Diagonal bottom up traversal of matrix in 45 degree
In this programming tutorial, we will explore the concept of diagonal traversal of a matrix at a 45-degree angle from the bottom to the top. We will walk you through the problem statement, provide a detailed explanation using a suitable example, present the algorithm, pseudocode, and implementation in C, discuss the output, and conclude with the time complexity analysis.

Problem Statement
Given a matrix, we are tasked with traversing its elements diagonally from the bottom to the top at a 45-degree angle and printing those elements.
Example
Consider the following matrix:
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
The expected output of the diagonal traversal
is:1 6 2 11 7 3 16 12 8 4 17 13 9 5 18 14 10 19 15 20
Idea to Solve
The problem can be solved by dividing it into two parts: the first half diagonal elements and the second half diagonal elements. For the first half, we iterate through each row and traverse the diagonal in a bottom-up manner. For the second half, we start from the second column and iterate through the rows while traversing the diagonal elements.
Algorithm
- Initialize a matrix
arr
with the given elements. - For the first half diagonal elements:
- Loop through each row
i
from 0 toROW - 1
. - Inside this loop, iterate through column
j
from 0 toi
and ensure thatj < COL
andi - j >= 0
. - Print the element at
arr[i - j][j]
.
- Loop through each row
- For the second half diagonal elements:
- Loop through each column
i
from 1 toCOL - 1
. - Inside this loop, initialize
j
asROW - 1
andk
asi
. - While
j >= 0
andk < COL
, print the element atarr[j][k]
, and decrementj
and incrementk
.
- Loop through each column
- End the program.
Pseudocode
function diagonalTraversal(matrix):
for i from 0 to ROW - 1:
for j from 0 to i:
if j < COL and i - j >= 0:
print matrix[i - j][j]
for i from 1 to COL - 1:
j = ROW - 1
k = i
while j >= 0 and k < COL:
print matrix[j][k]
j = j - 1
k = k + 1
Code Solution
/*
C Program
+ Diagonal traversal of matrix in 45 degree from bottom up
*/
#include<stdio.h>
#define ROW 4
#define COL 5
//Display element from bottom to top diagonal elements
void diagonal(int matrix[ROW][COL])
{
//First half elements
for (int i = 0; i < ROW; ++i)
{
for (int j = 0; j <=i && j < COL && i-j >=0; ++j)
{
printf(" %d",matrix[i-j][j] );
}
}
//Second half elements
for (int i = 1; i < COL; ++i)
{
for (int j = ROW-1 , k = i; j >= 0 && k < COL; --j, k++)
{
printf(" %d",matrix[j][k] );
}
}
}
int main(){
int arr[ROW][COL]= {
{1, 2, 3, 4, 5},
{6, 7, 8, 9, 10},
{11, 12, 13, 14, 15},
{16, 17, 18, 19, 20}
};
diagonal(arr);
return 0;
}
Output
1 6 2 11 7 3 16 12 8 4 17 13 9 5 18 14 10 19 15 20
/*
C++ Program
Diagonal traversal of matrix in 45 degree from bottom up
*/
#include<iostream>
#define ROW 4
#define COL 5
using namespace std;
class MyMatrix {
public:
//Display element from bottom to top diagonal elements
void diagonal(int matrix[ROW][COL])
{
//First half elements
for (int i = 0; i < ROW; ++i)
{
for (int j = 0; j <=i && j < COL && i-j >=0; ++j)
{
cout<<" "<<matrix[i-j][j] ;
}
}
//Second half elements
for (int i = 1; i < COL; ++i)
{
for (int j = ROW-1 , k = i; j >= 0 && k < COL; --j, k++)
{
cout<<" "<<matrix[j][k] ;
}
}
}
};
int main() {
MyMatrix obj;
int matrix[][COL] ={
{1, 2, 3, 4, 5},
{6, 7, 8, 9, 10},
{11, 12, 13, 14, 15},
{16, 17, 18, 19, 20}
};
obj.diagonal(matrix);
return 0;
}
Output
1 6 2 11 7 3 16 12 8 4 17 13 9 5 18 14 10 19 15 20
/*
Java Program
Diagonal traversal of matrix in 45 degree from bottom up
