uva online judge minesweeper problem solution python | uva 10189

The minesweeper problem (uva 10189) is a loop programming problem states that (from onlinejudge)

The goal of the game minesweeper  is to find where are all the mines within a m * n field. The game shows a number in a square which tells you how many mines there are adjacent to that square.

In this problem, given  a line indicating '*' as bomb and '.' as empty box we have to calculate the number of bombs placed adjecent to each box and replace '.' with number of bombs.

In above game example, we have to find number of bombs in adjecent boxes marked by black lines. Once number of bombs founded, replace value of center box with number of bombs. That's it.

Sample Input

   4 4 

   *... 

    .... 

    .*.. 

    .... 

    3 5 

    **... 

    ..... 

    .*...

    0 0

Sample Output

Field #1: 

    *100 

    2210 

    1*10 

    1110

Field #2: 

    **100 

    33200 

    1*100

Explanation

As the problem states, we have to loop for every box over its adjecent 3 * 3 area boxes. For each itteration check value at given box, if it is '*' i.e. bomb then skip else calculate number of bombs adjecent to it. 

Code

rows = int(input("Enter number of rows : "))

cols =  int(input("Enter number of cols : "))

a = list()

print("Enter patter : ")

for i in range(0,rows):

    a.append(list(input()))

    

for i in range(0,rows):

    for j in range(0,cols):

        count=0

        if(a[i][j]=='*'):

            continue

        else:

            for n in range(i-1,i+2):

                for m in range(j-1,j+2):

                    if(n<0 or m<0 or n>=rows or m>=cols):

                        continue

                    else:

                        if(a[n][m]=='*'):

                            count+=1

        a[i][j]=str(count)

        

for i in range(0,rows):

        print(a[i][:])

Here, inside main loop(i and j) we itterate with n and m loops with condition i-1, i+2. This is because for every box its adjecent boxes number will remain same i.e 3 * 3. Now, inside if condition we check if n and m is invalid or valid (side boxes i.e for i=0, j=0 its adjecent boxes are only three, and so on). If box is not exist then skip else increase count by 1. 

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