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Minimum cost to form a number X by adding up powers of 2
• Difficulty Level : Medium
• Last Updated : 07 May, 2021

Given an array arr[] of N integers and an integer X. Element arr[i] in array denotes the cost to use 2i. The task is to find the minimum cost to choose the numbers which add up to X
Examples:

Input: arr[] = { 20, 50, 60, 90 }, X = 7
Output: 120
22 + 21 + 20 = 4 + 2 + 1 = 7 with cost = 60 + 50 + 20 = 130
But we can use 22 + 3 * 20 = 4 + 3 * 1 = 7 with cost = 60 + 3 * 20 = 120 which is minimum possible.
Input: arr[] = { 10, 5, 50 }, X = 4
Output: 10

Approach: The problem can be solved using basic Dynamic Programming. The fact that every number can be formed using powers of 2 has been used here. We can initially calculate the minimal cost required to form powers of 2 themselves. The recurrence will be as follows:

a[i] = min(a[i], 2 * a[i – 1])

Once the array is re-computed, we can simply keep adding the cost according to the set bits in the number X.
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` `MinimumCost(``int` `a[], ``int` `n, ``int` `x)``{` `    ``// Re-compute the array``    ``for` `(``int` `i = 1; i < n; i++) {``        ``a[i] = min(a[i], 2 * a[i - 1]);``    ``}` `    ``int` `ind = 0;` `    ``int` `sum = 0;` `    ``// Add answers for set bits``    ``while` `(x) {` `        ``// If bit is set``        ``if` `(x & 1)``            ``sum += a[ind];` `        ``// Increase the counter``        ``ind++;` `        ``// Right shift the number``        ``x = x >> 1;``    ``}` `    ``return` `sum;``}` `// Driver code``int` `main()``{``    ``int` `a[] = { 20, 50, 60, 90 };``    ``int` `x = 7;``    ``int` `n = ``sizeof``(a) / ``sizeof``(a);``    ``cout << MinimumCost(a, n, x);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.io.*;` `class` `GFG``{``    ` `// Function to return the minimum cost``static` `int` `MinimumCost(``int` `a[], ``int` `n, ``int` `x)``{` `    ``// Re-compute the array``    ``for` `(``int` `i = ``1``; i < n; i++)``    ``{``        ``a[i] = Math.min(a[i], ``2` `* a[i - ``1``]);``    ``}` `    ``int` `ind = ``0``;` `    ``int` `sum = ``0``;` `    ``// Add answers for set bits``    ``while` `(x > ``0``)``    ``{` `        ``// If bit is set``        ``if` `(x != ``0` `)``            ``sum += a[ind];` `        ``// Increase the counter``        ``ind++;` `        ``// Right shift the number``        ``x = x >> ``1``;``    ``}` `    ``return` `sum;``}` `// Driver code``public` `static` `void` `main (String[] args)``{` `    ``int` `a[] = { ``20``, ``50``, ``60``, ``90` `};``    ``int` `x = ``7``;``    ``int` `n =a.length;``    ``System.out.println (MinimumCost(a, n, x));``}``}` `// This Code is contributed by akt_mit`

## Python3

 `# Python 3 implementation of the approach` `# Function to return the minimum cost``def` `MinimumCost(a, n, x):``    ` `    ``# Re-compute the array``    ``for` `i ``in` `range``(``1``, n, ``1``):``        ``a[i] ``=` `min``(a[i], ``2` `*` `a[i ``-` `1``])` `    ``ind ``=` `0` `    ``sum` `=` `0` `    ``# Add answers for set bits``    ``while` `(x):``        ` `        ``# If bit is set``        ``if` `(x & ``1``):``            ``sum` `+``=` `a[ind]` `        ``# Increase the counter``        ``ind ``+``=` `1` `        ``# Right shift the number``        ``x ``=` `x >> ``1` `    ``return` `sum` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``a ``=` `[``20``, ``50``, ``60``, ``90``]``    ``x ``=` `7``    ``n ``=` `len``(a)``    ``print``(MinimumCost(a, n, x))` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `// Function to return the minimum cost``public` `static` `int` `MinimumCost(``int` `[]a, ``int` `n, ``int` `x)``{` `    ``// Re-compute the array``    ``for` `(``int` `i = 1; i < n; i++)``    ``{``        ``a[i] = Math.Min(a[i], 2 * a[i - 1]);``    ``}` `    ``int` `ind = 0;` `    ``int` `sum = 0;` `    ``// Add answers for set bits``    ``while` `(x > 0)``    ``{` `        ``// If bit is set``        ``if` `(x != 0 )``            ``sum += a[ind];` `        ``// Increase the counter``        ``ind++;` `        ``// Right shift the number``        ``x = x >> 1;``    ``}` `    ``return` `sum;``}` `// Driver code``public` `static` `void` `Main ()``{` `    ``int` `[]a = { 20, 50, 60, 90 };``    ``int` `x = 7;``    ``int` `n =a.Length;``    ``Console.WriteLine(MinimumCost(a, n, x));``}``}` `// This Code is contributed by SoM15242`

## PHP

 `> 1;``    ``}` `    ``return` `\$sum``;``}` `    ``// Driver code``    ``\$a` `= ``array``( 20, 50, 60, 90 );``    ``\$x` `= 7;``    ``\$n` `= sizeof(``\$a``) / sizeof(``\$a``);``    ``echo` `MinimumCost(``\$a``, ``\$n``, ``\$x``);` `// This code is contributed by ajit.``?>`

## Javascript

 ``
Output:
`120`

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