# Minimum number of given powers of 2 required to represent a number

Given an integer x and an array arr[] each element of which is a power of 2. The task is to find the minimum number of integer powers of 2 from the array which when added give x. If it is not possible to represent x with the given array elements then print -1.

Examples:

Input: arr[] = {2, 4, 8, 2, 4}, x = 14
Output: 3
14 can be written as 8 + 4 + 2

Input: arr[] = {2, 4, 8, 2, 4}, x = 5
Output: -1
5 cannot be represented as the sum any elements from the given array.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: For each power of 2 let’s calculate the number of elements in the given array with the value equals this. Let’s call it cnt. It is obvious that we can obtain the value x greedily (because all fewer values of elements are divisors of all greater values of elements).

Now let’s iterate over all powers of 2 from 30 to 0. Let’s deg be the current degree. We can take min(x / 2deg, cntdeg) elements with the value equals 2deg. Let it be cur. Add cur to the answer and subtract 2deg * cur from x. Repeat the process until the x can no longer be reduced. If after iterating over all powers, x is still non-zero then print -1. Otherwise, print the answer.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the minimum number ` `// of given integer powers of 2 required ` `// to represent a number as sum of these powers ` `int` `power_of_two(``int` `n, ``int` `a[], ``int` `x) ` `{ ` ` `  `    ``// To store the count of powers of two ` `    ``vector<``int``> cnt(31); ` ` `  `    ``for` `(``int` `i = 0; i < n; ++i) { ` ` `  `        ``// __builtin_ctz(a[i]) returns the count  ` `        ``// of trailing 0s in a[i] ` `        ``++cnt[__builtin_ctz(a[i])]; ` `    ``} ` ` `  `    ``int` `ans = 0; ` `    ``for` `(``int` `i = 30; i >= 0 && x > 0; --i) { ` ` `  `        ``// If current power is available ` `        ``// in the array and can be used ` `        ``int` `need = min(x >> i, cnt[i]); ` ` `  `        ``// Update the answer ` `        ``ans += need; ` ` `  `        ``// Reduce the number ` `        ``x -= (1 << i) * need; ` `    ``} ` ` `  `    ``// If the original number is not reduced to 0 ` `    ``// It cannot be represented as the sum ` `    ``// of the given powers of 2 ` `    ``if` `(x > 0) ` `        ``ans = -1; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 2, 4, 4, 8 }, x = 6; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``cout << power_of_two(n, arr, x); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach  ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` `     `  `// __builtin_ctz(a[i]) returns the count  ` `// of trailing 0s in a[i]  ` `static` `int` `__builtin_ctz(``int` `a) ` `{ ` `    ``int` `count = ``0``; ` `    ``for``(``int` `i = ``0``; i < ``40``; i++) ` `    ``if``(((a >> i) & ``1``) == ``0``) ` `    ``{ ` `        ``count++; ` `    ``} ` `    ``else`  `        ``break``; ` `    ``return` `count; ` `} ` ` `  `// Function to return the minimum number  ` `// of given integer powers of 2 required  ` `// to represent a number as sum of these powers  ` `static` `int` `power_of_two(``int` `n, ``int` `a[], ``int` `x)  ` `{  ` ` `  `    ``// To store the count of powers of two  ` `    ``Vector cnt = ``new` `Vector();  ` `     `  `    ``for` `(``int` `i = ``0``; i < ``31``; ++i) ` `        ``cnt.add(``0``); ` ` `  `    ``for` `(``int` `i = ``0``; i < n; ++i)  ` `    ``{  ` ` `  `        ``// __builtin_ctz(a[i]) returns the count  ` `        ``// of trailing 0s in a[i]  ` `         `  `        ``cnt.set(__builtin_ctz(a[i]),  ` `        ``(cnt.get(__builtin_ctz(a[i]))==``null``) ? ` `        ``1` `: cnt.get(__builtin_ctz(a[i]))+``1``);  ` `    ``}  ` ` `  `    ``int` `ans = ``0``;  ` `    ``for` `(``int` `i = ``30``; i >= ``0` `&& x > ``0``; --i)  ` `    ``{  ` ` `  `        ``// If current power is available  ` `        ``// in the array and can be used  ` `        ``int` `need = Math.min(x >> i, cnt.get(i));  ` ` `  `        ``// Update the answer  ` `        ``ans += need;  ` ` `  `        ``// Reduce the number  ` `        ``x -= (``1` `<< i) * need;  ` `    ``}  ` ` `  `    ``// If the original number is not reduced to 0  ` `    ``// It cannot be represented as the sum  ` `    ``// of the given powers of 2  ` `    ``if` `(x > ``0``)  ` `        ``ans = -``1``;  ` ` `  `    ``return` `ans;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String args[]) ` `{  ` `    ``int` `arr[] = { ``2``, ``2``, ``4``, ``4``, ``8` `}, x = ``6``;  ` `    ``int` `n = arr.length;  ` `    ``System.out.println(power_of_two(n, arr, x));  ` `}  ` `} ` ` `  `// This code is contributed by Arnab Kundu `

