Minimize cost of operation to equalize tower heights
Given heights of n (n <=10000) towers as an array h[]; we need to bring every tower to same height by either adding or removing blocks in a tower. Every addition or removal operation costs a different value in a different tower. The objective is to minimize this cost.
Examples:
Input : Tower heights h[] = {1, 2, 3}
Costs of operations cost[] = {10, 100, 1000}
Output : 120
The heights can be equalized by either "Removing
one block from 3 and adding one in 1" or "Adding
two blocks in 1 and adding one in 2". Since the
cost of operation in tower 3 is 1000, the first
process would yield 1010 while the second one
yields 120. Since the second process yields the
lowest cost of operation, it is the required
output.
Input : h[] = {7,1,5 };
cost[] = { 1, 1, 1 };
Output : 6
abs(7-5) + abs(1-5) + abs(5-5) = 6 (taking 5
as final height)
Input : h[] = {2, 4, 8}
cost[] = {10, 100, 1000}
Output : 460
Naive Approach : A naive approach would be to find the cost of each and every possible operation set, and then find the minimum cost of operation. This would give O(n2) in the worst case.
Best Approach : The best approach here seems to be Binary search. We store the minimum height & alter our resultant height by a factor of 2 in each step.
C++
// C++ program to minimize the cost of operation// to bring all towers to same height.#include <bits/stdc++.h>using namespace std; // Returns the cost of entire operation in bringing // the height of all towers to equal height eq_hlong int costOfOperation(int n, int h[], int cost[], int eq_h){ // Initialize initial cost to 0 long int c = 0; // Adding cost for each tower for (int i = 0; i < n; i++) c += abs(h[i] - eq_h) * cost[i]; return c;} // Return the minimum possible cost of operation// to bring all the towers of different height// in height[0..n-1] to same height.long long int Bsearch(int n, int h[], int cost[]){ long int max_h = *max_element(h, h + n); long int ans = LONG_MAX; // Do binary search for minimum cost long int high = 1 + max_h; long int low = 0; while (high > low) { int mid = (low + high) >> 1; // Cost below mid long int bm = (mid > 0) ? costOfOperation(n, h, cost, mid - 1) : LONG_MAX; // Cost at mid long int m = costOfOperation(n, h, cost, mid); // Cost above mid long int am = costOfOperation(n, h, cost, mid + 1); // ans should hold the minimum cost of operation ans = min(ans, m); // Search lower cost if (bm <= m) high = mid; // Search higher cost else if (am <= m) low = mid + 1; // If am > m and bm > m // then ans is m else return m; } return ans;} // Driver codeint main(){ int h[] = { 1, 2, 3 }; int cost[] = { 10, 100, 1000 }; int n = sizeof(h)/sizeof(h[0]); cout << Bsearch(n, h, cost); return 0;} |
Java
// Java program to minimize the cost of operation// to bring all towers to same height.import java.util.Arrays;public class MinCostOp_Mintowerheight { // Returns the cost of entire operation in bringing // the height of all towers to equal height eq_h static long costOfOperation(int n, int h[], int cost[], int eq_h) { // Initialize initial cost to 0 long c = 0; // Adding cost for each tower for (int i = 0; i < n; i++) c += Math.abs(h[i] - eq_h) * cost[i]; return c; } // Return the minimum possible cost of operation // to bring all the towers of different height // in height[0..n-1] to same height. static long Bsearch(int n, int h[], int cost[]) { int max_h = Arrays.stream(h).max().getAsInt(); long ans = Long.MAX_VALUE; // Do binary search for minimum cost long high = 1 + max_h; long low = 0; while (high > low) { int mid = (int) ((low + high) >> 1); // Cost below mid long bm = (mid > 0) ? costOfOperation(n, h, cost, mid - 1) : Long.MAX_VALUE; // Cost at mid long m = costOfOperation(n, h, cost, mid); // Cost above mid long am = costOfOperation(n, h, cost, mid + 1); // Break if the answer becomes equal to // minimum cost m if (ans == m) break; // ans should hold the minimum cost of operation ans = Long.min(ans, m); // Search lower cost if (bm <= m) high = mid; // Search higher cost else if (am <= m) low = mid + 1; // If am > m and bm > m // then ans is m else return m; } return ans; } // Driver code public static void main(String args[]) { int h[] = { 1, 2, 3 }; int cost[] = { 10, 100, 1000 }; int n = h.length; System.out.println(Bsearch(n, h, cost)); }}// This code is contributed by Sumit Ghosh |
Python3
