Modify Submatrices Diagonally

-

 Modify Submatrices Diagonally 

PROBLEM STATEMENT :

The program must accept an integer matrix of size R*C and an integer K as the input. The values of R and C are always divisible by K. The given R*C matrix contains only 0s and 1s. The program must form an alphabet matrix of size (R/K)*(C/K) starting from "a" being present in the bottom left corner. Then the program must modify the submatrices of size K*K diagonally. In each integer submatrix of size K*K, all 1s must be replaced with the related upper case alphabet in the alphabet matrix and all 0s must be replaced with the related lower case alphabet in the alphabet matrix. If "z" has reached while filling the alphabet matrix then "a" must be considered as the next alphabet in circular fashion while forming the alphabet matrix. Finally, the program must print the integer matrix replaced with the corresponding alphabets as the output. 

Boundary Condition(s): 
2 <= R, C <= 50 

Input Format: 
The first line contains R and C separated by a space. 
The next R lines, each contains C integer values separated by a space. 
The (R+2)nd line contains K. 

Output Format: 
The first R lines, each contains C alphabets separated by a space. 

Example Input/Output 1: 
Input: () 

9 9 
0 0 1 0 0 0 0 1 1 
0 0 1 1 1 1 1 0 1 
0 1 1 1 1 1 0 1 1 
1 1 1 1 1 0 1 1 1 
0 1 1 0 0 0 1 1 0 
0 0 0 1 0 1 1 0 0 
0 0 1 1 0 1 1 1 1 
0 1 1 0 1 1 1 0 0 
0 0 1 1 1 1 0 1 1 
3 


Output: 
d d D g g g i I I 
d d D G G G I i I 
d D D G G G i I I 
B B B E E e H H H 
b B B e e e H H h 
b b b E e E H h h 
a a A C c C F F F 
a A A c C C F f f 
a a A C C C f F F

Explanation: 
Here R = 9, C = 9 and K = 3. 
So the size of the alphabet matrix to be formed is (9/3)*(9/3). 
The alphabet matrix of size 3*3 is given below. 
d g i 
b e h 
a c f 
In the above alphabet matrix, the alphabets starting from 'a' are filled diagonally in the top-left to bottom-right diagonals starting the bottom-left corner. 
After replacing the submatrices with their corresponding alphabets based on the given conditions, the matrix becomes 
d d D g g g i I I 
d d D G G G I i I 
d D D G G G i I I 
B B B E E e H H H 
b B B e e e H H h 
b b b E e E H h h 
a a A C c C F F F 
a A A c C C F f f 
a a A C C C f F F 



Example Input/Output 2: 
Input: ()

6 8 
1 0 1 0 0 1 0 0 
0 1 1 1 1 1 0 1 
0 0 0 0 1 0 1 0 
1 1 1 0 0 0 1 1 
1 0 1 1 1 0 0 0 
1 1 1 0 0 1 1 0 
2 


Output: 
D d G g j J 
l l d D G G 
J J l L b b 
e e H h K k 
B B E e h h 
K K A a C C 
F f i i A A 
C c f F I i


SOLUTION :

C (Programming Language)


      
#include<stdio.h>
#include<stdlib.h>

int main()
{
    char ch='a';
    int r,c,i,j,k,i1,j1;
    scanf("%d%d",&r,&c);
    int a[r][c];
    for(i=0;i<r;++i)
    {
        for(j=0;j<c;++j)
        {
            scanf("%d",&a[i][j]);
        }
    }
    scanf("%d",&k);
    i=r-k;
    j=0;
    while(j<c)
    {
        i1=i;
        j1=j;
        while(i1<r&&j1<c)
        {
            for(int i=i1;i<i1+k;++i)
            {
                for(int j=j1;j<j1+k;++j)
                {
                    if(a[i][j]==0) a[i][j]=ch;
                    else a[i][j]=toupper(ch);
                }
            }
            i1+=k;
            j1+=k;
            if(ch=='z') ch='a';
            else ch++;
        }
        if(i>0) i-=k;
        else j+=k;
    }
    for(i=0;i<r;++i)
    {
        for(j=0;j<c;++j)
        {
            printf("%c ",a[i][j]);
        }
        printf("\n");
    }
}


C++ (CPP)

       
#include <bits/stdc++.h>
 
using namespace std;

int main(int argc, char** argv)
{
    char ch='a';
    int r,c,i,j,i1,j1,k;
    cin>>r>>c;
    char a[r][c];
    for(i=0;i<r;++i)
    {
        for(j=0;j<c;++j)
        {
            cin>>a[i][j];
        }
    }
    cin>>k;
    i=r-k;
    j=0;
    while(j<c)
    {
        i1=i;
        j1=j;
        while(i1<r&&j1<c)
        {
            for(int i=i1;i<i1+k;++i)
            {
                for(int j=j1;j<j1+k;++j)
                {
                    if(a[i][j]=='0') a[i][j]=ch;
                    else a[i][j]=toupper(ch);
                }
            }
            if(ch=='z') ch='a';
            else ch++;
            i1+=k;
            j1+=k;
        }
        if(i>0) i-=k;
        else j+=k;
    }
    for(i=0;i<r;++i)
    {
        for(j=0;j<c;++j)
        {
            cout<<a[i][j]<<" ";
        }
        cout<<"\n";
    }
}


