Given a matrix mat[][] of size N*N, the task is to rotate the matrix by 45 degrees and print the matrix.
Examples:
Input: N = 6,
mat[][] = {{3, 4, 5, 1, 5, 9, 5},
{6, 9, 8, 7, 2, 5, 2},
{1, 5, 9, 7, 5, 3, 2},
{4, 7, 8, 9, 3, 5, 2},
{4, 5, 2, 9, 5, 6, 2},
{4, 5, 7, 2, 9, 8, 3}}
Output:
3
6 4
1 9 5
4 5 8 1
4 7 9 7 5
4 5 8 7 2 9
5 2 9 5 5
7 9 3 3
2 5 5
9 6
8Input: N = 4,
mat[][] = {{2, 5, 7, 2},
{9, 1, 4, 3},
{5, 8, 2, 3},
{6, 4, 6, 3}}Output:
2
9 5
5 1 7
6 8 4 2
4 2 3
6 3
3
Approach 1:
Follow the steps given below in order to solve the problem:
- Store the diagonal elements in a list using a counter variable.
- Print the number of spaces required to make the output look like the desired pattern.
- Print the list elements after reversing the list.
- Traverse through only diagonal elements to optimize the time taken by the operation.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to rotate matrix by 45 degree void matrix( int n, int m, vector<vector< int >> li)
{ // Counter Variable
int ctr = 0;
while (ctr < 2 * n - 1)
{
for ( int i = 0;
i < abs (n - ctr - 1);
i++)
{
cout << " " ;
}
vector< int > lst;
// Iterate [0, m]
for ( int i = 0; i < m; i++)
{
// Iterate [0, n]
for ( int j = 0; j < n; j++)
{
// Diagonal Elements
// Condition
if (i + j == ctr)
{
// Appending the
// Diagonal Elements
lst.push_back(li[i][j]);
}
}
}
// Printing reversed Diagonal
// Elements
for ( int i = lst.size() - 1; i >= 0; i--)
{
cout << lst[i] << " " ;
}
cout << endl;
ctr += 1;
}
} // Driver code int main()
{ // Dimensions of Matrix
int n = 8;
int m = n;
// Given matrix
vector<vector< int >> li{
{ 4, 5, 6, 9, 8, 7, 1, 4 },
{ 1, 5, 9, 7, 5, 3, 1, 6 },
{ 7, 5, 3, 1, 5, 9, 8, 0 },
{ 6, 5, 4, 7, 8, 9, 3, 7 },
{ 3, 5, 6, 4, 8, 9, 2, 1 },
{ 3, 1, 6, 4, 7, 9, 5, 0 },
{ 8, 0, 7, 2, 3, 1, 0, 8 },
{ 7, 5, 3, 1, 5, 9, 8, 5 } };
// Function call
matrix(n, m, li);
return 0;
} // This code is contributed by divyeshrabadiya07 |
// Java program for // the above approach import java.util.*;
class GFG{
// Function to rotate // matrix by 45 degree static void matrix( int n, int m,
int [][]li)
{ // Counter Variable
int ctr = 0 ;
while (ctr < 2 * n - 1 )
{
for ( int i = 0 ; i < Math.abs(n - ctr - 1 );
i++)
{
System.out.print( " " );
}
Vector<Integer> lst = new Vector<Integer>();
// Iterate [0, m]
for ( int i = 0 ; i < m; i++)
{
// Iterate [0, n]
for ( int j = 0 ; j < n; j++)
{
// Diagonal Elements
// Condition
if (i + j == ctr)
{
// Appending the
// Diagonal Elements
lst.add(li[i][j]);
}
}
}
// Printing reversed Diagonal
// Elements
for ( int i = lst.size() - 1 ; i >= 0 ; i--)
{
System.out.print(lst.get(i) + " " );
}
System.out.println();
ctr += 1 ;
}
} // Driver code public static void main(String[] args)
{ // Dimensions of Matrix
int n = 8 ;
int m = n;
// Given matrix
int [][] li = {{ 4 , 5 , 6 , 9 , 8 , 7 , 1 , 4 },
{ 1 , 5 , 9 , 7 , 5 , 3 , 1 , 6 },
{ 7 , 5 , 3 , 1 , 5 , 9 , 8 , 0 },
{ 6 , 5 , 4 , 7 , 8 , 9 , 3 , 7 },
{ 3 , 5 , 6 , 4 , 8 , 9 , 2 , 1 },
{ 3 , 1 , 6 , 4 , 7 , 9 , 5 , 0 },
{ 8 , 0 , 7 , 2 , 3 , 1 , 0 , 8 },
