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

• Last Updated : 13 Oct, 2021

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:
(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:

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 ` `using` `namespace` `std;`   `// Function to return the minimum cells that are` `// connected via the minimum length path` `int` `Minimum_Cells(vector > 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 > 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 > v = { { 0, 0 }, { 1, 1 }, { 2, 2 } };`   `    ``cout << Minimum_Cells(v);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.util.*;` `import` `java.awt.Point;` `public` `class` `Main` `{` `    ``// Function to return the minimum cells that are` `    ``// connected via the minimum length path` `    ``static` `int` `Minimum_Cells(Vector v)` `    ``{` `        ``int``[] col = ``new` `int``[``3``];` `        ``int` `i, j;` `        ``for` `(i = ``0``; i < ``3``; i++) {` `            ``int` `column_number = v.get(i).y;` `       `  `            ``// Array to store column number` `            ``// of the given cells` `            ``col[i] = column_number;` `        ``}` `       `  `        ``Arrays.sort(col);` `       `  `        ``// Middle row number` `        ``int` `MidRow = v.get(``1``).x;` `       `  `        ``// Set pair to store required cells` `        ``Set s = ``new` `HashSet();` `       `  `        ``// 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.add(``new` `Point(MidRow, i));` `        ``}` `       `  `        ``for` `(i = ``0``; i < ``3``; i++) {` `            ``if` `(v.get(i).x == MidRow)` `                ``continue``;` `       `  `            ``// Final step to store all the column number` `            ``// of 1st and 3rd cell upto MidRow` `            ``for` `(j = Math.min(v.get(i).x, MidRow);` `                 ``j <= Math.max(v.get(i).x, MidRow); j++) {` `                ``s.add(``new` `Point(j, v.get(i).x));` `            ``}` `        ``}` `       `  `        ``return` `s.size();` `    ``}` `    `  `  ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `      `  `        ``// vector pair to store X, Y, Z` `        ``Vector v = ``new` `Vector();` `        ``v.add(``new` `Point(``0``, ``0``));` `        ``v.add(``new` `Point(``1``, ``1``));` `        ``v.add(``new` `Point(``2``, ``2``));` `       `  `        ``System.out.print(Minimum_Cells(v));` `    ``}` `}`   `// This code is contributed by mukesh07.`

## 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 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`

## C#

 `// C# implementation of the approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG {` `    `  `    ``// Function to return the minimum cells that are` `    ``// connected via the minimum length path` `    ``static` `int` `Minimum_Cells(List> v)` `    ``{` `        ``int``[] col = ``new` `int``[3];` `        ``int` `i, j;` `        ``for` `(i = 0; i < 3; i++) {` `            ``int` `column_number = v[i].Item2;` `      `  `            ``// Array to store column number` `            ``// of the given cells` `            ``col[i] = column_number;` `        ``}` `      `  `        ``Array.Sort(col);` `      `  `        ``// Sort cells in ascending` `        ``// order of row number` `        ``v.Sort();` `      `  `        ``// Middle row number` `        ``int` `MidRow = v[1].Item1;` `      `  `        ``// Set pair to store required cells` `        ``HashSet> s = ``new` `HashSet>();` `      `  `        ``// 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.Add(``new` `Tuple<``int``,``int``>(MidRow, i));` `        ``}` `      `  `        ``for` `(i = 0; i < 3; i++) {` `            ``if` `(v[i].Item1 == MidRow)` `                ``continue``;` `      `  `            ``// Final step to store all the column number` `            ``// of 1st and 3rd cell upto MidRow` `            ``for` `(j = Math.Min(v[i].Item1, MidRow);` `                 ``j <= Math.Max(v[i].Item1, MidRow); j++) {` `                ``s.Add(``new` `Tuple<``int``,``int``>(j, v[i].Item1));` `            ``}` `        ``}` `      `  `        ``return` `s.Count;` `    ``}`   `  ``static` `void` `Main() ` `  ``{` `    `  `    ``// vector pair to store X, Y, Z` `    ``List> v = ``new` `List>();` `    ``v.Add(``new` `Tuple<``int``,``int``>(0, 0));` `    ``v.Add(``new` `Tuple<``int``,``int``>(1, 1));` `    ``v.Add(``new` `Tuple<``int``,``int``>(2, 2));` `  `  `    ``Console.Write(Minimum_Cells(v));` `  ``}` `}`   `// This code is contributed by divyeshrabadiya07.`

## Javascript

 ``

Output:

`5`

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