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Find k numbers which are powers of 2 and have sum N | Set 1

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  • Difficulty Level : Medium
  • Last Updated : 20 Jun, 2022

Given two numbers N and K. The task is to print K numbers which are powers of 2 and their sum is N. Print -1 if not possible. 
Examples: 
 

Input: N = 9, K = 4
Output: 4 2 2 1
4 + 2 + 2 + 1 = 9 

Input: N = 4, K = 5
Output: -1 

 

Approach: The below algorithm can be followed to solve the above problem: 
 

  • If the K is less than the number of set bits in N or more than the number N, then it is not possible.
  • Insert the powers of two at set bits into Priority Queue.
  • Iterate in the Priority Queue till we get K elements, pop() the topmost element and
  • push()                                                                                                                                                                                                       element/2 twice into the Priority Queue again.
  • Once K elements are achieved, print them.

Below is the implementation of the above approach:
 

C++




// CPP program to find k numbers that
// are power of 2 and have sum equal
// to N
#include <bits/stdc++.h>
using namespace std;
 
// function to print numbers
void printNum(int n, int k)
{
    // Count the number of set bits
    int x = __builtin_popcount(n);
 
    // Not-possible condition
    if (k < x || k > n) {
        cout << "-1";
        return;
    }
 
    // Stores the number
    priority_queue<int> pq;
 
    // Get the set bits
    int two = 1;
    while (n) {
        if (n & 1) {
            pq.push(two);
        }
 
        two = two * 2;
        n = n >> 1;
    }
 
    // Iterate till we get K elements
    while (pq.size() < k) {
 
        // Get the topmost element
        int el = pq.top();
        pq.pop();
 
        // Push the elements/2 into
        // priority queue
        pq.push(el / 2);
        pq.push(el / 2);
    }
 
    // Print all elements
    int ind = 0;
    while (ind < k) {
        cout << pq.top() << " ";
        pq.pop();
        ind++;
    }
}
 
// Driver Code
int main()
{
    int n = 9, k = 4;
    printNum(n, k);
    return 0;
}


Java




// Java program to find k numbers that
// are power of 2 and have sum equal
// to N
import java.io.*;
import java.util.*;
 
class GFG
{
 
    // function to print numbers
    static void printNum(int n, int k)
    {
 
        // Count the number of set bits
        String str = Integer.toBinaryString(n);
        int x = 0;
        for (int i = 0; i < str.length(); i++)
            if (str.charAt(i) == '1')
                x++;
 
        // Not-possible condition
        if (k < x || k > n)
        {
            System.out.println("-1");
            return;
        }
 
        // Stores the number
        PriorityQueue<Integer> pq =
        new PriorityQueue<>(Comparator.reverseOrder());
 
        // Get the set bits
        int two = 1;
        while (n > 0)
        {
            if ((n & 1) == 1)
                pq.add(two);
            two *= 2;
            n = n >> 1;
        }
 
        // Iterate till we get K elements
        while (pq.size() < k)
        {
 
            // Get the topmost element
            int el = pq.poll();
 
            // Push the elements/2 into
            // priority queue
            pq.add(el / 2);
            pq.add(el / 2);
        }
 
        // Print all elements
        int ind = 0;
        while (ind < k)
        {
            System.out.print(pq.poll() + " ");
            ind++;
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 9, k = 4;
        printNum(n, k);
    }
}
 
// This code is contributed by
// sanjeev2552


Python




# Python program to find k numbers that
# are power of 2 and have sum equal
# to N
 
# function to print numbers
def printNum(n, k):
     
    # Count the number of set bits
    x = 0
    m = n
    while (m):
        x += m & 1
        m >>= 1
     
    # Not-possible condition
    if k < x or k > n:
        print("-1")
        return
     
    # Stores the number
    pq = []
     
    # Get the set bits
    two = 1
    while (n):
        if (n & 1):
            pq.append(two)
             
        two = two * 2
        n = n >> 1
         
    # Iterate till we get K elements
    while (len(pq) < k):
     
        # Get the topmost element
        el = pq[-1]
        pq.pop()
 
        # append the elements/2 into
        # priority queue
        pq.append(el // 2)
        pq.append(el // 2)
         
    # Print all elements
    ind = 0
    pq.sort()
    while (ind < k):
        print(pq[-1], end = " ")
        pq.pop()
        ind += 1
 
# Driver Code
n = 9
k = 4
printNum(n, k)
 
# This code is contributed by SHUBHAMSINGH10


C#




// C# program to find k numbers that
// are power of 2 and have sum equal
// to N
 
using System;
using System.Collections.Generic;
 
public class GFG {
 
  // function to print numbers
  static void printNum(int n, int k)
  {
 
    // Count the number of set bits
    int x = 0;
    int m = n;
    while (m > 0) {
      x += m & 1;
      m >>= 1;
    }
 
    // Not-possible condition
    if (k < x || k > n) {
      Console.WriteLine("-1");
      return;
    }
 
    // Stores the number
    List<int> pq = new List<int>();
 
    // Get the set bits
    int two = 1;
    while (n > 0) {
      if ((n & 1) != 0)
        pq.Add(two);
 
      two = two * 2;
      n = n >> 1;
    }
 
    // Iterate till we get K elements
    while (pq.Count < k) {
 
      // Get the topmost element
      int el = pq[pq.Count - 1];
      pq.RemoveAt(pq.Count - 1);
 
      // append the elements/2 into
      // priority queue
      pq.Add(el / 2);
      pq.Add(el / 2);
    }
 
    // Print all elements
    int ind = 0;
    pq.Sort();
    while (ind < k) {
      Console.Write(pq[pq.Count - 1] + " ");
      pq.RemoveAt(pq.Count - 1);
      ind += 1;
    }
  }
 
  // Driver code
  public static void Main(string[] args)
  {
    int n = 9;
    int k = 4;
    printNum(n, k);
  }
}
 
// This code is contributed by phasing17


Javascript




<script>
 
// JavaScript program to find k numbers that
// are power of 2 and have sum equal
// to N
 
// function to print numbers
function printNum(n, k)
{
     
    // Count the number of set bits
    let x = 0
    let m = n
    while (m){
        x += m & 1
        m >>= 1
    }
     
    // Not-possible condition
    if(k < x || k > n){
        document.write("-1","</br>")
        return
    }
     
    // Stores the number
    let pq = []
     
    // Get the set bits
    let two = 1
    while (n){
        if (n & 1)
            pq.push(two)
             
        two = two * 2
        n = n >> 1
    }
         
    // Iterate till we get K elements
    while (pq.length < k){
     
        // Get the topmost element
        let el = pq[pq.length -1]
        pq.pop()
 
        // push the elements/2 into
        // priority queue
        pq.push(Math.floor(el / 2))
        pq.push(Math.floor(el / 2))
    }
         
    // Print all elements
    let ind = 0
    pq.sort()
    while (ind < k){
        document.write(pq[pq.length -1]," ")
        pq.pop()
        ind += 1
    }
}
 
// Driver Code
let n = 9
let k = 4
printNum(n, k)
 
// This code is contributed by shinjanpatra
 
</script>


Output: 

4 2 2 1

 

Time Complexity: O(N*logN), as we are using a loop to traverse N times and in each traversal we are using priority queue operation which will cost logN time.

Auxiliary Space: O(N), as we are using extra space for the priority queue.

Represent n as the sum of exactly k powers of two | Set 2
 


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