Bomb Blast - No Water

-

 Bomb Blast - No Water 

PROBLEM STATEMENT :

A matrix of size R*C containing positive integer values is passed as the input. An integer value which is a multiple of 5 is a bomb and will destroy all four adjacent cells (in the following order - left, right, top and bottom). But if the adjacent cell is water (the integer value will be a multiple of 3), then the bomb will stop destroying the adjacent cells. If a cell is destroyed then the value of that cell becomes 0. After the bomb is blasted then the cell value becomes 0. If the cell value is both a multiple of 5 and 3, then consider that as water. Finally, the program must print the values in the matrix as the output. 

Note: The bombs are triggered from right to left starting from the last row. 

Boundary Condition(s): 

1 <= R, C <= 100 

Input Format: 

The first line contains R and C separated by a space. 

The next R lines contain C integer values each. 

Output Format: 

The first R lines contain C integer values each. 

Example Input/Output 1: 
Input: () 

3 3 
5 5 5 
61 20 1 
15 6 10 
Output: 
0 0 0 
0 0 0 
15 6 0

Explanation: 

After the 1st bomb 10 in the last row is triggered, the left value is 6 and is water. So the explosion stops. Hence the matrix state is 

5 5 5 
61 20 1 
15 6 0 

Now 15 is both a multiple of 5 and 3. Hence it is considered as water. 

In the 2nd row from the bottom, 20 is a bomb and it destroys the left cell with 61, right cell with 1, the top cell with 5. The bottom cell is water and hence remains unchanged. Hence the matrix state is 

5 0 5 
0 0 0 
15 6 0 

In the 3rd row from the bottom, two bombs(with the values 5) explode and hence the final state is 

0 0 0 
0 0 0 
15 6 0 

Example Input/Output 2: 
Input: () 

3 3 
5 56 51 
63 22 1 
15 6 10 
Output: 
0 0 51 
63 22 1 
15 6 0


SOLUTION :

C (Programming Language)


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

int R,C;
int mat[100][100];

int isValidBomb(int N)
{
    return N%5==0 && N%3!=0 ? 1 : 0;
}

int isNotWater(int row,int col)
{
    if(row<=R-1 && col<=C-1 && row>=0 && col>=0)
    {
        if(mat[row][col]%3 == 0) return 0;
        else return 1;
    }
    return -1;
}
int main()
{
    scanf("%d %d",&R,&C);
    for(int i=0;i<R;i++)
    {
        for(int j=0;j<C;j++)
        {
            scanf("%d ",&mat[i][j]);
        }
    }
    for(int i=R-1;i>=0;i--)
    {
        for(int j=C-1;j>=0;j--)
        {
            if(isValidBomb(mat[i][j]))
            {
                mat[i][j]=0;
                if(isNotWater(i,j-1)==1) mat[i][j-1]=0;
                else if(isNotWater(i,j-1)==0) continue;
                if(isNotWater(i,j+1)==1) mat[i][j+1]=0;
                else if(isNotWater(i,j+1)==0) continue;
                if(isNotWater(i-1,j)==1) mat[i-1][j]=0;
                else if(isNotWater(i-1,j)==0)continue;
                if(isNotWater(i+1,j)==1) mat[i+1][j]=0;
                else if(isNotWater(i+1,j)==0) continue;
            }
        }
    }
    for(int i=0;i<R;i++)
    {
        for(int j=0;j<C;j++)
        {
            printf("%d ",mat[i][j]);
        }
        printf("\n");
    }
}


C++ (CPP)

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

int main(int argc, char** argv)
{
    
    int r , c ; 
    
    cin >> r >> c ;
    
    int mat[r][c];
    
    for(int i = 0 ; i < r ; i++)
    {
        for(int j = 0 ; j<c ; j++)
           cin >> mat[i][j];
    }
    
    for(int i = r-1 ; i>=0 ; i--)
    {
        for(int j = c-1 ; j>=0 ; j--)
        {
            
            if(mat[i][j]%3 && mat[i][j]%5==0)
            {
                mat[i][j] = 0;
                
                if(j-1>=0)
                {
                    if(mat[i][j-1]%3==0)
                      continue;
                    else
                      mat[i][j-1]=0;
                }
                
                if(j+1<c)
                {
                    if(mat[i][j+1]%3==0)
                      continue;
                    else
                      mat[i][j+1]=0;
                }
                
                if(i-1>=0)
                {
                    if(mat[i-1][j]%3==0)
                       continue;
                    else
                       mat[i-1][j]=0;
                }
                
                if(i+1<r)
                {
                    if(mat[i+1][j]%3==0)
                      continue;
                    else
                       mat[i+1][j]=0;
                }
                
            }
        }
    }
    
    for(int i = 0 ; i<r ; i++)
    {
        for(int j = 0 ; j<c ; j++)
           cout << mat[i][j] << " ";
           
        cout << endl;
    }
}


JAVA

       
import java.util.*;
public class Hello {

    public static void main(String[] args) {
		//Your Code Here
		Scanner sc=new Scanner(System.in);
		int i,j,k,l,n,r,c;
		r=sc.nextInt();
		c=sc.nextInt();
		int a[][]=new int[r][c];
		for(i=0;i<r;i++){
		    for(j=0;j<c;j++){
		        a[i][j]=sc.nextInt();
		    }
		}
		for(i=r-1;i>=0;i--){
		    for(j=c-1;j>=0;j--){
		        if(a[i][j]%5==0 && a[i][j]%3!=0){
		            a[i][j]=0;
		            if(j-1>=0){
		                if(a[i][j-1]%3==0)    continue;
		                a[i][j-1]=0;
		            }
		            if(j+1<c){
		                if(a[i][j+1]%3==0)  continue;
		                a[i][j+1]=0;
		            }
		            if(i-1>=0){
		                if(a[i-1][j]%3==0)  continue;
		                a[i-1][j]=0;
		            }
		            if(i+1<r){
		                if(a[i+1][j]%3==0)    continue;
		                a[i+1][j]=0;
		            }
		        }
		    }
		}
		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())
l1=[0,0,-1,1]
l2=[-1,1,0,0]
l=[list(map(int,input().split())) for i in range(r)]
for i in range(r-1,-1,-1):
    for j in range(c-1,-1,-1):
        if l[i][j]%5==0 and l[i][j]%3!=0:
            l[i][j]=0
            for k in range(4):
                i1=i+l1[k]
                j1=j+l2[k]
                if i1>=0 and i1<r and j1>=0 and j1<c:
                    if l[i1][j1]%3==0:
                        break
                    else:
                        l[i1][j1]=0
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