You're welcome

Your're Welcome

You're Welcome

Using C++

int main(){ int n; cin >> n; vector grid(n); for(int i = 0; i < n; i++){ cin >> grid[i]; } std::string::iterator it,jt1,jt2; int c=0; char temp; for(int i = 1; i < n-1; i++) { c=0; int d=0,e=0; for (it=grid[i].begin(); it!=grid[i].end(); it++) { c++; if(c>1 && c*jt1) {

                            for(jt2=grid[i+1].begin(); jt2!=grid[i+1].end(); jt2++)
                            {
                                e++;
                                if(e==c)
                                {
                                    if(temp>*jt2)
                                        temp='X';
                                }
                            }
                        }
                    }
                }          
            }   
            *it=temp;
        }
    } 
}
for(int i = 0; i < n; i++)
   cout<< grid[i]<<endl;
return 0;

}

for scanning use int and For Placing X put it in string row grid[]

You can't. What you can do is, do the comparisons and output at the same time, so when you find a cavity, output an X instead of the digit. While you could have read the input as characters and compared and assigned the X, using integers is generally easier to understand, so I would prefer reading the input as integer. But if you want to stick to keeping the processing and output separate, use a 2D character array.

Why would one store X if you can store a sentinel value(say Integer.MAX_INT), and print X while printing, when a sentinel is encountered

make it string array

I suggest you to change the number when their is any cavity to any negative number (let us change cavity a[i][j] to value -1) and then while printing you print the value 'X' in place of that negative number.

for loops will exclude all elements at the edge (Only - elements in the following diagram)

+++++
+---+
+---+
+---+
+++++

check all the inner elements' left, right, upper and lower elements. If these elements are smaller than the center element, change it to Integer.MAX_INT.

While printing, simply print X if Integer.MAX_INT is encountered. ;)

Basically there is no need to convert them in any language, since comparison of characters leads to same conclusion. Any language will consider "12" > "13" as falsy. And "9" > "8" as truthy.

helpful. i ws nt getting the scenario exactly

Thanks brother you saved my time. Initially I just took for the adjacent ones in the row only.

if i take input as string the how will you print the numbers inside it

What is the point though? It ISN'T an integer, it is a string of integers...

Ha! So true. I converted the ints to strings to convert them into lists, and solved the problem. I never noticed that I never converted each char in the array back to an int for comparison, but everything worked! Plus, I saw something about 75 lines above?

n = int(input())

a = []

for i in range(n):

a.append(list(input()))

for i in range(1,n-1):

for j in range(1,n-1):
    if a[i][j] > a[i+1][j] and a[i][j] > a[i-1][j] and a[i][j] > a[i][j+1] and a[i][j] > a[i][j+1]:
        a[i][j]="X"

for i in range(n):

print("".join(str(x) for x in a[i]))

Can you tell me what the error is?

Yeah, test case #2 breaks the specified input format.

%1d signifies read 1 digit at a time. If you write %d, one entire row of digits will be read at once, which is not what you want.

my solution is same as you describing and it is running fine on dev-c++ but here its not working.

The code does not assign the line of string into a 1D array. In the code, charArray is a 2D array (an array of arrays). The line which has "charArray[i] = lines.toCharArray" converts the string to an array of characters, and this array of characters is then stored in charArray[i]. Hence, filling the i-th position of the 2D array.

Even my second test is failing, I think there is some faulty test scenario. Failed Test case Info :

Input : first line 1 and second line 9 output : 9