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77 files changed

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‎2D Arrays/rotate_by_90.cpp‎

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// Given a square matrix of size N x N. The task is to rotate it by 90 degrees in anti-clockwise direction without using any extra space.
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#include<bits/stdc++.h>
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#include <stdio.h>
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using namespace std;
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int main()
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{
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int t; cin>>t;
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while(t--)
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{
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int n;
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cin>>n;
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int matrix[n][n];
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for(int i = 0; i < n; i++)
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{
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for(int j = 0; j < n; j++)
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{
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cin>>matrix[i][j];
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}
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}
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//rotating the matrix by 90 , anticlockwise
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// the matrix will have floor(n/2) cycles
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for(int i=0; i<n/2; i++)
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{
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//parsing through each cycle and rotating
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for(int j=i; j< n-i-1; j++)
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{
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//stored current element in temporary variable temp
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int temp= matrix[i][j];
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//shifted right element to top
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matrix[i][j]= matrix[j][n-i-1];
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//shifted bottom element to right
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matrix[j][n-i-1]= matrix[n-1-i][n-1-j];
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//shifted left element to bottom
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matrix[n-1-i][n-1-j] = matrix[n-1-j][i];
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//assign temp to left
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matrix[n-1-j][i]= temp;
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}
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}
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for(int i = 0; i < n; i++)
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for(int j=0; j<n; j++)
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cout<< matrix[i][j];
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cout<<endl;
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}
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return 0;
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}

‎2D Arrays/transpose.java‎

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/*
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Title : Find the transpose of the matrix .
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User needs to provide the number of rows and columns .
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And the user needs to provide the elements of the matrix .
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User will get the transpose of the matrix .
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i.e Row elements become the column elements and vice- versa .
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**** Maximum 10 number of rows and columns can be only entered .
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*/
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import java.util.Scanner;
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import java.lang.*;
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class transpose
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{
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final static int MAXLIMIT=10 ;
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private int row;
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private int col;
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int i ,j ;
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private double a[][];
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transpose(int rows , int cols) //Constructor
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{
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row=rows ;
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col=cols ;
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a=new double[row][col] ;
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}
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Scanner in=new Scanner(System.in);
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public void get ()
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{
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System.out.println("\n Enter the elements : \n");
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for(i=0;i<row;i++)
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{
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for(j=0;j<col;j++)
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{
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a[i][j]=in.nextDouble();
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}
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}
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}
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public void display ()
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{
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System.out.println( "\n Matrix A " + row + " * " + col + " : \n") ;
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for(i=0;i<row;i++)
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{
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for(j=0;j<col;j++)
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{
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System.out.print(a[i][j]+ "\t") ;
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}
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System.out.println("\n") ;
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}
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}
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public void trans()
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{
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System.out.println( "\n\n The transpose of the given matrix : \n");
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System.out.println( "\n Matrix AT " + col + " * " + row+ " : \n") ;
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for(i=0;i<col;i++)
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{
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for(j=0;j<row;j++)
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{
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System.out.print(a[j][i]+ "\t") ;
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}
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System.out.println("\n") ;
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}
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}
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public static void main (String [] args )
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{
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int rows , cols ;
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int opt;
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Scanner in=new Scanner(System.in);
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do
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{ opt=0;
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System.out.println("\n Enter the total number of rows and columns respectively : \n");
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rows=in.nextInt();
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cols=in.nextInt();
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if(rows>MAXLIMIT|| cols>MAXLIMIT|| rows<1||cols<1)
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{
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System.out.println("\n Invalid enteries ! \n Remember , Maximum limit of rows and columns is 10 .\n");
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System.out.println(" Do you wish to continue ? \n Press 1 if yes otherwise any to stop . \n");
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opt=in.nextInt();
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}
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else
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{
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transpose A = new transpose(rows , cols);
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A.get();
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A.display();
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A.trans();
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}
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} while(opt==1);
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}
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}
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/*
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Enter the total number of rows and columns respectively :
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13
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1
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Invalid enteries !
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Remember , Maximum limit of rows and columns is 10 .
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Do you wish to continue ?
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Press 1 if yes otherwise any to stop .
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1
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Enter the total number of rows and columns respectively :
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3
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2
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Enter the elements :
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1
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2
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3
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5
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5
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5
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Matrix A 3 * 2 :
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1.0 2.0
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3.0 5.0
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5.0 5.0
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The transpose of the given matrix : \n");
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Matrix AT 2 * 3 :
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1.0 3.0 5.0
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2.0 5.0 5.0
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*/