*/
public class MyMatrix {
//Display element from bottom to top diagonal elements
public void diagonal(int [][]matrix)
{
//Get the size of matrix
int row = matrix.length;
int col = matrix[0].length;
//First half elements
for (int i = 0; i < row; ++i)
{
for (int j = 0; j <=i && j < col && i-j >=0; ++j)
{
System.out.print(" "+matrix[i-j][j] );
}
}
//Second half elements
for (int i = 1; i < col; ++i)
{
for (int j = row-1 , k = i; j >= 0 && k < col; --j, k++)
{
System.out.print(" "+matrix[j][k] );
}
}
}
public static void main(String[] args) {
MyMatrix obj = new MyMatrix();
//Define matrix
int [][]matrix ={
{1, 2, 3, 4, 5},
{6, 7, 8, 9, 10},
{11, 12, 13, 14, 15},
{16, 17, 18, 19, 20}
};
obj.diagonal(matrix);
}
}
Output
1 6 2 11 7 3 16 12 8 4 17 13 9 5 18 14 10 19 15 20
/*
C# Program
Diagonal traversal of matrix in 45 degree from bottom up
*/
using System;
public class MyMatrix {
//Display element from bottom to top diagonal elements
public void diagonal(int[,] matrix) {
//Get the size of matrix
int row = matrix.GetLength(0);
int col = matrix.GetLength(1);
//First half elements
for (int i = 0; i < row; ++i) {
for (int j = 0; j <= i && j < col && i - j >= 0; ++j) {
Console.Write(" " + matrix[i - j,j]);
}
}
//Second half elements
for (int i = 1; i < col; ++i) {
for (int j = row - 1, k = i; j >= 0 && k < col; --j, k++) {
Console.Write(" " + matrix[j,k]);
}
}
}
public static void Main(String[] args) {
MyMatrix obj = new MyMatrix();
//Define matrix
int[,] matrix = {
{
1,
2,
3,
4,
5
},
{
6,
7,
8,
9,
10
},
{
11,
12,
13,
14,
15
},
{
16,
17,
18,
19,
20
}
};
obj.diagonal(matrix);
}
}
Output
1 6 2 11 7 3 16 12 8 4 17 13 9 5 18 14 10 19 15 20
<?php
/*
Php Program
Diagonal traversal of matrix in 45 degree from bottom up
*/
class MyMatrix {
//Display element from bottom to top diagonal elements
public function diagonal($matrix) {
//Get the size of matrix
$row = count($matrix);
$col = count($matrix[0]);
//First half elements
for ($i = 0; $i < $row; ++$i) {
for ($j = 0; $j <= $i && $j < $col && $i - $j >= 0; ++$j) {
echo(" ". $matrix[$i - $j][$j]);
}
}
//Second half elements
for ($i = 1; $i < $col; ++$i) {
for ($j = $row - 1, $k = $i; $j >= 0 && $k < $col; --$j, $k++) {
echo(" ". $matrix[$j][$k]);
}
}
}
};
function main() {
$obj = new MyMatrix();
//Define matrix
$matrix = array(array(1, 2, 3, 4, 5), array(6, 7, 8, 9, 10), array(11, 12, 13, 14, 15), array(16, 17, 18, 19, 20));
$obj->diagonal($matrix);
}
main();
Output
1 6 2 11 7 3 16 12 8 4 17 13 9 5 18 14 10 19 15 20
/*
Node Js Program
Diagonal traversal of matrix in 45 degree from bottom up
*/
class MyMatrix {
//Display element from bottom to top diagonal elements
diagonal(matrix) {
//Get the size of matrix
var row = matrix.length;
var col = matrix[0].length;
//First half elements
for (var i = 0; i < row; ++i) {
for (var j = 0; j <= i && j < col && i - j >= 0; ++j) {
process.stdout.write(" " + matrix[i - j][j]);
}
}
//Second half elements
for (var i = 1; i < col; ++i) {
for (var j = row - 1, k = i; j >= 0 && k < col; --j, k++) {
process.stdout.write(" " + matrix[j][k]);
}
}
}
}
function main(args) {
var obj = new MyMatrix();
//Define matrix
var matrix = [
[1, 2, 3, 4, 5],
[6, 7, 8, 9, 10],
[11, 12, 13, 14, 15],
[16, 17, 18, 19, 20]
];
obj.diagonal(matrix);
}
main();
Output
1 6 2 11 7 3 16 12 8 4 17 13 9 5 18 14 10 19 15 20
# Python 3 Program
# Diagonal traversal of matrix in 45 degree from bottom up
class MyMatrix :
# Display element from bottom to top diagonal elements
def diagonal(self, matrix) :
row = len(matrix)
