Print matrix in diagonal pattern

Given a matrix of n*n size, the task is to print its elements in a diagonal pattern.

Input : mat[3][3] = {{1, 2, 3},
                     {4, 5, 6},
                     {7, 8, 9}}
Output : 1 2 4 7 5 3 6 8 9.
Explanation: Start from 1 
Then from upward to downward diagonally i.e. 2 and 4
Then from downward to upward diagonally i.e 7, 5, 3 
Then from up to down diagonally i.e  6, 8 
Then down to up i.e. end at 9.

Input :  mat[4][4] =  {{1,  2,  3,  10},
                      {4,  5,  6,  11},
                      {7,  8,  9,  12},
                      {13, 14, 15, 16}}
Output:  1 2 4 7 5 3 10 6 8 13 14 9 11 12 15 16 .
Explanation: Start from 1 
Then from upward to downward diagonally i.e. 2 and 4
Then from downward to upward diagonally i.e 7, 5, 3 
Then from upward to downward diagonally i.e. 10 6 8 13
Then from downward to upward diagonally i.e 14 9 11
Then from upward to downward diagonally i.e. 12 15
then end at 16

Approach: From the diagram it can be seen that every element is either printed diagonally upward or diagonally downward. Start from the index (0,0) and print the elements diagonally upward then change the direction, change the column and print diagonally downwards. This cycle continues until the last element is reached.

Algorithm:

  1. Create variables i=0, j=0 to store the current indices of row and column
  2. Run a loop from 0 to n*n, where n is side of the matrix.
  3. Use a flag isUp to decide whether the direction is upwards or downwards. Set isUp = true initially the direction is going upward.
  4. If isUp = 1 then start printing elements by incrementing column index and decrementing the row index.
  5. Similarly if isUp = 0, then decrement the column index and increment the row index.
  6. Move to the next column or row (next starting row and column
  7. Do this till all the elements get traversed.

Implementation:

C++



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// C++ program to print matrix in diagonal order
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100;
  
void printMatrixDiagonal(int mat[MAX][MAX], int n)
{
    // Initialize indexes of element to be printed next
    int i = 0, j = 0;
  
    // Direction is initially from down to up
    bool isUp = true;
  
    // Traverse the matrix till all elements get traversed
    for (int k = 0; k < n * n;) {
        // If isUp = true then traverse from downward
        // to upward
        if (isUp) {
            for (; i >= 0 && j < n; j++, i--) {
                cout << mat[i][j] << " ";
                k++;
            }
  
            // Set i and j according to direction
            if (i < 0 && j <= n - 1)
                i = 0;
            if (j == n)
                i = i + 2, j--;
        }
  
        // If isUp = 0 then traverse up to down
        else {
            for (; j >= 0 && i < n; i++, j--) {
                cout << mat[i][j] << " ";
                k++;
            }
  
            // Set i and j according to direction
            if (j < 0 && i <= n - 1)
                j = 0;
            if (i == n)
                j = j + 2, i--;
        }
  
        // Revert the isUp to change the direction
        isUp = !isUp;
    }
}
  
int main()
{
    int mat[MAX][MAX] = { { 1, 2, 3 },
                          { 4, 5, 6 },
                          { 7, 8, 9 } };
  
    int n = 3;
    printMatrixDiagonal(mat, n);
    return 0;
}

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Java

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// Java program to print matrix in diagonal order
class GFG {
    static final int MAX = 100;
  
    static void printMatrixDiagonal(int mat[][], int n)
    {
        // Initialize indexes of element to be printed next
        int i = 0, j = 0;
  
        // Direction is initially from down to up
        boolean isUp = true;
  
        // Traverse the matrix till all elements get traversed
        for (int k = 0; k < n * n;) {
            // If isUp = true then traverse from downward
            // to upward
            if (isUp) {
                for (; i >= 0 && j < n; j++, i--) {
                    System.out.print(mat[i][j] + " ");
                    k++;
                }
  
                // Set i and j according to direction
                if (i < 0 && j <= n - 1)
                    i = 0;
                if (j == n) {
                    i = i + 2;
                    j--;
                }
            }
  
            // If isUp = 0 then traverse up to down
            else {
                for (; j >= 0 && i < n; i++, j--) {
                    System.out.print(mat[i][j] + " ");
                    k++;
                }
  
                // Set i and j according to direction
                if (j < 0 && i <= n - 1)
                    j = 0;
                if (i == n) {
                    j = j + 2;
                    i--;
                }
            }
  
            // Revert the isUp to change the direction
            isUp = !isUp;
        }
    }
  
    // Driver code
    public static void main(String[] args)
    {
        int mat[][] = { { 1, 2, 3 },
                        { 4, 5, 6 },
                        { 7, 8, 9 } };
  
        int n = 3;
        printMatrixDiagonal(mat, n);
    }
}
// This code is contributed by Anant Agarwal.

