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Suppose we have two numbers N and K. The task is to print K numbers, which are the power of 2 and their sum is N. If it is not possible, then return -1. Suppose N = 9 and K = 4, then the output will be 4 2 2 1, whose sum is 9, and a number of elements is 4, and each of them is a power of 2.

We have to follow these steps to solve this problem −

If k is less than the number of set bits in N or more than the number N, then return -1

Add the powers of two at set bits into the Priority queue

Initiate the priority queue till we get K elements, then remove the element from the priority queue

Insert the removed element/2 twice into the priority queue again

If k elements are achieved, then print them.

#include<iostream> #include<algorithm> #include<queue> using namespace std; void displayKnumbers(int n, int k) { int set_bit_count = __builtin_popcount(n); if (k < set_bit_count || k > n) { cout << "-1"; return; } priority_queue<int> queue; int two = 1; while (n) { if (n & 1) { queue.push(two); } two = two * 2; n = n >> 1; } while (queue.size() < k) { int element = queue.top(); queue.pop(); queue.push(element / 2); queue.push(element / 2); } int ind = 0; while (ind < k) { cout << queue.top() << " "; queue.pop(); ind++; } } int main() { int n = 30, k = 5; cout << "Numbers are: "; displayKnumbers(n, k); }

Numbers are: 8 8 8 4 2

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