# Minimum number of operations on an array to make all elements 0

• Last Updated : 11 May, 2021

Given an array arr[] of N integers and an integer cost, the task is to calculate the cost of making all the elements of the array 0 with the given operation. In a single operation, an index 0 â‰¤ i < N and an integer X > 0 can be chosen such that 0 â‰¤ i + X < N then elements can be updated as arr[i] = arr[i] – 1 and arr[i + X] = arr[i + X] + 1. If i + X â‰¥ N then only arr[i] will be updated but with twice the regular cost. Print the minimum cost required.
Examples:

Input: arr[] = {1, 2, 4, 5}, cost = 1
Output: 31
Move 1: i = 0, X = 3, arr[] = {0, 2, 4, 6} (cost = 1)
Moves 2 and 3: i = 1, X = 2, arr[] = {0, 0, 4, 8} (cost = 2)
Moves 4, 5, 6 and 7: i = 2, X = 1, arr[] = {0, 0, 0, 12} (cost = 4)
Move 8: i = 3, X > 0, arr[] = {0, 0, 0, 0} (cost = 24)
Total cost = 1 + 2 + 4 + 24 = 31
Input: arr[] = {1, 1, 0, 5}, cost = 2
Output: 32

Approach: To minimize the cost, for every index i always choose X such that i + X = N – 1 i.e. the last element then minimum cost can be calculated as:

• Store the sum of the elements from arr[0] to arr[n – 2] in sum then update totalCost = cost * sum and arr[n – 1] = arr[n – 1] + sum.
• Now the cost of making all the elements 0 except the last one has been calculated. And the cost of making the last element 0 can be calculated as totalCost = totalCost + (2 * cost * arr[n – 1]).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to return the minimum cost``int` `minCost(``int` `n, ``int` `arr[], ``int` `cost)``{``    ``int` `sum = 0, totalCost = 0;` `    ``// Sum of all the array elements``    ``// except the last element``    ``for` `(``int` `i = 0; i < n - 1; i++)``        ``sum += arr[i];` `    ``// Cost of making all the array elements 0``    ``// except the last element``    ``totalCost += cost * sum;` `    ``// Update the last element``    ``arr[n - 1] += sum;` `    ``// Cost of making the last element 0``    ``totalCost += (2 * cost * arr[n - 1]);` `    ``return` `totalCost;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 1, 2, 4, 5 };``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);``    ``int` `cost = 1;``    ``cout << minCost(n, arr, cost);``}`

## Java

 `// Java implementation of the approach``public` `class` `GfG``{` `    ``// Function to return the minimum cost``    ``static` `int` `minCost(``int` `n, ``int` `arr[], ``int` `cost)``    ``{``        ``int` `sum = ``0``, totalCost = ``0``;``    ` `        ``// Sum of all the array elements``        ``// except the last element``        ``for` `(``int` `i = ``0``; i < n - ``1``; i++)``            ``sum += arr[i];``    ` `        ``// Cost of making all the array elements 0``        ``// except the last element``        ``totalCost += cost * sum;``    ` `        ``// Update the last element``        ``arr[n - ``1``] += sum;``    ` `        ``// Cost of making the last element 0``        ``totalCost += (``2` `* cost * arr[n - ``1``]);``    ` `        ``return` `totalCost;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String []args)``    ``{``        ` `        ``int` `arr[] = { ``1``, ``2``, ``4``, ``5` `};``        ``int` `n = arr.length;``        ``int` `cost = ``1``;``        ``System.out.println(minCost(n, arr, cost));``    ``}``}` `// This code is contributed by Rituraj Jain`

## Python3

 `# Python3 implementation of the approach` `# Function to return the minimum cost``def` `minCost(n, arr, cost):` `    ``Sum``, totalCost ``=` `0``, ``0` `    ``# Sum of all the array elements``    ``# except the last element``    ``for` `i ``in` `range``(``0``, n ``-` `1``):``        ``Sum` `+``=` `arr[i]` `    ``# Cost of making all the array elements 0``    ``# except the last element``    ``totalCost ``+``=` `cost ``*` `Sum` `    ``# Update the last element``    ``arr[n ``-` `1``] ``+``=` `Sum` `    ``# Cost of making the last element 0``    ``totalCost ``+``=` `(``2` `*` `cost ``*` `arr[n ``-` `1``])` `    ``return` `totalCost` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``arr ``=` `[``1``, ``2``, ``4``, ``5``]``    ``n ``=` `len``(arr)``    ``cost ``=` `1``    ``print``(minCost(n, arr, cost))` `# This code is contributed by Rituraj Jain`

## C#

 `// C# implementation of the approach``using` `System ;` `class` `GfG``{` `    ``// Function to return the minimum cost``    ``static` `int` `minCost(``int` `n, ``int` `[]arr, ``int` `cost)``    ``{``        ``int` `sum = 0, totalCost = 0;``    ` `        ``// Sum of all the array elements``        ``// except the last element``        ``for` `(``int` `i = 0; i < n - 1; i++)``            ``sum += arr[i];``    ` `        ``// Cost of making all the array elements 0``        ``// except the last element``        ``totalCost += cost * sum;``    ` `        ``// Update the last element``        ``arr[n - 1] += sum;``    ` `        ``// Cost of making the last element 0``        ``totalCost += (2 * cost * arr[n - 1]);``    ` `        ``return` `totalCost;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ` `        ``int` `[]arr = { 1, 2, 4, 5 };``        ``int` `n = arr.Length;``        ``int` `cost = 1;``        ``Console.WriteLine(minCost(n, arr, cost));``    ``}``}` `// This code is contributed by Ryuga`

## PHP

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## Javascript

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Output:
`31`

Time Complexity: O(n)

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