Print matrix in diagonal pattern
Last Updated :
11 Sep, 2023
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:
- Create variables i=0, j=0 to store the current indices of row and column
- Run a loop from 0 to n*n, where n is side of the matrix.
- Use a flag isUp to decide whether the direction is upwards or downwards. Set isUp = true initially the direction is going upward.
- If isUp = 1 then start printing elements by incrementing column index and decrementing the row index.
- Similarly if isUp = 0, then decrement the column index and increment the row index.
- Move to the next column or row (next starting row and column
- Do this till all the elements get traversed.
Implementation:
C++
// 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;
}
// Java program to print matrix in diagonal order
import java.io.*;
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.
# 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
// 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.
<?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 Surbhi Tyagi
?>
<script>
// Javascript program to print
// matrix in diagonal order
function printMatrixDiagonal(arr, len) {
// Initialize indices to traverse
// through the array
let i = 0, j = 0;
// Direction is initially from
// down to up
let isUp = true;
// Traverse the matrix till all
// elements get traversed
for (let k = 0; k < len * len;) {
// If isUp = true traverse from
// bottom to top
if (isUp) {
for (;i >= 0 && j < len; i--, j++) {
document.write(arr[i][j] + ' ');
k++;
}
// Set i and j according to direction
if (i < 0 && j < len) i = 0;
if (j === len) i = i + 2, j--;
}
// If isUp = false then traverse
// from top to bottom
else {
for (;j >= 0 && i < len; i++, j--) {
document.write(arr[i][j] + ' ');
k++;
}
// Set i and j according to direction
if (j < 0 && i < len) j = 0;
if (i === len) j = j + 2, i--;
}
// Inverse the value of isUp to
// change the direction
isUp = !isUp
}
}
// function call
let arr = [[1, 2, 3], [4, 5, 6], [7, 8, 9]];
let arrLength = arr.length;
printMatrixDiagonal(arr, arrLength);
// This code is contributed by karthiksrinivasprasad
</script>
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. - Auxiliary Space: 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++
// 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
// Java program to print matrix in diagonal order
import java.io.*;
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]);
}
}
}
}
}
# Python3 program to print matrix 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
// 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
<script>
// Javascript program to print an n * n matrix in diagonal order
function MatrixDiag(arr){
// n - size of the array
// mode - to switch from top/bottom traversal
// it - iterator count - increases until it
// reaches n and then decreases
// lower - to ensure we move to the
// next row when columns go out of bounds
let n = arr.length, mode = 0, it = 0, lower = 0;
// A 4 * 4 matrix has 7 diagonals.
// Hence (2 * n -1) iterations
for(let t=0; t< (2 * n - 1); t++){
let t1 = t;
if(t1 >= n){
mode++;
t1 = n - 1;
it--;
lower++;
} else {
lower = 0;
it++;
}
for(let i = t1; i>= lower; i--){
if((t1 + mode) % 2 === 0){
console.log(arr[i][t1 + lower - i])
} else{
console.log(arr[t1 + lower - i][i])
}
}
}
}
// function call
let arr = [
[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]
];
MatrixDiag(arr)
// This code is contributed by karthiksrinivasprasad
</script>
Output: 1
2
5
9
6
3
4
7
10
13
14
11
8
12
15
16
Time complexity: O(n2).
Auxiliary Space: O(1).
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