# Python 3 program to minimize the cost of # operation to bring all towers to same height.import sys # Returns the cost of entire operation in # bringing the height of all towers to# equal height eq_hdef costOfOperation(n, h,cost, eq_h): # Initialize initial cost to 0 c = 0 # Adding cost for each tower for i in range(0, n, 1): c += abs(h[i] - eq_h) * cost[i] return c # Return the minimum possible cost of # operation to bring all the towers of # different height in height[0..n-1] # to same height.def Bsearch(n, h, cost): max_h = h[0] for i in range(len(h)): if(h[i] > max_h): max_h = h[i] ans = sys.maxsize # Do binary search for minimum cost high = 1 + max_h low = 0 while (high > low): mid = (low + high) >> 1 # Cost below mid if(mid > 0): bm = costOfOperation(n, h, cost, mid - 1) else: bm = sys.maxsize # Cost at mid m = costOfOperation(n, h, cost, mid) # Cost above mid am = costOfOperation(n, h, cost, mid + 1) # Break if the answer becomes equal # to minimum cost m if (ans == m): break # ans should hold the minimum cost # of operation ans = min(ans, m) # Search lower cost if (bm <= m): high = mid # Search higher cost elif(am <= m): low = mid + 1 # If am > m and bm > m # then ans is m else : return m; return ans # Driver codeif __name__ == '__main__': h = [1, 2, 3] cost = [10, 100, 1000] n = len(h) print(Bsearch(n, h, cost)) # This code is contributed by# Sahil_shelangia |
C#
// C# program to minimize the // cost of operation to bring// all towers to same height.using System;using System.Linq; public class MinCostOp_Mintowerheight{ // Returns the cost of entire // operation in bringing the height // of all towers to equal height eq_h static long costOfOperation(int n, int []h, int []cost, int eq_h) { // Initialize initial cost to 0 long c = 0; // Adding cost for each tower for (int i = 0; i < n; i++) c += Math.Abs(h[i] - eq_h) * cost[i]; return c; } // Return the minimum possible // cost of operation to bring // all the towers of different height // in height[0..n-1] to same height. static long Bsearch(int n, int []h, int []cost) { int max_h = h.Max(); long ans = long.MaxValue; // Do binary search for minimum cost long high = 1 + max_h; long low = 0; while (high > low) { int mid = (int) ((low + high) >> 1); // Cost below mid long bm = (mid > 0) ? costOfOperation(n, h, cost, mid - 1) : long.MaxValue; // Cost at mid long m = costOfOperation(n, h, cost, mid); // Cost above mid long am = costOfOperation(n, h, cost, mid + 1); // Break if the answer becomes // equal to minimum cost m if (ans == m) break; // ans should hold the minimum // cost of operation ans = Math.Min(ans, m); // Search lower cost if (bm <= m) high = mid; // Search higher cost else if (am <= m) low = mid + 1; // If am > m and bm > m // then ans is m else return m; } return ans; } // Driver code public static void Main(String []args) { int []h = { 1, 2, 3 }; int []cost = { 10, 100, 1000 }; int n = h.Length; Console.WriteLine(Bsearch(n, h, cost)); }} // This code contributed by Rajput-Ji |
PHP
<?php// PHP program to minimize the cost // of operation to bring all towers // to same height. // Returns the cost of entire operation // in bringing the height of all towers// to equal height eq_hfunction costOfOperation($n, $h, $cost, $eq_h){ // Initialize initial cost to 0 $c = 0; // Adding cost for each tower for ($i = 0; $i < $n; $i++) $c += abs($h[$i] - $eq_h) * $cost[$i]; return $c;} // Return the minimum possible cost of operation// to bring all the towers of different height// in height[0..n-1] to same height.function Bsearch($n, $h, $cost){ $max_h = max($h); $ans = PHP_INT_MAX; // Do binary search for minimum cost $high = 1 + $max_h; $low = 0; while ($high > $low) { $mid = ($low + $high) >> 1; // Cost below mid $bm = ($mid > 0) ? costOfOperation($n, $h, $cost, $mid - 1) : PHP_INT_MAX; // Cost at mid $m = costOfOperation($n, $h, $cost, $mid); // Cost above mid $am = costOfOperation($n, $h, $cost, $mid + 1); // Break if the answer becomes equal to // minimum cost m if ($ans == $m) break; // ans should hold the minimum // cost of operation $ans = min($ans, $m); // Search lower cost if ($bm <= $m) $high = $mid; // Search higher cost else if ($am <= $m) $low = $mid + 1; // If am > m and bm > m // then ans is m else return $m; } return $ans;} // Driver code$h = array( 1, 2, 3 );$cost = array( 10, 100, 1000 );$n = count($h);echo Bsearch($n, $h, $cost); // This code is contributed by mits?> |
Output:
120
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