# Minimum cost to cover the given positions in a N*M grid

Last Updated : 09 Sep, 2022

Given a n*m grid and the position of some poles to be painted in the grid, the task is to find the minimum cost to paint all the poles. There is no cost involved in moving from one row to the other, whereas moving to an adjacent column has 1 rupee cost associated with it.

Examples:

```Input: n = 2, m = 2, noOfPos = 2
pos[0] = 0, 0
pos[1] = 0, 1

Output: 1
The grid is of 2*2 size and there are two poles at {0, 0} and {0, 1}.
So we will start at {0, 0} and paint the pole and then go to
the next column to paint the pole at {0, 1} position which will
cost 1 rupee to move one column.

Input: n = 2, m = 2, noOfPos = 2
pos[0] = {0, 0}
pos[1] = {1, 0}
Output: 0
Both poles are in the same column. So, no need to move to another column. ```

Approach: As there is the only cost of moving in columns, if we go to any column we will paint all the poles in that column and then move forward. So, basically, the answer will be the difference between the two farthest columns.

Below is the required implementation:

## C++

 `// C++ implementation of the above approach`   `#include ` `#include `   `using` `namespace` `std;`   `    ``// Function to find the cost to paint all poles` `    ``void` `find(``int` `n,``int` `m,``int` `p,``int` `q[2][2])` `    ``{` `        ``// To store all the columns,create list` `        ``list <``int``> z ;` `        ``int` `i ;` `        `  `        ``for``(i = 0;i < p;i++)` `            ``z.push_back(q[i][1]);` `        `  `        ``// sort in ascending order` `        ``z.sort();` `        `  `        ``// z.back() gives max value` `        ``// z.front() gives min value` `        ``cout << z.back() - z.front() <

## Java

 `// Java implementation of the above approach `   `import` `java.util.*;` `class` `solution` `{`   `    ``// Function to find the cost to paint all poles ` `    ``static` `void` `find(``int` `n,``int` `m,``int` `p,``int` `q[][]) ` `    ``{ ` `        ``// To store all the columns,create list ` `        ``Vector z= ``new` `Vector() ; ` `        ``int` `i ; ` `        `  `        ``for``(i = ``0``;i < p;i++) ` `            ``z.add(q[i][``1``]); ` `        `  `        ``// sort in ascending order ` `        ``Collections.sort(z); ` `        `  `        ``// z.back() gives max value ` `        ``// z.front() gives min value ` `        ``System.out.print(z.get(z.size()-``1``) - z.get(``0``) ) ; ` `    ``} ` `    `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[])` `    ``{ ` `        ``int` `n = ``2``; ` `        ``int` `m = ``2``; ` `        ``int` `p = ``2``; ` `        `  `        ``int` `q[][] = {{``0``,``0``},{``0``,``1``}} ; ` `        `  `        ``find(n, m, p, q); ` `        `  `         `  `    ``} `       `}` `//contributed by Arnab Kundu`

## Python3

 `# Function to find the cost to paint all poles` `import` `math as ma`   `def` `find(n, m, p, q):`   `    ``# To store all the columns` `    ``z ``=``[]` `    ``for` `i ``in` `range``(p):` `        ``z.append(q[i][``1``])`   `    ``print``(``max``(z)``-``min``(z))` `    `  `n, m, p ``=` `2``, ``2``, ``2` `q ``=``[(``0``, ``0``), (``0``, ``1``)]` `find(n, m, p, q)`

## C#

 `// C# implementation of the above approach ` `using` `System;` `using` `System.Collections.Generic;`   `class` `GFG` `{`   `    ``// Function to find the cost to paint all poles ` `    ``static` `void` `find(``int` `n, ``int` `m, ``int` `p, ``int` `[,]q) ` `    ``{ ` `        ``// To store all the columns,create list ` `        ``List <``int``> z = ``new` `List<``int``>(); ` `        ``int` `i; ` `        `  `        ``for``(i = 0; i < p; i++) ` `            ``z.Add(q[i, 1]); ` `        `  `        ``// sort in ascending order ` `        ``z.Sort(); ` `        `  `        ``// z.back() gives max value ` `        ``// z.front() gives min value ` `        ``Console.Write(z[z.Count-1] - z[0]); ` `    ``} ` `    `  `    ``// Driver code ` `    ``public` `static` `void` `Main(String []args)` `    ``{ ` `        ``int` `n = 2; ` `        ``int` `m = 2; ` `        ``int` `p = 2; ` `        `  `        ``int` `[,]q = {{0, 0}, {0, 1}}; ` `        `  `        ``find(n, m, p, q); ` `    ``} ` `}`   `// This code is contributed by PrinciRaj1992`

## PHP

 ` `

## Javascript

 ``

Output

```1
```

Complexity Analysis:

• Time Complexity: O(plogp)
• Auxiliary Space: O(p)

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