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Python 3

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# Python 3 program to print matrix in diagonal order
MAX = 100
  
def printMatrixDiagonal(mat, n):
    # Initialize indexes of element to be printed next
    i = 0
    j = 0
    k = 0
    # Direction is initially from down to up
    isUp = True
  
     # Traverse the matrix till all elements get traversed
    while k<n * n:
         # If isUp = True then traverse from downward
         # to upward
        if isUp:
            while i >= 0 and j<n :
                print(str(mat[i][j]), end = " ")
                k += 1
                j += 1
                i -= 1
  
              # Set i and j according to direction
            if i < 0 and j <= n - 1:
                i = 0
            if j == n:
                i = i + 2
                j -= 1
  
         # If isUp = 0 then traverse up to down
        else:
            while j >= 0 and i<n :
                print(mat[i][j], end = " ")
                k += 1
                i += 1
                j -= 1
  
              # Set i and j according to direction
            if j < 0 and i <= n - 1:
                j = 0
            if i == n:
                j = j + 2
                i -= 1
  
         # Revert the isUp to change the direction
        isUp = not isUp
  
# Driver program
if __name__ == "__main__":
    mat = [[1, 2, 3],
        [4, 5, 6],
        [7, 8, 9] ]
  
   n = 3
   printMatrixDiagonal(mat, n)
  
# This code is contributed by Chitra Nayal

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C#

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// C# program to print matrix in diagonal order
using System;
class GFG {
    static int MAX = 100;
  
    static void printMatrixDiagonal(int[, ] mat, int n)
    {
        // Initialize indexes of element to be printed next
        int i = 0, j = 0;
  
        // Direction is initially from down to up
        bool isUp = true;
  
        // Traverse the matrix till all elements get traversed
        for (int k = 0; k < n * n;) {
            // If isUp = true then traverse from downward
            // to upward
            if (isUp) {
                for (; i >= 0 && j < n; j++, i--) {
                    Console.Write(mat[i, j] + " ");
                    k++;
                }
  
                // Set i and j according to direction
                if (i < 0 && j <= n - 1)
                    i = 0;
                if (j == n) {
                    i = i + 2;
                    j--;
                }
            }
  
            // If isUp = 0 then traverse up to down
            else {
                for (; j >= 0 && i < n; i++, j--) {
                    Console.Write(mat[i, j] + " ");
                    k++;
                }
  
                // Set i and j according to direction
                if (j < 0 && i <= n - 1)
                    j = 0;
                if (i == n) {
                    j = j + 2;
                    i--;
                }
            }
  
            // Revert the isUp to change the direction
            isUp = !isUp;
        }
    }
  
    // Driver code
    public static void Main()
    {
        int[, ] mat = { { 1, 2, 3 },
                        { 4, 5, 6 },
                        { 7, 8, 9 } };
  
        int n = 3;
        printMatrixDiagonal(mat, n);
    }
}
// This code is contributed by vt_m.

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PHP

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<?php
// php program to print matrix
// in diagonal order
  
$MAX = 100;
  
function printMatrixDiagonal($mat, $n)
{
      
    // Initialize indexes of element
    // to be printed next
    $i = 0;
    $j = 0 ;
  
    // Direction is initially
    // from down to up
    $isUp = true;
  
    // Traverse the matrix till
    // all elements get traversed
    for ($k = 0;$k < $n * $n😉
    {
        // If isUp = true then traverse 
        // from downward to upward
        if ($isUp)
        {
            for ( ;$i >= 0 && $j < $n;$j++, $i--)
            {
                echo $mat[$i][$j]." ";
                $k++;
            }
  
            // Set i and j according 
            // to direction
            if ($i < 0 && $j <= $n - 1)
                $i = 0;
            if ($j == $n)
            {
                $i = $i + 2;
                $j--;
            }
        }
  
        // If isUp = 0 then 
        // traverse up to down
        else
        {
            for ( ; $j >= 0 && 
                 $i<$n ; $i++, $j--)
            {
                echo $mat[$i][$j]." ";
                $k++;
            }
  
            // Set i and j according
            // to direction
            if ($j < 0 && $i <= $n - 1)
                $j = 0;
            if ($i == $n)
            {
                $j = $j + 2;
                $i--;
            }
        }
  
        // Revert the isUp to
        // change the direction
        $isUp = !$isUp;
    }
}
  
    // Driver code
    $mat= array(array(1, 2, 3),
          array(4, 5, 6),
          array(7, 8, 9));
  
    $n = 3;
    printMatrixDiagonal($mat, $n);
  
// This code is contributed by mits 
?>

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Output:

1 2 4 7 5 3 6 8 9

Complexity Analysis:

  • Time Complexity: O(n*n).
    To traverse the matrix O(n*n) time complexity is needed.
  • Space Compelxity: O(1).
    As no extra space is required.

