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_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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