Submatrices - Diagonal Elements Sum :

-

Submatrices - Diagonal Elements Sum : 

PROBLEM STATEMENT : 

The program must accept an integer matrix M of size R*C and an integer N as the input. The program must print the sum of diagonal elements of all the N*N non-overlapping submatrices in M as the output. 

Note: The value of N is always greater than 1 and less than or equal to the minimum value between R and C. 

Boundary Condition(s): 
2 <= R, C <= 50 
0 <= Matrix element value <= 1000 

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)th line contains N. 

Output Format: 
The first line contains the integer values representing the sum of diagonal elements of all N*N non-overlapping submatrices in M. 

Example Input/Output 1: 
Input: 
5 6 
6 3 6 3 2 1
4 5 7 1 3 8
6 9 8 4 3 2 
4 1 8 8 9 1 
8 1 5 7 9 0 
3 

Output: 
31 13 

Explanation: 
There are 2 non-overlapping submatrices of size 3*3 in the given 5*6 matrix. 
In the first submatrix, the sum of diagonal elements is 31 (6+5+8+6+6). 
In the second submatrix, the sum of diagonal elements is 13 (3+3+2+1+4). 
Hence the output is 31 13 

Example Input/Output 2: 
Input: 
6 6 
6 3 6 3 2 1 
4 5 7 1 3 8 
6 9 8 4 3 2 
4 1 8 8 9 1 
8 1 5 7 9 0 
1 0 6 9 2 1 
3 

Output: 
31 13 20 28


SOLUTIONS :

C (Programming Language)

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

int main()
{
    int r,c,i,j,i1,j1,n;
    scanf("%d %d\n",&r,&c);
    int m[r][c];
    for(i=0;i<r;++i)
    {
        for(j=0;j<c;++j)
        {
            scanf("%d ",&m[i][j]);
        }
    }
    scanf("%d",&n);
    for(i=0;i<r;i+=n)
    {
        for(j=0;j<c;j+=n)
        {
            if(i+n>r||j+n>c) continue;
            int s=0;
            for(i1=i;i1<i+n;i1++)
            {
                for(j1=j;j1<j+n;j1++)
                {
                    if(i1%n==j1%n||(i1+j1)%n==n-1)
                    {
                        s+=m[i1][j1];
                    }
                }
            }
            printf("%d ",s);
        }
    }
}


C++ (CPP)

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

int main(int argc, char** argv)
{
    int r,c,i,j,i1,j1,n;
    cin>>r>>c;
    int m[r][c];
    for(i=0;i<r;++i)
    {
        for(j=0;j<c;++j)
        {
            cin>>m[i][j];
        }
    }
    cin>>n;
    for(i=0;i<r;i+=n)
    {
        for(j=0;j<c;j+=n)
        {
            int s=0;
            if(i+n>r||j+n>c) continue;
            for(i1=i;i1<i+n;i1++)
            {
                for(j1=j;j1<j+n;j1++)
                {
                    if(i1%n==j1%n||(i1+j1)%n==n-1)
                    {
                        s+=m[i1][j1];
                    }
                }
            }
            cout<<s<<" ";
        }
    }
}



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,n;
		int[][] m=new int[r][c];
		for(i=0;i<r;++i)
		{
		    for(j=0;j<c;++j)
		    {
		        m[i][j]=sc.nextInt();
		    }
		}
		n=sc.nextInt();
		for(i=0;i<r;i+=n)
		{
		    for(j=0;j<c;j+=n)
		    {
		        if(i+n>r||j+n>c) continue;
		        int s=0;
		        for(i1=i;i1<i+n;i1++)
		        {
		            for(j1=j;j1<j+n;j1++)
		            {
		                if(i1%n==j1%n||(i1+j1)%n==n-1)
		                {
		                    s+=m[i1][j1];
		                }
		            }
		        }
		        System.out.print(s+" ");
		    }
		}
	}
}


PYTHON

       
r,c=map(int,input().split())
m=[list(map(int,input().split())) for i in range(r)]
n=int(input())
for i in range(0,r,n):
    for j in range(0,c,n):
        if i+n<=r and j+n<=c:
            s=sum([m[i1][j1] for i1 in range(i,i+n) for j1 in range(j,j+n) if i1%n==j1%n or (i1+j1)%n==n-1])
            print(s,end=' ')


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