Alternate Implementation: This is another simple and compact implementation of the same approach as mentioned above.

C++

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// C++ program to print matrix in diagonal order
#include <bits/stdc++.h>
using namespace std;
  
int main()
{
    // Initialize matrix
    int mat[][4] = { { 1, 2, 3, 4 },
                     { 5, 6, 7, 8 },
                     { 9, 10, 11, 12 },
                     { 13, 14, 15, 16 } };
    // n - size
    // mode - switch to derive up/down traversal
    // it - iterator count - increases until it
    // reaches n and then decreases
    int n = 4, mode = 0, it = 0, lower = 0;
  
    // 2n will be the number of iterations
    for (int t = 0; t < (2 * n - 1); t++) {
        int t1 = t;
        if (t1 >= n) {
            mode++;
            t1 = n - 1;
            it--;
            lower++;
        }
        else {
            lower = 0;
            it++;
        }
        for (int i = t1; i >= lower; i--) {
            if ((t1 + mode) % 2 == 0) {
                cout << (mat[i][t1 + lower - i]) << endl;
            }
            else {
                cout << (mat[t1 + lower - i][i]) << endl;
            }
        }
    }
    return 0;
}
  
// This code is contributed by princiraj1992

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Java

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// Java program to print matrix in diagonal order
public class MatrixDiag {
  
    public static void main(String[] args)
    {
        // Initialize matrix
        int[][] mat = { { 1, 2, 3, 4 },
                        { 5, 6, 7, 8 },
                        { 9, 10, 11, 12 },
                        { 13, 14, 15, 16 } };
        // n - size
        // mode - switch to derive up/down traversal
        // it - iterator count - increases until it
        // reaches n and then decreases
        int n = 4, mode = 0, it = 0, lower = 0;
  
        // 2n will be the number of iterations
        for (int t = 0; t < (2 * n - 1); t++) {
            int t1 = t;
            if (t1 >= n) {
                mode++;
                t1 = n - 1;
                it--;
                lower++;
            }
            else {
                lower = 0;
                it++;
            }
            for (int i = t1; i >= lower; i--) {
                if ((t1 + mode) % 2 == 0) {
                    System.out.println(mat[i][t1 + lower - i]);
                }
                else {
                    System.out.println(mat[t1 + lower - i][i]);
                }
            }
        }
    }
}

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Python3

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# Python3 program to prmatrix in diagonal order
  
# Driver code
  
# Initialize matrix
mat = [[ 1, 2, 3, 4 ],
       [ 5, 6, 7, 8 ],
       [ 9, 10, 11, 12 ],
       [ 13, 14, 15, 16 ]];
         
# n - size
# mode - switch to derive up/down traversal
# it - iterator count - increases until it
# reaches n and then decreases
n = 4
mode = 0
it = 0
lower = 0
  
# 2n will be the number of iterations
for t in range(2 * n - 1):
    t1 = t
    if (t1 >= n):
        mode += 1
        t1 = n - 1
        it -= 1
        lower += 1
    else:
        lower = 0
        it += 1
  
    for i in range(t1, lower - 1, -1):
        if ((t1 + mode) % 2 == 0):
            print((mat[i][t1 + lower - i]))
        else:
            print(mat[t1 + lower - i][i])
  
# This code is contributed by princiraj1992

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C#

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// C# program to print matrix in diagonal order
using System;
  
public class MatrixDiag {
    // Driver code
    public static void Main(String[] args)
    {
        // Initialize matrix
        int[, ] mat = { { 1, 2, 3, 4 },
                        { 5, 6, 7, 8 },
                        { 9, 10, 11, 12 },
                        { 13, 14, 15, 16 } };
        // n - size
        // mode - switch to derive up/down traversal
        // it - iterator count - increases until it
        // reaches n and then decreases
        int n = 4, mode = 0, it = 0, lower = 0;
  
        // 2n will be the number of iterations
        for (int t = 0; t < (2 * n - 1); t++) {
            int t1 = t;
            if (t1 >= n) {
                mode++;
                t1 = n - 1;
                it--;
                lower++;
            }
            else {
                lower = 0;
                it++;
            }
            for (int i = t1; i >= lower; i--) {
                if ((t1 + mode) % 2 == 0) {
                    Console.WriteLine(mat[i, t1 + lower - i]);
                }
                else {
                    Console.WriteLine(mat[t1 + lower - i, i]);
                }
            }
        }
    }
}
  
// This code contributed by Rajput-Ji

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Output:

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

This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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