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C++ program to find minimum number of operations needed to make all cells at r row c columns black
Suppose we have two numbers r, c and a grid of size n x m. Some cells are in black and remaining are white. In one operation, we can select some black cells and can do exactly one of these two −
- Color all cells in its row black, or
- color all cells in its column black.
We have to find the minimum number of operations needed to make the cells in row r and column c black. If impossible, return -1.
So, if the input is like
| W | B | W | W | W |
| B | B | B | W | B |
| W | W | B | B | B |
r = 0 and c = 3
then the output will be 1, because we can change the first row to make this like −
| B | B | B | B | B |
| B | B | B | W | B |
| W | W | B | B | B |
Steps
To solve this, we will follow these steps −
n := row count of grid m := column count of grid ans := inf 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 matrix[i, j] is same as 'B', then: ans := minimum of ans and (1 if i and r are different, otherwise 0) + (1 if j and c are different, otherwise 0) if ans > 2, then: return -1 Otherwise return ans
Example
Let us see the following implementation to get better understanding −
#include <bits/stdc++.h>
using namespace std;
int solve(vector<vector<char>> matrix, int r, int c) {
int n = matrix.size();
int m = matrix[0].size();
int ans = 999999;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
if (matrix[i][j] == 'B') {
ans = min(ans, (i != r) + (j != c));
}
}
}
if (ans > 2) {
return -1;
}
else
return ans;
}
int main() {
vector<vector<char>> matrix = { { 'W', 'B', 'W', 'W', 'W' }, { 'B', 'B', 'B', 'W', 'B' }, { 'W', 'W', 'B', 'B', 'B' } };
int r = 0, c = 3;
cout << solve(matrix, r, c) << endl;
}Input
{ { 'W', 'B', 'W', 'W', 'W' }, { 'B', 'B', 'B', 'W', 'B' }, { 'W', 'W', 'B', 'B', 'B' } }, 0, 3Output
1
Updated on: 03-Mar-2022
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