{ 7 , 5 , 3 , 1 , 5 , 9 , 8 , 5 }};
// Function call
matrix(n, m, li);
} } // This code is contributed by Princi Singh |
# Python3 program for the above approach # Function to rotate matrix by 45 degree def matrix(n, m, li):
# Counter Variable
ctr = 0
while (ctr < 2 * n - 1 ):
print ( " " * abs (n - ctr - 1 ), end = "")
lst = []
# Iterate [0, m]
for i in range (m):
# Iterate [0, n]
for j in range (n):
# Diagonal Elements
# Condition
if i + j = = ctr:
# Appending the
# Diagonal Elements
lst.append(li[i][j])
# Printing reversed Diagonal
# Elements
lst.reverse()
print ( * lst)
ctr + = 1
# Driver Code # Dimensions of Matrix n = 8
m = n
# Given matrix li = [[ 4 , 5 , 6 , 9 , 8 , 7 , 1 , 4 ],
[ 1 , 5 , 9 , 7 , 5 , 3 , 1 , 6 ],
[ 7 , 5 , 3 , 1 , 5 , 9 , 8 , 0 ],
[ 6 , 5 , 4 , 7 , 8 , 9 , 3 , 7 ],
[ 3 , 5 , 6 , 4 , 8 , 9 , 2 , 1 ],
[ 3 , 1 , 6 , 4 , 7 , 9 , 5 , 0 ],
[ 8 , 0 , 7 , 2 , 3 , 1 , 0 , 8 ],
[ 7 , 5 , 3 , 1 , 5 , 9 , 8 , 5 ]]
# Function Call matrix(n, m, li) |
// C# program for // the above approach using System;
using System.Collections;
class GFG{
// Function to rotate // matrix by 45 degree static void matrix( int n, int m,
int [,]li)
{ // Counter Variable
int ctr = 0;
while (ctr < 2 * n - 1)
{
for ( int i = 0;
i < Math.Abs(n - ctr - 1);
i++)
{
Console.Write( " " );
}
ArrayList lst = new ArrayList();
// Iterate [0, m]
for ( int i = 0; i < m; i++)
{
// Iterate [0, n]
for ( int j = 0; j < n; j++)
{
// Diagonal Elements
// Condition
if (i + j == ctr)
{
// Appending the
// Diagonal Elements
lst.Add(li[i, j]);
}
}
}
// Printing reversed Diagonal
// Elements
for ( int i = lst.Count - 1;
i >= 0; i--)
{
Console.Write(( int )lst[i] + " " );
}
Console.Write( "\n" );
ctr += 1;
}
} // Driver code public static void Main( string [] args)
{ // Dimensions of Matrix
int n = 8;
int m = n;
// Given matrix
int [,] li = {{4, 5, 6, 9, 8, 7, 1, 4},
{1, 5, 9, 7, 5, 3, 1, 6},
{7, 5, 3, 1, 5, 9, 8, 0},
{6, 5, 4, 7, 8, 9, 3, 7},
{3, 5, 6, 4, 8, 9, 2, 1},
{3, 1, 6, 4, 7, 9, 5, 0},
{8, 0, 7, 2, 3, 1, 0, 8},
{7, 5, 3, 1, 5, 9, 8, 5}};
// Function call
matrix(n, m, li);
} } // This code is contributed by Rutvik_56 |
<script> // Javascript program to rotate a matrix by 45 degrees // A function to rotate a matrix // mat[][] of size R x C. // Initially, m = R and n = C function matrix(m, n, mat)
{ let ctr = 0;
while (ctr < 2*n-1)
{
for (let i = 0; i < Math.abs(n-ctr-1); i++)
{
document.write( " " );
}
var list = [];
// Iterate [0, m]
for (let i = 0; i < m; i++)
{
// Iterate [0, n]
for (let j = 0; j < n; j++)
{
// Diagonal Elements
// Condition
if (i + j == ctr)
{
// Appending the
// Diagonal Elements
list.push(mat[i][j]);
}
}
}
// Print rotated matrix
for (let i = list.length-1; i >= 0; i--)
{
document.write(list[i] + " " );
}
document.write( "<br>" );
ctr+=1;
}
} // Driver code // Test Case 1 let R = 8; let C = 8; let a = [ [ 4, 5, 6, 9, 8, 7, 1, 4 ], [ 1, 5, 9, 7, 5, 3, 1, 6 ],
[ 7, 5, 3, 1, 5, 9, 8, 0 ],
[ 6, 5, 4, 7, 8, 9, 3, 7 ],
[ 3, 5, 6, 4, 8, 9, 2, 1 ],
[ 3, 1, 6, 4, 7, 9, 5, 0 ],
[ 8, 0, 7, 2, 3, 1, 0, 8 ],
