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. 

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 / 2^deg, cnt_deg) elements with the value equals 2^deg. Let it be cur. Add cur to the answer and subtract 2^deg * 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 <bits/stdc++.h>
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[0]);

    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<Integer> cnt = new Vector<Integer>();

    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[32];

    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

Javascript

<script>

// JavaScript implementation of the approach
// Function to return the minimum number
// of given integer powers of 2 required
// to represent a number as sum of these powers
function power_of_two( n, a, x)
{

    // To store the count of powers of two
    let cnt = [];
    for(let i = 0;i<31;i++)
        cnt.push(0);
    for (let i = 0; i < n; ++i) {

        // __builtin_ctz(a[i]) returns the count
        // of trailing 0s in a[i]
        let count = 0;
        let xx = a[i];
        while ((xx & 1) == 0){
            xx = xx >> 1
            count += 1
        }
        cnt[count]+=1;
    }

    let ans = 0;
    for (let i = 30; i >= 0 && x > 0; --i) {

        // If current power is available
        // in the array and can be used
        let 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
let arr = [ 2, 2, 4, 4, 8 ], x = 6;
let n = arr.length;
document.write( power_of_two(n, arr, x));

</script>

Output

2

Time Complexity: O(N)

Auxiliary Space: O(32), since no extra space has been taken.