Minimum Numbers of cells that are connected with the smallest path between 3 given cells

Given coordinates of 3 cells (X1, Y1), (X2, Y2) and (X3, Y3) of a matrix. The task is to find the minimum path which connects all three of these cells and print the count of all the cells that are connected through this path.
Note: Only possible moves are up, down, left and right.

Examples:

Input: X1 = 0, Y1 = 0, X2 = 1, Y2 = 1, X3 = 2 and Y3 = 2
Output: 5
(0, 0), (1, 0), (1, 1), (1, 2), (2, 2) are the required cells.

Input: X1 = 0, Y1 = 0, X2 = 2, Y2 = 0, X3 = 1 and Y3 = 1
Output: 4

Approach: First sort the cells from the one with minimum row number at first to one with maximum row number at last. Also, store minimum column number and maximum column number among these three cells in variable MinCol and MaxCol respectively.
After that, store row number of the middle cell(from sorted cells) in variable MidRow and mark all the cells of this MidRow from MinCol to MaxCol.
Now our final step will be to mark all the column number of 1st and 3rd cell till they reach MidRow.
Here, marking means we will store the required cells coordinate in a set. Thus, our answer will be size of this set.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the minimum cells that are
// connected via the minimum length path
int Minimum_Cells(vector<pair<int, int> > v)
{
    int col[3], i, j;
    for (i = 0; i < 3; i++) {
        int column_number = v[i].second;
  
        // Array to store column number
        // of the given cells
        col[i] = column_number;
    }
  
    sort(col, col + 3);
  
    // Sort cells in ascending
    // order of row number
    sort(v.begin(), v.end());
  
    // Middle row number
    int MidRow = v[1].first;
  
    // Set pair to store required cells
    set<pair<int, int> > s;
  
    // Range of column number
    int Maxcol = col[2], MinCol = col[0];
  
    // Store all cells of middle row
    // within column number range
    for (i = MinCol; i <= Maxcol; i++) {
        s.insert({ MidRow, i });
    }
  
    for (i = 0; i < 3; i++) {
        if (v[i].first == MidRow)
            continue;
  
        // Final step to store all the column number
        // of 1st and 3rd cell upto MidRow
        for (j = min(v[i].first, MidRow);
             j <= max(v[i].first, MidRow); j++) {
            s.insert({ j, v[i].second });
        }
    }
  
    return s.size();
}
  
// Driver Function
int main()
{
    // vector pair to store X, Y, Z
    vector<pair<int, int> > v = { { 0, 0 }, { 1, 1 }, { 2, 2 } };
  
    cout << Minimum_Cells(v);
  
    return 0;
}

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Python3

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# Python3 implementation of the approach 
  
# Function to return the minimum cells that 
# are connected via the minimum length path 
def Minimum_Cells(v) :
  
    col = [0] * 3
    for i in range(3) : 
        column_number = v[i][1]
  
        # Array to store column number 
        # of the given cells 
        col[i] = column_number 
      
    col.sort()
  
    # Sort cells in ascending order 
    # of row number 
    v.sort()
  
    # Middle row number 
    MidRow = v[1][0]
  
    # Set pair to store required cells 
    s = set()
  
    # Range of column number 
    Maxcol = col[2]
    MinCol = col[0]
  
    # Store all cells of middle row 
    # within column number range 
    for i in range(MinCol, int(Maxcol) + 1) : 
        s.add((MidRow, i))
  
    for i in range(3) : 
        if (v[i][0] == MidRow) :
            continue
  
        # Final step to store all the column 
        # number of 1st and 3rd cell upto MidRow 
        for j in range(min(v[i][0], MidRow), 
                       max(v[i][0], MidRow) + 1) :
            s.add((j, v[i][1])); 
              
    return len(s)
  
# Driver Code
if __name__ == "__main__"
  
    # vector pair to store X, Y, Z 
    v = [(0, 0 ), ( 1, 1 ), ( 2, 2 )]
  
    print(Minimum_Cells(v))
  
# This code is contributed by Ryuga

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Output:

5


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Improved By : Ryuga