[ 7, 5, 3, 1, 5, 9, 8, 5 ] ];
matrix(R, C, a); // This code is contributed by Aarti_Rathi </script> |
4 1 5 7 5 6 6 5 9 9 3 5 3 7 8 3 5 4 1 5 7 8 1 6 7 5 3 1 7 0 6 4 8 9 1 4 5 7 4 8 9 8 6 3 2 7 9 3 0 1 3 9 2 7 5 1 5 1 9 0 0 8 8 5
Time Complexity: O(N3)
Auxiliary Space: O(N)
Approach 2:
(by rythmrana2)
Follow the given steps to print the matrix rotated by 45 degree:
- print the spaces required.
- print the element this way –
traverse the matrix the way you want to print it –
- We will print the matrix by dividing it into two parts using two for loops , the first loop will print the first triangle of the matrix and the second loop will print the second triangle of the matrix
- for the first half of matrix, take a loop that goes from the first row to the last row of the matrix.
- at each row , take a while loop which prints the first element of the row and the second element of the row above the current row and the third element of the row which is twice above the current row, going this way we will print the first half triangle of the matrix.
- take two counters, c for counting how many more elements the current diagonal have and counter inc for accessing those elements.
- similarly do this for the second half of the matrix.
Below is the implementation of the above approach:
#include <bits/stdc++.h> using namespace std;
// function to print the matrix rotated at 45 degree. void rotate( int m, int n, int * a)
{ // for printing the first half of matrix
for ( int i = 0; i < m; i++) {
// print spaces
for ( int k = 0; k < m - i; k++) {
cout << " " ;
}
// counter to keep a track of the number of elements
// left for the current diagonal
int c = min(m - 1, min(i, n - 1));
// to access the diagonal elements
int inc = 0;
// while loop prints the diagonal elements of the
// current loop.
while (i <= m - 1 && c != -1) {
cout << *(a + (i - inc) * n + inc) << " " ;
// increasing this variable to reach the
// remaining elements of the diagonal
inc++;
// decreasing the counter to as an element is
// printed
c--;
}
cout << endl;
}
// for printing second half of the matrix
for ( int j = 1; j < n; j++) {
// print spaces
if (m < n) {
for ( int k = 0; k <= j - 1; k++) {
cout << " " ;
}
}
else {
for ( int k = 0; k <= j; k++) {
cout << " " ;
}
}
// counter to keep a track of the number of elements
// left for the current diagonal
int c2 = min(m - 1, n - j - 1);
// to access the diagonal elements
int inc2 = 0;
// while loop prints the diagonal elements of the
// current loop.
while (j <= n - 1 && c2 != -1) {
cout << *((a + (m - 1 - inc2) * n) + j + inc2)
<< " " ;
inc2++;
c2--;
}
cout << endl;
}
} // Driver code int main()
{ int m = 6;
int n = m;
int a[m][n] = {
{3,4,5,1,5,9},
{6,9,8,7,2,5},
{1,5,9,7,5,3},
{4,7,8,9,3,5},
{4,5,2,9,5,6},
{4,5,7,2,9,8}
};
rotate(m, n, ( int *)a);
return 0;
} // THIS CODE IS CONTRIBUTED BY YASH AGARWAL(YASHAGARWAL2852002) |
/*package whatever //do not write package name here */ import java.io.*;
class GFG {
// function to print the matrix rotated at 45 degree.
