Lonely Pixel I in C++

C++Server Side ProgrammingProgramming

Suppose we have a picture consisting of black and white pixels, we have to find the number of black lonely pixels. Here the picture is represented by a 2D char array consisting of 'B' and 'W', for the black and white pixels respectively.

A black lonely pixel is actually 'B' that located at a specific position where the same row and same column don't have any other black pixels.

If the input is like −

WWB
WBW
BWW

Output will be 3. Because all the three 'B's are black lonely pixels.

To solve this, we will follow these steps −

  • n := size of picture

  • m := (if n is non-zero, then column size, otherwise 0)

  • Define two arrays row and col of size n

  • ret := 0, firstRow := 0

  • for initialize i := 0, when i < n, update (increase i by 1), do −

    • for initialize j := 0, when j < m, update (increase j by 1), do −

      • if picture[i, j] is same as 'B', then −

        • if picture[0, j] < 'Y' and picture[0, j] is not equal to 'V', then −

          • (increase picture[0, j] by 1)

        • if i is same as 0, then −

          • (increase firstRow by 1)

        • otherwise when picture[i, 0] < 'Y' and picture[i, 0] is not equal to 'V', then −

          • (increase picture[i, 0] by 1)

  • for initialize i := 0, when i < n, update (increase i by 1), do −

    • for initialize j := 0, when j < m, update (increase j by 1), do −

      • if picture[i, j] < 'W' and (picture[0, j] is same as 'C' or picture[0, j] is same as 'X'), then −

        • if i is same as 0, then −

          • ret := (if ret + firstRow is same as 1, then 1, otherwise 0)

        • otherwise when picture[i, 0] is same as 'C' or picture[i, 0] is same as 'X', then −

          • (increase ret by 1)

  • return ret

Example 

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   int findLonelyPixel(vector<vector<char>>& picture) {
      int n = picture.size();
      int m = n ? picture[0].size() : 0;
      vector<int< row(n);
      vector<int< col(m);
      int ret = 0;
      int firstRow = 0;
      for (int i = 0; i < n; i++) {
         for (int j = 0; j < m; j++) {
            if (picture[i][j] == 'B') {
               if (picture[0][j] < 'Y' && picture[0][j] != 'V'){
                  picture[0][j]++;
               }
               if (i == 0)
                  firstRow++;
               else if (picture[i][0] < 'Y' && picture[i][0] != 'V') {
                  picture[i][0]++;
               }
            }
         }
      }
      for (int i = 0; i < n; i++) {
         for (int j = 0; j < m; j++) {
            if (picture[i][j] < 'W' && (picture[0][j] == 'C' || picture[0][j] == 'X')) {
               if (i == 0)
                  ret += firstRow == 1 ? 1 : 0;
               else if (picture[i][0] == 'C' || picture[i][0] == 'X')
                  ret++;
            }
         }
      }
      return ret;
   }
};
main(){
   Solution ob;
   vector<vector<char>> v = {{'W','W','B'},{'W','B','W'},{'B','W','W'}};
   cout << (ob.findLonelyPixel(v));
}

Input

{{'W','W','B'},{'W','B','W'},{'B','W','W'}}

Output

3
raja
Updated on 19-Nov-2020 10:17:47

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