## python

 `# Python3 implementation of the approach ` ` `  `# Function to return the minimum number ` `# of given eger powers of 2 required ` `# to represent a number as sum of these powers ` `def` `power_of_two( n, a, x): ` ` `  ` `  `    ``# To store the count of powers of two ` `    ``cnt``=``[``0` `for` `i ``in` `range``(``31``)] ` ` `  `    ``for` `i ``in` `range``(n): ` `        ``# __builtin_ctz(a[i]) returns the count ` `        ``# of trailing 0s in a[i] ` `        ``count ``=` `0` `        ``xx ``=` `a[i] ` `        ``while` `((xx & ``1``) ``=``=` `0``): ` `            ``xx ``=` `xx >> ``1` `            ``count ``+``=` `1` ` `  `        ``cnt[count]``+``=``1` ` `  `    ``ans ``=` `0` `    ``for` `i ``in` `range``(``30``,``-``1``,``-``1``): ` `        ``if` `x<``=``0``: ` `            ``continue` ` `  `        ``# If current power is available ` `        ``# in the array and can be used ` `        ``need ``=` `min``(x >> i, cnt[i]) ` ` `  `        ``# Update the answer ` `        ``ans ``+``=` `need ` ` `  `        ``# Reduce the number ` `        ``x ``-``=` `(``1` `<< i) ``*` `need ` ` `  ` `  `    ``# If the original number is not reduced to 0 ` `    ``# It cannot be represented as the sum ` `    ``# of the given powers of 2 ` `    ``if` `(x > ``0``): ` `        ``ans ``=` `-``1` ` `  `    ``return` `ans ` ` `  ` `  `# Driver code ` ` `  `arr``=``[``2``, ``2``, ``4``, ``4``, ``8` `] ` `x ``=` `6` `n ``=` `len``(arr) ` ` `  `print``(power_of_two(n, arr, x)) ` ` `  `# This code is contributed by mohit kumar 29 `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// __builtin_ctz(a[i]) returns the count  ` `// of trailing 0s in a[i]  ` `static` `int` `__builtin_ctz(``int` `a) ` `{ ` `    ``int` `count = 0; ` `    ``for``(``int` `i = 0; i < 40; i++) ` `    ``if``(((a >> i) & 1) == 0) ` `    ``{ ` `        ``count++; ` `    ``} ` `    ``else` `        ``break``; ` `    ``return` `count; ` `} ` ` `  `// Function to return the minimum number  ` `// of given integer powers of 2 required  ` `// to represent a number as sum of these powers  ` `static` `int` `power_of_two(``int` `n, ``int` `[]a, ``int` `x)  ` `{  ` ` `  `    ``// To store the count of powers of two  ` `    ``int``[] cnt = ``new` `int``;  ` ` `  `    ``for` `(``int` `i = 0; i < n; ++i)  ` `    ``{  ` ` `  `        ``// __builtin_ctz(a[i]) returns the count  ` `        ``// of trailing 0s in a[i]  ` `         `  `        ``cnt[__builtin_ctz(a[i])] =  ` `        ``cnt[__builtin_ctz(a[i])] ==  ` `        ``0?1 : cnt[__builtin_ctz(a[i])] + 1;  ` `    ``}  ` ` `  `    ``int` `ans = 0;  ` `    ``for` `(``int` `i = 30; i >= 0 && x > 0; --i)  ` `    ``{  ` ` `  `        ``// If current power is available  ` `        ``// in the array and can be used  ` `        ``int` `need = Math.Min(x >> i, cnt[i]);  ` ` `  `        ``// Update the answer  ` `        ``ans += need;  ` ` `  `        ``// Reduce the number  ` `        ``x -= (1 << i) * need;  ` `    ``}  ` ` `  `    ``// If the original number is not reduced to 0  ` `    ``// It cannot be represented as the sum  ` `    ``// of the given powers of 2  ` `    ``if` `(x > 0)  ` `        ``ans = -1;  ` ` `  `    ``return` `ans;  ` `}  ` ` `  `// Driver code  ` `static` `void` `Main() ` `{  ` `    ``int` `[]arr = { 2, 2, 4, 4, 8 }; ` `    ``int` `x = 6;  ` `    ``int` `n = arr.Length;  ` `    ``Console.WriteLine(power_of_two(n, arr, x));  ` `}  ` `} ` ` `  `// This code is contributed by mits `

Output:

```2
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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