static void rotate( int m, int n, int [] a) {
// printing first half of matrix
for ( int i = 0 ; i < m; i++) {
// print space
for ( int k = 0 ; k < m - i; k++) {
System.out.print( " " );
}
// counter to keep a track of the number of elements left
int c = Math.min(m - 1 , Math.min(i, n - 1 ));
//for the diagonal elements
int inc = 0 ;
// prints the diagonal elements of the loop.
while (i <= m - 1 && c != - 1 ) {
System.out.print(a[(i - inc) * n + inc] + " " );
// increasing the variable to reach the remaining elements
inc++;
// decreasing the counter to as an element is printed
c--;
}
System.out.println();
}
// for printing second half of the matrix
for ( int j = 1 ; j < n; j++) {
// print spaces
if (m < n) {
for ( int k = 0 ; k <= j - 1 ; k++) {
System.out.print( " " );
}
} else {
for ( int k = 0 ; k <= j; k++) {
System.out.print( " " );
}
}
// counter to keep a track of the number of elements left
int c2 = Math.min(m - 1 , n - j - 1 );
// to access the diagonal elements
int inc2 = 0 ;
// while loop prints the diagonal elements
while (j <= n - 1 && c2 != - 1 ) {
System.out.print(a[(m - 1 - inc2) * n + j + inc2] + " " );
inc2++;
c2--;
}
System.out.println();
}
}
// Driver code
public static void main(String[] args) {
int m = 6 ;
int n = m;
int [][] a = {
{ 3 , 4 , 5 , 1 , 5 , 9 },
{ 6 , 9 , 8 , 7 , 2 , 5 },
{ 1 , 5 , 9 , 7 , 5 , 3 },
{ 4 , 7 , 8 , 9 , 3 , 5 },
{ 4 , 5 , 2 , 9 , 5 , 6 },
{ 4 , 5 , 7 , 2 , 9 , 8 }
};
int [] flattenedArray = new int [m * n];
int idx = 0 ;
for ( int i = 0 ; i < m; i++) {
for ( int j = 0 ; j < n; j++) {
flattenedArray[idx++] = a[i][j];
}
}
rotate(m, n, flattenedArray);
}
} |
# Python program to rotate a matrix by 45 degrees # A function to rotate a matrix # mat[][] of size R x C. # Initially, m = R and n = C def matrix(m, n, a):
# for printing the first half of matrix
for i in range (m):
# print spaces
for k in range (m - i):
print ( " " , end = "")
# counter to keep a track of the number of elements
# left for the current diagonal
c = min (m - 1 , min (i, n - 1 ))
# to access the diagonal elements
inc = 0
# while loop prints the diagonal elements of the
# current loop.
while (i < = m - 1 and c ! = - 1 ):
print (a[i - inc][inc], end = " " )
# increasing this variable to reach the
# remaining elements of the diagonal
inc + = 1
# decreasing the counter to as an element is
# printed
c = c - 1
print ("")
# for printing second half of the matrix
for j in range ( 1 , n):
# print spaces
if (m < n):
for k in range (j):
print ( " " , end = "")
else :
for k in range (j + 1 ):
print ( " " , end = "")
# counter to keep a track of the number of elements
# left for the current diagonal
c2 = min (m - 1 , n - j - 1 )
# to access the diagonal elements
inc2 = 0
# while loop prints the diagonal elements of the
# current loop.
while (j < = n - 1 and c2 ! = - 1 ):
print (a[m - 1 - inc2][j + inc2], end = " " )
inc2 + = 1
c2 - = 1
print ("")
# Driver code # Test Case 1 m = 6
n = 6
a = [[ 3 , 4 , 5 , 1 , 5 , 9 ],
[ 6 , 9 , 8 , 7 , 2 , 5 ],
[ 1 , 5 , 9 , 7 , 5 , 3 ],
[ 4 , 7 , 8 , 9 , 3 , 5 ],
[ 4 , 5 , 2 , 9 , 5 , 6 ],
[ 4 , 5 , 7 , 2 , 9 , 8 ]]
matrix(m, n, a) |
// C# program for the above approach using System;
public class Program
{ // function to print the matrix rotated at 45 degree.
