C++ Program to find out the number of operations to maximize the number of even-numbered cells in a grid

Suppose, we are given a grid of dimensions h * w. Every cell in the grid has a specific value assigned to it. We have to maximize the cells having an even value. To do that, we can select a cell that has not been selected before, and then decrease the value by 1 of the current cell and increase the value by 1 of another cell that is located vertically or horizontally adjacent to the current cell. We print out the number of operations and the cell numbers of the increase and decrease operations. The output will be in the format below −

number of operations
1st (decreased cell position) - (increased cell position)
....
nth (decreased cell position) - (increased cell position)

So, if the input is like h = 3, w = 3, grid = {{2, 3, 4}, {2, 0, 1}, {1, 2, 3}}, then the output will be

4
(0, 1) - (0, 2)
(2, 0) - (2, 1)
(2, 1) - (2, 2)
(0, 2) - (1, 2)

Steps

To solve this, we will follow these steps −

Define a new array result that contains a tuple
for initialize i := 0, when i < h, update (increase i by 1), do:
   tp := 0
   for initialize j := 0, when j < w, update (increase j by 1), do:
      if tp > 0, then:
         insert tuple(i, j - 1, i, j) at the end of result
         grid[i, j] := grid[i, j] + tp
      if grid[i, j] mod 2 is same as 1 and j < w-1, then:
         grid[i, j] := grid[i, j] - 1
         tp := 1
      Otherwise
         tp := 0
tp := 0
for initialize i := 0, when i < h, update (increase i by 1), do:
   if tp > 0, then:
  insert tuple(i - 1, w - 1, i, w - 1) at the end of result
grid[i, w - 1] := grid[i, w - 1] + tp
if grid[i, w - 1] mod 2 is same as 1, then:
grid[i, w - 1] := grid[i, w - 1] - 1
tp := 1
Otherwise
tp := 0
print(size of result)
for initialize i := 0, when i < size of result, update (increase i by 1), do:
print('(' + first value of result[i] + ',' + second value of result[i] + '- (' + third value of result[i] + ',' + fourth value of result[i])

Example

Let us see the following implementation to get better understanding −

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

void solve(int h, int w, vector<vector<int>>grid){
   vector<tuple<int,int,int,int>> result;
   for(int i = 0; i < h; i++){
      int tp = 0;
      for(int j = 0; j < w; j++){
         if(tp > 0){
            result.push_back(make_tuple(i, j-1, i, j));
            grid[i][j] += tp;
         }
         if(grid[i][j]%2 == 1 && j < w-1){
            grid[i][j] -= 1;
            tp = 1;
         }
         else
            tp = 0;
      }
   }
   int tp = 0;
   for(int i = 0; i < h; i++){
      if(tp > 0){
         result.push_back(make_tuple(i-1, w-1, i, w-1));
         grid[i][w-1] += tp;
      }
      if(grid[i][w-1]%2 == 1){
         grid[i][w-1] -= 1;
         tp = 1;
      }
      else
         tp = 0;
   }
   cout << (int)result.size() << endl;
   for(int i = 0; i < (int)result.size(); i++){
      cout << "(" << get<0>(result[i]) << ", " << get<1>(result[i])
      << ")" << " - (" << get<2>(result[i]) << ", " << get<3>(result[i]) << ")";
      cout << '\n';
   }
}
int main() {
   int h = 3, w = 3 ;
   vector<vector<int>> grid = {{2, 3, 4}, {2, 0, 1}, {1, 2, 3}};
   solve(h, w, grid);
   return 0;
}

Input

3, 3, {{2, 3, 4}, {2, 0, 1}, {1, 2, 3}}

Output

4
(0, 1) - (0, 2)
(2, 0) - (2, 1)
(2, 1) - (2, 2)
(0, 2) - (1, 2)

Arnab Chakraborty

Updated on 02-Mar-2022 10:56:05

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