Image Smoother in C++

C++Server Side ProgrammingProgramming

Suppose we have a 2D matrix M representing the gray scale of an image, we have to design a smoother to make the gray scale of each pixel becomes the average gray scale (rounding down) of all the 8 surrounding pixels and itself. If a cell has less than 8 surrounding cells, convert all possible pixels.

So, if the input is like

111
101
111

then the output will be

000
000
000

To solve this, we will follow these steps −

  • R := row count of M

  • C := column count of

  • Define an array d = { -1, 0, 1 }

  • Define one 2D array res of size (R x C)

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

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

      • sum := 0, count := 0

      • for initialize k := 0, when k < 3, update (increase k by 1), do −

        • for initialize l := 0, when l − 3, update (increase l by 1), do −

          • m := i + d[k], n := j + d[l]

          • if m >= 0 and m < R and n >= 0 and n < C, then −

            • increase count by 1 and sum = sum + M[m, n]

      • res[i, j] := sum / count

  • return res

Example 

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<vector<auto> > v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << "[";
      for(int j = 0; j <v[i].size(); j++){
         cout << v[i][j] << ", ";
      }
      cout << "],";
   }
   cout << "]"<<endl;
}
class Solution {
public:
   vector<vector<int>> imageSmoother(vector<vector<int>>& M) {
      int R = M.size();
      int C = M[0].size();
      vector<int> d{ -1, 0, 1 };
      vector<vector<int> > res(R, vector<int>(C, 0));
      for (int i = 0; i < R; ++i) {
         for (int j = 0; j < C; ++j) {
            int sum = 0, count = 0;
            for (int k = 0; k < 3; ++k) {
               for (int l = 0; l < 3; ++l) {
                  int m = i + d[k], n = j + d[l];
                     if (m >= 0 && m < R && n >= 0 && n < C) ++count, sum += M[m][n];
                  }
               }
               res[i][j] = sum / count;
            }
         }
         return res;
      }
};
main(){
   Solution ob;
   vector<vector<int>> v = {{1,1,1},{1,0,1},{1,1,1}};
   print_vector(ob.imageSmoother(v));
}

Input

{{1,1,1},{1,0,1},{1,1,1}}

Output

[[0, 0, 0, ],[0, 0, 0, ],[0, 0, 0, ],]
raja
Published on 11-Jun-2020 15:51:24
Advertisements