public static void Rotate( int m, int n, int [] a)
{
// for printing the first half of matrix
for ( int i = 0; i < m; i++)
{
// print spaces
for ( int k = 0; k < m - i; k++)
{
Console.Write( " " );
}
// counter to keep a track of the number of elements
// left for the current diagonal
int c = Math.Min(m - 1, Math.Min(i, n - 1));
// to access the diagonal elements
int inc = 0;
// while loop prints the diagonal elements of the
// current loop.
while (i <= m - 1 && c != -1)
{
Console.Write(a[(i - inc) * n + inc] + " " );
// increasing this variable to reach the
// remaining elements of the diagonal
inc++;
// decreasing the counter to as an element is
// printed
c--;
}
Console.WriteLine();
}
// for printing second half of the matrix
for ( int j = 1; j < n; j++)
{
// print spaces
if (m < n)
{
for ( int k = 0; k <= j - 1; k++)
{
Console.Write( " " );
}
}
else
{
for ( int k = 0; k <= j; k++)
{
Console.Write( " " );
}
}
// counter to keep a track of the number of elements
// left for the current diagonal
int c2 = Math.Min(m - 1, n - j - 1);
// to access the diagonal elements
int inc2 = 0;
// while loop prints the diagonal elements of the
// current loop.
while (j <= n - 1 && c2 != -1)
{
Console.Write(a[(m - 1 - inc2) * n + j + inc2] + " " );
inc2++;
c2--;
}
Console.WriteLine();
}
}
// Driver code
public static void Main()
{
int m = 6;
int n = m;
int [] a = {
3, 4, 5, 1, 5, 9,
6, 9, 8, 7, 2, 5,
1, 5, 9, 7, 5, 3,
4, 7, 8, 9, 3, 5,
4, 5, 2, 9, 5, 6,
4, 5, 7, 2, 9, 8
};
Rotate(m, n, a);
}
} // This code is contributed by codebraxnzt |
// Javascript program to rotate a matrix by 45 degrees // A function to rotate a matrix // mat[][] of size R x C. // Initially, m = R and n = C function matrix(m, n, a)
{ // for printing the first half of matrix
for (let i = 0; i < m; i++){
// print spaces
for (let k = 0; k < m - i; k++) {
console.log( " " );
}
// counter to keep a track of the number of elements
// left for the current diagonal
let c = Math.min(m - 1, Math.min(i, n - 1));
// to access the diagonal elements
let inc = 0;
// while loop prints the diagonal elements of the
// current loop.
while (i <= m - 1 && c != -1) {
console.log(a[i-inc][inc] + " " );
// increasing this variable to reach the
// remaining elements of the diagonal
inc++;
// decreasing the counter to as an element is
// printed
c--;
}
console.log( "\n" );
}
// for printing second half of the matrix
for (let j = 1; j < n; j++) {
// print spaces
if (m < n) {
for (let k = 0; k <= j - 1; k++) {
console.log( " " );
}
}
else {
for (let k = 0; k <= j; k++) {
console.log( " " );
}
}
// counter to keep a track of the number of elements
// left for the current diagonal
let c2 = Math.min(m - 1, n - j - 1);
// to access the diagonal elements
let inc2 = 0;
// while loop prints the diagonal elements of the
// current loop.
while (j <= n - 1 && c2 != -1) {
console.log(a[m-1-inc2][j+inc2] + " " );
inc2++;
c2--;
}
console.log( "\n" );
}
} // Driver code // Test Case 1 let m = 6; let n = 6; let a = [ [ 3,4,5,1,5,9 ], [ 6,9,8,7,2,5 ],
[ 1,5,9,7,5,3 ],
[ 4,7,8,9,3,5 ],
[ 4,5,2,9,5,6 ],
[ 4,5,7,2,9,8 ]];
matrix(m, n, a); |
3 6 4 1 9 5 4 5 8 1 4 7 9 7 5 4 5 8 7 2 9 5 2 9 5 5 7 9 3 3 2 5 5 9 6 8
Time Complexity: O(N2)
Auxiliary Space: O(1)