Zigzag (or diagonal) traversal of Matrix

Difficulty Level : Medium
Last Updated : 20 Jan, 2021

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

Another example:

diagonal-matrix

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.

C++

#include <stdio.h> 
#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

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 
?> 

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

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

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 
?> 

Output

Thanks to Gaurav Ahirwar for suggesting this method.

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 
    // diagnols 
    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; 
}

Output

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My Personal Notes

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

C++

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C++

Java

Python3

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Another Approach:

C++