给定一个由正整数和三个整数A 、 R 、 M 组成的数组arr[] ,其中
- 将 1 添加到数组元素的成本是A ,
- 从数组的元素中减去 1 的成本是R并且
- 将一个元素加 1 并同时从另一个元素中减去 1 的成本是M 。
任务是找到使数组中所有元素相等的最小总成本。
例子:
Input: arr[] = {5, 5, 3, 6, 5}, A = 1, R = 2, M = 4
Output: 4
Explanation:
Operation 1: Add two times to element 3, Array – {5, 5, 5, 6, 5}, Cost = 2
Operation 2: Subtract one time to element 6, Array – {5, 5, 5, 5, 5}, Cost = 4
Therefore, minimum cost is 4.
Input: arr[] = {5, 5, 3, 6, 5}, A = 1, R = 2, M = 2
Output: 3
方法:想法是:
- 找到M和A + R的最小值,因为 M 可以同时进行这两个操作。
- 然后,将前缀总和存储在一个数组中以在恒定时间内找到总和。
- 现在为每个元素计算使成本等于当前元素并找到它们中的最小值。
- 当我们使所有元素等于数组的平均值时,也可以存在最小的答案。
- 因此,最后还要计算使所有元素等于元素的近似平均总和的成本。
下面是上述方法的实现:
C++
// C++ implementation to find the
// minimum cost to make all array
// elements equal
#include
using namespace std;
// Function that returns the cost of
// making all elements equal to current element
int costCalculation(
int current, int arr[],
int n, int pref[],
int a, int r,
int minimum)
{
// Compute the lower bound
// of current element
int index
= lower_bound(
arr, arr + n, current)
- arr;
// Calculate the requirement
// of add operation
int left
= index * current - pref[index];
// Calculate the requirement
// of subtract operation
int right
= pref[n] - pref[index]
- (n - index)
* current;
// Compute minimum of left and right
int res = min(left, right);
left -= res;
right -= res;
// Computing the total cost of add
// and subtract operations
int total = res * minimum;
total += left * a;
total += right * r;
return total;
}
// Function that prints minimum cost
// of making all elements equal
void solve(int arr[], int n,
int a, int r, int m)
{
// Sort the given array
sort(arr, arr + n);
// Calculate minimum from a + r and m
int minimum = min(a + r, m);
int pref[n + 1] = { 0 };
// Compute prefix sum
// and store in pref
// array
for (int i = 0; i < n; i++)
pref[i + 1] = pref[i] + arr[i];
int ans = 10000;
// Find the minimum cost
// from the given elements
for (int i = 0; i < n; i++)
ans
= min(
ans,
costCalculation(
arr[i], arr, n,
pref, a, r,
minimum));
// Finding the minimum cost
// from the other cases where
// minimum cost can occur
ans
= min(
ans,
costCalculation(
pref[n] / n, arr,
n, pref, a,
r, minimum));
ans
= min(
ans,
costCalculation(
pref[n] / n + 1,
arr, n, pref,
a, r, minimum));
// Printing the minimum cost of making
// all elements equal
cout << ans << "\n";
}
// Driver Code
int main()
{
int arr[] = { 5, 5, 3, 6, 5 };
int A = 1, R = 2, M = 4;
int size = sizeof(arr) / sizeof(arr[0]);
// Function Call
solve(arr, size, A, R, M);
return 0;
}
Java
// Java implementation to find the
// minimum cost to make all array
// elements equal
import java.lang.*;
import java.util.*;
class GFG{
public static int lowerBound(int[] array, int length,
int value)
{
int low = 0;
int high = length;
while (low < high)
{
final int mid = (low + high) / 2;
// Checks if the value is less
// than middle element of the array
if (value <= array[mid])
{
high = mid;
}
else
{
low = mid + 1;
}
}
return low;
}
// Function that returns the cost of making
// all elements equal to current element
public static int costCalculation(int current, int arr[],
int n, int pref[],
int a, int r,
int minimum)
{
// Compute the lower bound
// of current element
int index = lowerBound(arr, arr.length, current);
// Calculate the requirement
// of add operation
int left = index * current - pref[index];
// Calculate the requirement
// of subtract operation
int right = pref[n] -
pref[index]- (n - index)* current;
// Compute minimum of left and right
int res = Math.min(left, right);
left -= res;
right -= res;
// Computing the total cost of add
// and subtract operations
int total = res * minimum;
total += left * a;
total += right * r;
return total;
}
// Function that prints minimum cost
// of making all elements equal
public static void solve(int arr[], int n,
int a, int r, int m)
{
// Sort the given array
Arrays.sort(arr);
// Calculate minimum from a + r and m
int minimum = Math.min(a + r, m);
int []pref = new int [n + 1];
Arrays.fill(pref, 0);
// Compute prefix sum and
// store in pref array
for(int i = 0; i < n; i++)
pref[i + 1] = pref[i] + arr[i];
int ans = 10000;
// Find the minimum cost
// from the given elements
for(int i = 0; i < n; i++)
ans = Math.min(ans, costCalculation(arr[i], arr,
n, pref,
a, r, minimum));
// Finding the minimum cost
// from the other cases where
// minimum cost can occur
ans = Math.min(ans, costCalculation(pref[n] / n, arr,
n, pref, a, r,
minimum));
ans = Math.min(ans, costCalculation(pref[n] / n + 1,
arr, n, pref,
a, r, minimum));
// Printing the minimum cost of making