‎2D Arrays/transpose_of_matrix.cpp‎

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//Write a program to find the transpose of a square matrix of size N*N. The transpose of a matrix is obtained by changing rows to columns and columns to rows.
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#include <bits/stdc++.h>
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using namespace std;
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//Function to find transpose of a matrix.
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void transpose(vector<vector<int> > &matrix, int n)
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{
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for(int i=0; i< n; i++)
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{
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for(int j=i; j< n; j++)
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{
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swap(matrix[i][j], matrix[j][i]);
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}
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}
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}
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int main()
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{
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int t;
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cin>>t;
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while(t--)
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{
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int n;
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cin>>n;
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vector<vector<int> > matrix(n);
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for(int i=0; i<n; i++)
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{
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matrix[i].assign(n, 0);
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for( int j=0; j<n; j++)
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{
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cin>>matrix[i][j];
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}
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}
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transpose(matrix,n);
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for (int i = 0; i < n; ++i)
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{
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for (int j = 0; j < n; ++j)
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cout<<matrix[i][j]<<" ";
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cout<<endl;
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}
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}
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return 0;
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}

‎Arrays/Hourglass_Sum.py‎

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'''
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Given a 6X6 2D Array, A:
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1 1 1 0 0 0
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0 1 0 0 0 0
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1 1 1 0 0 0
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0 0 0 0 0 0
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0 0 0 0 0 0
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0 0 0 0 0 0
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An hourglass in A is a subset of values with indices falling in this pattern
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in A's graphical representation:
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a b c
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d
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e f g
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There are 16 hourglasses in A, and an hourglass sum is the sum of an hourglass'
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values. Like here, the maximum hourglass sum is 7 for the hourglass in the top
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left corner.
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Aim: Check all such hourglasses and print the maximum hourglass sum for the
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2D array entered by user.
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'''
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# getting the 2D array as input
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arr = []
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for _ in range(6):
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arr.append(list(map(int, input().rstrip().split())))
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res = []
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# looping through the 2D array
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for x in range(0, 4):
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for y in range(0, 4):
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# selecting combinations that make an hourglass
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s = sum(arr[x][y:y+3]) + arr[x+1][y+1] + sum(arr[x+2][y:y+3])
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res.append(s)
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# printing out the maximum hourglass sum
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print(max(res))
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'''
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COMPLEXITY:
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Time Complexity -> O(N^2)
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Space Complexity -> O(N)
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Sample Input:
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0 0 1 2 3 0
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0 0 0 5 0 0
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1 0 0 7 1 0
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0 0 0 0 0 0
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0 1 1 1 0 1
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0 0 4 0 0 4
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Sample Output:
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19
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Explanation:
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The hourglass with the maximum sum is,
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1 2 3
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5
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0 7 1
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with the sum being 1 +たす 2 +たす 3 +たす 5 +たす 0 +たす 7 +たす 1 =わ 19
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'''

‎Arrays/Leap_Year.py‎

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'''
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Aim: To check if the entered year is a leap year or not.
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'''
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def is_leap(year):
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# a year is a leap one if it's divisible by 4 and if it's not
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# divisible by 100 or is divisible by 400
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print((year%4==0 and (year%100!=0 or year%400==0)))
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# getting the input
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year = int(input().strip())
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# displaying the result
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is_leap(year)
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'''
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COMPLEXITY:
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Time Complexity -> O(1)
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Space Complexity -> O(1)
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Sample Input:
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2000
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Sample Output:
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True
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Explanation:
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2000 % 400 = 0
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2000 % 4 = 0
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Hence, 2000 was a leap year.
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'''

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