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

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
```

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

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:

 `// CPP program to find k numbers that  ` `// are power of 2 and have sum equal  ` `// to N ` `#include ` `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 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 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 program to find k numbers that  ` `# are power of 2 and have sum equal  ` `# to N  ` ` `  `# function to prnumbers  ` `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``) ` `         `  `    ``# Prall 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 `

Output:
```4 2 2 1
```

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

Striver(underscore)79 at Codechef and codeforces D

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Article Tags :
Practice Tags :