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 numbersvoid 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 Codeint 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 = 9k = 4printNum(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 Nusing 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 = 9let k = 4printNum(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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