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

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 <iostream>
#include <list>

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() <<endl ;
    }

    // Driver code
    int main()
    {
        int n = 2;
        int m = 2;
        int p = 2;

        int q[2][2] = {{0,0},{0,1}} ;

        find(n, m, p, q);

       return 0;

    // This code is contributed by ANKITRAI1
    }

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 <Integer> z= new Vector<Integer>() ;
        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

<?php
// PHP implementation of the above approach

// Function to find the cost to
// paint all poles
function find($n, $m, $p, $q)
{
    // To store all the columns,
    // create an array
    $z = array();

    for($i = 0; $i < $p; $i++)
        array_push($z, $q[$i][1]);

    // sort in ascending order
    sort($z);

    echo max($z) - min($z);
}

// Driver code
$n = 2;
$m = 2;
$p = 2;

$q = array(array(0, 0),
            array(0, 1));

find($n, $m, $p, $q);

// This code is contributed by ita_c
?>

Javascript

<script>

// Javascript implementation of the above approach
// Function to find the cost to paint all poles
function find(n,m,p,q)
{
    // To store all the columns,create list
    var z = [];
    var i ;

    for(i = 0;i < p;i++)
        z.push(q[i][1]);

    // sort in ascending order
    z.sort();

    // z.back() gives max value
    // z.front() gives min value
    document.write( z[z.length-1] - z[0]);
}

// Driver code
var n = 2;
var m = 2;
var p = 2;

var q = [[0,0],[0,1]];

find(n, m, p, q);


</script>

Output

1

Complexity Analysis:

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