col = len(matrix[0])
# First half elements
i = 0
while (i < row) :
j = 0
while (j <= i and j < col and i - j >= 0) :
print(" ", matrix[i - j][j], end = "")
j += 1
i += 1
# Second half elements
i = 1
while (i < col) :
j = row - 1
k = i
while (j >= 0 and k < col) :
print(" ", matrix[j][k], end = "")
j -= 1
k += 1
i += 1
def main() :
obj = MyMatrix()
matrix = [
[1, 2, 3, 4, 5],
[6, 7, 8, 9, 10],
[11, 12, 13, 14, 15],
[16, 17, 18, 19, 20]
]
obj.diagonal(matrix)
if __name__ == "__main__":
main()
Output
1 6 2 11 7 3 16 12 8 4 17 13 9 5 18 14 10 19 15 20
# Ruby Program
# Diagonal traversal of matrix in 45 degree from bottom up
class MyMatrix
# Display element from bottom to top diagonal elements
def diagonal(matrix)
row = matrix.length
col = matrix[0].length
# First half elements
i = 0
while (i < row)
j = 0
while (j <= i and j < col and i - j >= 0)
print(" ", matrix[i - j][j])
j += 1
end
i += 1
end
# Second half elements
i = 1
while (i < col)
j = row - 1
k = i
while (j >= 0 and k < col)
print(" ", matrix[j][k])
j -= 1
k += 1
end
i += 1
end
end
end
def main()
obj = MyMatrix.new()
matrix = [
[1, 2, 3, 4, 5],
[6, 7, 8, 9, 10],
[11, 12, 13, 14, 15],
[16, 17, 18, 19, 20]
]
obj.diagonal(matrix)
end
main()
Output
1 6 2 11 7 3 16 12 8 4 17 13 9 5 18 14 10 19 15 20
/*
Scala Program
Diagonal traversal of matrix in 45 degree from bottom up
*/
class MyMatrix {
//Display element from bottom to top diagonal elements
def diagonal(matrix: Array[Array[Int]]): Unit = {
val row: Int = matrix.length;
val col: Int = matrix(0).length;
//First half elements
var i: Int = 0;
var j: Int = 0;
while (i < row) {
j = 0;
while (j <= i && j < col && i - j >= 0) {
print(" " + matrix(i - j)(j));
j += 1;
}
i += 1;
}
//Second half elements
i = 1;
while (i < col) {
j = row - 1;
var k: Int = i;
while (j >= 0 && k < col) {
print(" " + matrix(j)(k));
j -= 1;
k += 1;
}
i += 1;
}
}
}
object Main {
def main(args: Array[String]): Unit = {
val obj: MyMatrix = new MyMatrix();
val matrix: Array[Array[Int]] = Array(
Array(1, 2, 3, 4, 5),
Array(6, 7, 8, 9, 10),
Array(11, 12, 13, 14, 15),
Array(16, 17, 18, 19, 20));
obj.diagonal(matrix);
}
}
Output
1 6 2 11 7 3 16 12 8 4 17 13 9 5 18 14 10 19 15 20
/*
Swift 4 Program
Diagonal traversal of matrix in 45 degree from bottom up
*/
class MyMatrix {
//Display element from bottom to top diagonal elements
func diagonal(_ matrix: [
[Int]
]) {
let row: Int = matrix.count;
let col: Int = matrix[0].count;
//First half elements
var i: Int = 0;
var j: Int = 0;
while (i < row) {
j = 0;
while (j <= i && j < col && i - j >= 0) {
print(" ", matrix[i - j][j], terminator: "");
j += 1;
}
i += 1;
}
//Second half elements
i = 1;
while (i < col) {
j = row - 1;
var k: Int = i;
while (j >= 0 && k < col) {
print(" ", matrix[j][k], terminator: "");
j -= 1;
k += 1;
}
i += 1;
}
}
}
func main() {
let obj: MyMatrix = MyMatrix();
let matrix: [
[Int]
] = [
[1, 2, 3, 4, 5],
[6, 7, 8, 9, 10],
[11, 12, 13, 14, 15],
[16, 17, 18, 19, 20]
];
obj.diagonal(matrix);
}
main();
Output
1 6 2 11 7 3 16 12 8 4 17 13 9 5 18 14 10 19 15 20
Time Complexity Analysis
For the first half traversal, we have two nested loops: the outer loop runs ROW
times, and the
inner loop runs at most COL
times. For the second half traversal, again we have two nested loops:
the outer loop runs COL - 1
times, and the inner loop runs at most ROW
times.
Therefore, the time complexity is dominated by the number of elements in the matrix, which is
ROW * COL
, resulting in O(ROW * COL) time complexity.
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