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++
// 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 pathint 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 Functionint 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;} |
Python3
# 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 Codeif __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 |
5
