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 polesimport math as madef 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, 2q =[(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 polesfunction 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 polesfunction 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 codevar 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)
