Minimum size of set having either element in range [0, X] or an odd power of 2 with sum N

Given two positive integers N and X, the task is to find the size of the smallest set of integers such that the sum of all elements of the set is N and each set element is either in the range [0, X] or is an odd power of 2. If it is not possible to find such a size of the set then print “-1”.

Examples:

Input: N = 11, X = 2
Output: 3
Explanation: The set {1, 2, 8} is the set of minimum number of elements such that the sum of elements is 11 and each element is either in range [0, 2] (i.e, 1 and 2) or is an odd power of 2 (i.e., 8 = 2³).

Input: N = 3, X = 0
Output: -1
Explanation : No valid set exist.

Approach: The given problem can be solved using the below steps:

• Maintain a variable size that stores the minimum possible size of a valid set and initialize it with 0.
• Iterate until the value of N is greater than X  and perform the following steps:
  • Subtract the largest odd power i of 2 that is less than or equal to N from N.
  • Increment the value of size by 1.
• If the value of N is positive, then increment the value of size by 1.
• After completing the above steps, print the value of size as the required result.

Below is the implementation of the above approach:

C++

// CPP program for the above approach
#include<bits/stdc++.h>
using namespace std;

// Function to find the highest odd power
// of 2 in the range [0, N]
int highestPowerof2(int n)
{

    int p = int(log2(n));

    // If P is even, subtract 1
    if(p % 2 == 0)
        p -= 1;

    return int(pow(2, p));
}

// Function to find the minimum operations
// to make N
int minStep(int N, int X)
{

   // If N is odd and X = 0, then no
    // valid set exist
    if(N % 2 and X == 0)
        return -1;

    // Stores the minimum possible size
    // of the valid set
    int size = 0;

    // Loop to subtract highest odd power
    // of 2 while X < N, step 2
    while(X < N){
        N -= highestPowerof2(N);
        size += 1;
     }

    // If N > 0, then increment the value
    // of answer by 1
    if(N)
        size += 1;

    // Return the resultant size of set
    return size;

}

// Driver Code
int main(){
    int N = 11;
    int X = 2;
    cout<<(minStep(N, X));

}

// This code is contributed by ipg2016107.

Java

// Java program for the above approach
import java.io.*;

class GFG {

// Function to find the highest odd power
// of 2 in the range [0, N]
static int highestPowerof2(int n)
{

    int p = (int)Math.floor(Math.log(n)/Math.log(2.0));

    // If P is even, subtract 1
    if(p % 2 == 0)
        p -= 1;

    int result = (int)(Math.pow(2,p));

        return result;
}

// Function to find the minimum operations
// to make N
static int minStep(int N, int X)
{

   // If N is odd and X = 0, then no
    // valid set exist
    if (N % 2 != 0 && X == 0)
        return -1;

    // Stores the minimum possible size
    // of the valid set
    int size = 0;

    // Loop to subtract highest odd power
    // of 2 while X < N, step 2
    while(X < N){
        N -= highestPowerof2(N);
        size += 1;
     }

    // If N > 0, then increment the value
    // of answer by 1
    if (N != 0)
        size += 1;

    // Return the resultant size of set
    return size;

}

// Driver Code
public static void main (String[] args)
{
    int N = 11;
    int X = 2;
    System.out.println(minStep(N, X));
}
}

// This code is contributed by shivanisinghss2110

Python3

# Python program for the above approach
import math

# Function to find the highest odd power
# of 2 in the range [0, N]
def highestPowerof2(n):

    p = int(math.log(n, 2))

    # If P is even, subtract 1
    if p % 2 == 0:
        p -= 1

    return int(pow(2, p))


# Function to find the minimum operations
# to make N
def minStep(N, X):

    # If N is odd and X = 0, then no
    # valid set exist
    if N % 2 and X == 0:
        return -1

    # Stores the minimum possible size
    # of the valid set
    size = 0

    # Loop to subtract highest odd power
    # of 2 while X < N, step 2
    while X < N:
        N -= highestPowerof2(N)
        size += 1

    # If N > 0, then increment the value
    # of answer by 1
    if N:
        size += 1

    # Return the resultant size of set
    return size


# Driver Code
if __name__ == '__main__':
    N = 11
    X = 2
    print(minStep(N, X))

C#

// C# program for the above approach
using System;

class GFG {

// Function to find the highest odd power
// of 2 in the range [0, N]
static int highestPowerof2(int n)
{

    int p = (int)Math.Floor(Math.Log(n)/Math.Log(2.0));

    // If P is even, subtract 1
    if(p % 2 == 0)
        p -= 1;

    int result = (int)(Math.Pow(2,p));

        return result;
}

// Function to find the minimum operations
// to make N
static int minStep(int N, int X)
{

// If N is odd and X = 0, then no
    // valid set exist
    if (N % 2 != 0 && X == 0)
        return -1;

    // Stores the minimum possible size
    // of the valid set
    int size = 0;

    // Loop to subtract highest odd power
    // of 2 while X < N, step 2
    while(X < N){
        N -= highestPowerof2(N);
        size += 1;
    }

    // If N > 0, then increment the value
    // of answer by 1
    if (N != 0)
        size += 1;

    // Return the resultant size of set
    return size;

}

// Driver Code
public static void Main (String[] args)
{
    int N = 11;
    int X = 2;
    Console.Write(minStep(N, X));
}
}

// This code is contributed by shivanisinghss2110

Javascript

<script>

// JavaScript program for the above approach

// Function to find the highest odd power
// of 2 in the range [0, N]
function highestPowerof2(n)
{
    let p = Math.floor(Math.log2(n));

    // If P is even, subtract 1
    if (p % 2 == 0)
    {
        p -= 1
    }

    return Math.pow(2, p)
}

// Function to find the minimum operations
// to make N
function minStep(N, X)
{

    // If N is odd and X = 0, then no
    // valid set exist
    if (N % 2 != 0 && X == 0)
        return -1

    // Stores the minimum possible size
    // of the valid set
    let size = 0

    // Loop to subtract highest odd power
    // of 2 while X < N, step 2
    while (X < N)
    {
        N -= highestPowerof2(N)
        size += 1
    }

    // If N > 0, then increment the value
    // of answer by 1
    if (N != 0)
        size += 1

    // Return the resultant size of set
    return size;
}

// Driver Code
let N = 11
let X = 2

document.write(minStep(N, X))

// This code is contributed by Potta Lokesh
</script>

Output:

3

Time Complexity: O(log N)
Auxiliary Space: O(1)