# 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(n^{2}) 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_h long 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 code int 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_h def 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 code if __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_h function 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 ?>

## Javascript

<script> // JavaScript 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_h function costOfOperation(n, h, cost, eq_h) { // Initialize initial cost to 0 let c = 0; // Adding cost for each tower for (let 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. function Bsearch(n, h, cost) { let max_h = [...h].sort((a, b) => a - b)[h.length - 1]; let ans = Number.MAX_SAFE_INTEGER; // Do binary search for minimum cost let high = 1 + max_h; let low = 0; while (high > low) { let mid = (low + high) >> 1; // Cost below mid let bm = (mid > 0) ? costOfOperation(n, h, cost, mid - 1) : PHP_INT_MAX; // Cost at mid let m = costOfOperation(n, h, cost, mid); // Cost above mid let 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 let h = new Array( 1, 2, 3 ); let cost = new Array( 10, 100, 1000 ); let n = h.length; document.write(Bsearch(n, h, cost)); // This code is contributed by gfgking </script>

**Output:**

120

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