1);
    for(int i = 0; i < n + m - 1; i++)
    {
      ans.add(new ArrayList<Integer>());
    }

    for (int i = 0; i < n; i++)
    {
      for (int j = 0; j < m; j++)
      {
        (ans.get(i+j)).add(arr[i][j]);
      }
    }

    for (int i = 0; i < ans.size(); i++)
    {
      for (int j = ans.get(i).size() - 1; j >= 0; j--)
      {    System.out.print(ans.get(i).get(j)+ " ");
      }
      System.out.println();
    }
  }

  // Driver code
  public static void main (String[] args) {
    int n = 5, m = 4;
    int[][] arr={
      { 1, 2, 3, 4 },
      { 5, 6, 7, 8 },
      { 9, 10, 11, 12 },
      { 13, 14, 15, 16 },
      { 17, 18, 19, 20 },
    };

    // Function call
    diagonalOrder(arr, n, m);
  }
}

// This code is contributed by Manu Pathria

Python3

R = 5
C = 5
def diagonalOrder(arr, n, m):

    # we will use a 2D vector to
    # store the diagonals of our array
    # the 2D vector will have (n+m-1)
    # rows that is equal to the number of
    # diagonals
    ans = [[] for i in range(n + m - 1)]

    for i in range(m):
        for j in range(n):
            ans[i + j].append(arr[j][i])

    for i in range(len(ans)):
        for j in range(len(ans[i])):
            print(ans[i][j], end = " ")
        print()

# Driver Code
# we have a matrix of n rows
# and m columns
n = 5
m = 4

# Function call
arr = [[1, 2, 3, 4],[ 5, 6, 7, 8],[9, 10, 11, 12 ],[13, 14, 15, 16 ],[ 17, 18, 19, 20]]
diagonalOrder(arr, n, m)

# This code is contributed by rag2127

C#

using System;
using System.Collections.Generic;

public class GFG
{

  public static int R = 5, C = 4;
  public static void diagonalOrder(int[,] arr, int n, int m)
  {

    // we will use a 2D vector to
    // store the diagonals of our array
    // the 2D vector will have (n+m-1)
    // rows that is equal to the number of
    // diagonals
    List<List<int>> ans = new List<List<int>>(n+m-1);
    for(int i = 0; i < n + m - 1; i++)
    {
      ans.Add(new List<int>());
    }

    for (int i = 0; i < n; i++)
    {
      for (int j = 0; j < m; j++)
      {
        (ans[i + j]).Add(arr[i, j]);
      }
    }

    for (int i = 0; i < ans.Count; i++)
    {
      for (int j = ans[i].Count - 1; j >= 0; j--)
      {
        Console.Write(ans[i][j] + " ");
      }
      Console.WriteLine();
    }

  }
  // Driver code
  static public void Main ()
  {
    int n = 5, m = 4;
    int[,] arr={
      { 1, 2, 3, 4 },
      { 5, 6, 7, 8 },
      { 9, 10, 11, 12 },
      { 13, 14, 15, 16 },
      { 17, 18, 19, 20 },
    };

    // Function call
    diagonalOrder(arr, n, m);
  }
}

// This code is contributed by avanitrachhadiya2155

Javascript

<script>

var R = 5;
var C = 4;

function diagonalOrder(arr, n, m)
{
    // we will use a 2D vector to
    // store the diagonals of our array
    // the 2D vector will have (n+m-1)
    // rows that is equal to the number of
    // diagonals
    var ans = Array.from(Array(n+m-1), ()=>Array());

    for (var i = 0; i < n; i++)
    {
        for (var j = 0; j < m; j++)
        {
            ans[i + j].push(arr[i][j]);
        }
    }

    for (var i = 0; i < ans.length; i++)
    {
      for (var j = ans[i].length - 1; j >= 0; j--)
      {
        document.write(ans[i][j] + " ");
      }
      document.write("<br>");
    }
}

// Driver Code
// we have a matrix of n rows
// and m columns
var n = 5, m = 4;
var arr = [
    [ 1, 2, 3, 4 ],
    [ 5, 6, 7, 8 ],
    [ 9, 10, 11, 12 ],
    [ 13, 14, 15, 16 ],
    [ 17, 18, 19, 20 ],
];

// Function call
diagonalOrder(arr, n, m);

// This code is contributed by rrrtnx.
</script>

Output

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

Time Complexity: O(row x col) 
Auxiliary Space: O(row + col)