JAVA

       
import java.util.*;
public class Hello {

    public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int r=sc.nextInt(),c=sc.nextInt(),i,j,i1,j1,k,x,y;
		char[][] a=new char[r][c];
		char ch='a';
		for(i=0;i<r;++i)
		{
		    for(j=0;j<c;++j)
		    {
		        a[i][j]=sc.next().charAt(0);
		    }
		}
		k=sc.nextInt();
		i=r-k;
		j=0;
		while(j<c)
		{
		    i1=i;
		    j1=j;
		    while(i1<r&&j1<c)
		    {
		        for(x=i1;x<i1+k;x++)
		        {
		            for(y=j1;y<j1+k;y++)
		            {
		                if(a[x][y]=='0') a[x][y]=ch;
		                else a[x][y]=Character.toUpperCase(ch);
		            }
		        }
		        if(ch=='z') ch='a';
		        else ch++;
		        i1+=k;
		        j1+=k;
		    }
		    if(i>0) i-=k;
		    else j+=k;
		}
		for(i=0;i<r;++i)
		{
		    for(j=0;j<c;++j)
		    {
		        System.out.print(a[i][j]+" ");
		    }
		    System.out.println();
		}
	}
}


PYTHON

       
r,c=map(int,input().split())
x=['a']
l=[list(map(int,input().split())) for i in range(r)]
copied=[l[i][:] for i in range(r)]
k=int(input())
i=r-k;j=0
while j<c:
    i1=i
    j1=j
    while i1<r and j1<c:
        for a in range(i1,i1+k):
            l[a][j1:j1+k]=x*k
        i1+=k;j1+=k
        if x[0]=='z':
            x[0]='a'
        else:
            x[0]=chr(ord(x[0])+1)
    if i>0:
        i-=k
    else:
        j+=k
for i in range(r):
    for j in range(c):
        if copied[i][j]==1:
            l[i][j]=l[i][j].upper()
for i in l:
    print(*i)


Never Stop Learning !!


Comments

Post a Comment

Popular posts from this blog

DP (1) - Count number of ways to cover a distance

-
Given a distance n, count total number  of ways to cover the distance with 1, 2 and 3 steps. Input: n = 4 Output: 7 Explantion: Below are the four ways  1 step + 1 step + 1 step + 1 step  1 step + 2 step + 1 step  2 step + 1 step + 1 step   1 step + 1 step + 2 step  2 step + 2 step  3 step + 1 step  1 step + 3 step // C++ program to count number of ways in DP  to cover a distance with 1, 2 and 3 steps  #include<iostream>  using namespace std;  int main() {      int n;     std::cin>>n;     int count[n+1];         // Initialize base values. There is one way to cover 0 and 1      // distances and two ways to cover 2 distance.     count[0]  = 1,  count[1] = 1,  count[2] = 2;         // Fill the count array in from bottom to up manner      for (int i=3; i<=n; i++)  ...
Read more

Zero Insert After K Times One

-
 Zero Insert After K Times One  PROBLEM STATEMENT : Given a bitstream of length N consisting of 0s and 1s, insert 0 after 1 has appeared K times consecutively.  Input Format:  The first line contains N and K separated by a space.  The second line contains the bitstream with 1s and 0s with each value separated by a space.  Output Format:   In the first line, the bitstream with the 0 inserted after 1s has appeared K times with each value separated by a space.  Boundary Conditions:   1 <= K <= 1000  2 <= N <= 1000  Example Input/Output 1: Input: ( ) 12 2 1 0 1 1 0 1 1 0 1 1 1 1 Output: 1 0 1 1 0 0 1 1 0 0 1 1 0 1 1 0 SOLUTION : C (Programming Language) Copy Code #include<stdio.h> #include<stdlib.h> int main() { int n,k,cnt=0,i,j,a[2002]; scanf("%d%d",&n,&k); for(i=0;i<n;++i) { scanf("%d",&a[i]); } //Count of 1 in first step for(i=0;i<k;++i...
Read more

Left Number Twice Right

-
 Left Number Twice Right  PROBLEM STATEMENT : A set of N numbers (separated by one or more spaces) is passed as input to the program. The program must identify the count of numbers C where the number on the left is twice the number on the right.  Boundary Conditions:   3 <= N <= 50  The value of the numbers can be from -99999999 to 99999999  Input Format: The first line will contain the N numbers separated by one or more spaces.  Output Format: The first line contains the value of C.  Example Input/Output 1:   Input: 10 20 5 40 15 -2 -30 -1 60  Output: 2 Explanation: The numbers meeting the criteria are 20, -30  Example Input/Output 2:  Input: 5 90 10 2 5 -4 10 6 5 3  Output: 3 Explanation: The numbers meeting the criteria are 2, 6, 5 SOLUTION : C (Programming Language) Copy Code #include<stdio.h> #include<stdlib.h> int main() { int a[50],n=-1,i,cnt=0; while(scanf("%d",&a[++n...
Read more