// all elements equal
System.out.println(ans);
}
// Driver Code
public static void main(String args[])
{
int arr[] = { 5, 5, 3, 6, 5 };
int A = 1, R = 2, M = 4;
int size = arr.length ;
// Function Call
solve(arr, size, A, R, M);
}
}
// This code is contributed by SoumikMondal
Python3
# Python3 implementation to find the
# minimum cost to make all array
# elements equal
def lowerBound(array, length, value):
low = 0
high = length
while (low < high):
mid = (low + high) // 2
# Checks if the value is less
# than middle element of the array
if (value <= array[mid]):
high = mid
else:
low = mid + 1
return low
# Function that returns the cost of making
# all elements equal to current element
def costCalculation(current, arr,
n, pref, a, r, minimum):
# Compute the lower bound
# of current element
index = lowerBound(arr, len(arr), current)
# Calculate the requirement
# of add operation
left = index * current - pref[index]
# Calculate the requirement
# of subtract operation
right = (pref[n] - pref[index] -
(n - index) * current)
# Compute minimum of left and right
res = min(left, right)
left -= res
right -= res
# Computing the total cost of add
# and subtract operations
total = res * minimum
total += left * a
total += right * r
return total
# Function that prints minimum cost
# of making all elements equal
def solve(arr, n, a, r, m):
# Sort the given array
arr.sort()
# Calculate minimum from a + r and m
minimum = min(a + r, m)
pref = [0] * (n + 1)
# Compute prefix sum and
# store in pref array
for i in range(n):
pref[i + 1] = pref[i] + arr[i]
ans = 10000
# Find the minimum cost
# from the given elements
for i in range(n):
ans = min(ans, costCalculation(arr[i], arr,
n, pref, a,
r, minimum))
# Finding the minimum cost
# from the other cases where
# minimum cost can occur
ans = min(ans, costCalculation(pref[n] // n,
arr, n, pref,
a, r, minimum))
ans = min(ans, costCalculation(pref[n] // n + 1,
arr, n, pref, a,
r, minimum))
# Printing the minimum cost of making
# all elements equal
print(ans)
# Driver Code
if __name__ == "__main__":
arr = [ 5, 5, 3, 6, 5 ]
A = 1
R = 2
M = 4
size = len(arr)
# Function call
solve(arr, size, A, R, M)
# This code is contributed by chitranayal
C#
// C# implementation to find the
// minimum cost to make all array
// elements equal
using System;
class GFG{
public static int lowerBound(int[] array,
int length,
int value)
{
int low = 0;
int high = length;
while (low < high)
{
int mid = (low + high) / 2;
// Checks if the value is less
// than middle element of the array
if (value <= array[mid])
{
high = mid;
}
else
{
low = mid + 1;
}
}
return low;
}
// Function that returns the cost of making
// all elements equal to current element
public static int costCalculation(int current,
int []arr, int n,
int []pref, int a,
int r, int minimum)
{
// Compute the lower bound
// of current element
int index = lowerBound(arr, arr.Length, current);
// Calculate the requirement
// of add operation
int left = index * current - pref[index];
// Calculate the requirement
// of subtract operation
int right = pref[n] - pref[index] -
(n - index) *
current;
// Compute minimum of left and right
int res = Math.Min(left, right);
left -= res;
right -= res;
// Computing the total cost of add
// and subtract operations
int total = res * minimum;
total += left * a;
total += right * r;
return total;
}
// Function that prints minimum cost
// of making all elements equal
public static void solve(int []arr, int n,
int a, int r, int m)
{
// Sort the given array
Array.Sort(arr);
// Calculate minimum from a + r and m
int minimum = Math.Min(a + r, m);
int []pref = new int [n + 1];
Array.Fill(pref, 0);
// Compute prefix sum and
// store in pref array
for(int i = 0; i < n; i++)
pref[i + 1] = pref[i] + arr[i];
int ans = 10000;
// Find the minimum cost
// from the given elements
for(int i = 0; i < n; i++)
ans = Math.Min(ans, costCalculation(arr[i], arr,
n, pref, a,
r, minimum));
// Finding the minimum cost
// from the other cases where
// minimum cost can occur
ans = Math.Min(ans, costCalculation(pref[n] / n, arr,
n, pref, a, r,
minimum));
ans = Math.Min(ans, costCalculation(pref[n] / n + 1,
arr, n, pref,
a, r, minimum));
// Printing the minimum cost of making
// all elements equal
Console.WriteLine(ans);
}
// Driver Code
public static void Main(string []args)
{
int []arr = { 5, 5, 3, 6, 5 };
int A = 1, R = 2, M = 4;
int size = arr.Length ;
// Function Call
solve(arr, size, A, R, M);
}
}
// This code is contributed by SoumikMondal
Javascript
输出:
4
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