Given two integers n and k, the task is to find whether it is possible to represent n as the sum of exactly k powers of 2. If possible then print k positive integers such that they are powers of 2 and their sum is exactly equal to n else print Impossible.
Examples:
Input: n = 9, k = 4
Output: 1 2 2 4
1, 2 and 4 are all powers of 2 and 1 + 2 + 2 + 4 = 9.Input: n = 3, k = 7
Output: Impossible
It is impossible since 3 cannot be represented as sum of 7 numbers which are powers of 2.
We have discussed one approach to solve this problem in Find k numbers which are powers of 2 and have sum N. In this post, a different approach is being discussed.
Approach:
- Create an array arr[] of size k with all elements initialized to 1 and create a variable sum = k.
- Now starting from the last element of arr[]
- If sum + arr[i] ? n then update sum = sum + arr[i] and arr[i] = arr[i] * 2.
- Else skip the current element.
- If sum = n then the contents of arr[] are the required elements.
- Else it is impossible to represent n as exactly k powers of 2.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <iostream>using namespace std;// Function to print k numbers which are powers of two// and whose sum is equal to nvoid FindAllElements(int n, int k){ // Initialising the sum with k int sum = k; // Initialising an array A with k elements // and filling all elements with 1 int A[k]; fill(A, A + k, 1); for (int i = k - 1; i >= 0; --i) { // Iterating A[] from k-1 to 0 while (sum + A[i] <= n) { // Update sum and A[i] // till sum + A[i] is less than equal to n sum += A[i]; A[i] *= 2; } } // Impossible to find the combination if (sum != n) { cout << "Impossible"; } // Possible solution is stored in A[] else { for (int i = 0; i < k; ++i) cout << A[i] << ' '; }}// Driver codeint main(){ int n = 12; int k = 6; FindAllElements(n, k); return 0;} |
Java
// Java implementation of the above approach import java.util.Arrays;public class GfG { // Function to print k numbers which are powers of two // and whose sum is equal to n public static void FindAllElements(int n, int k) { // Initialising the sum with k int sum = k; // Initialising an array A with k elements // and filling all elements with 1 int[] A = new int[k]; Arrays.fill(A, 0, k, 1); for (int i = k - 1; i >= 0; --i) { // Iterating A[] from k-1 to 0 while (sum + A[i] <= n) { // Update sum and A[i] // till sum + A[i] is less than equal to n sum += A[i]; A[i] *= 2; } } // Impossible to find the combination if (sum != n) { System.out.print("Impossible"); } // Possible solution is stored in A[] else { for (int i = 0; i < k; ++i) System.out.print(A[i] + " "); } } public static void main(String []args){ int n = 12; int k = 6; FindAllElements(n, k); }} // This code is contributed by Rituraj Jain |
Python3
# Python 3 implementation of the above approach# Function to print k numbers which are # powers of two and whose sum is equal to ndef FindAllElements(n, k): # Initialising the sum with k sum = k # Initialising an array A with k elements # and filling all elements with 1 A = [1 for i in range(k)] i = k - 1 while(i >= 0): # Iterating A[] from k-1 to 0 while (sum + A[i] <= n): # Update sum and A[i] till # sum + A[i] is less than equal to n sum += A[i] A[i] *= 2 i -= 1 # Impossible to find the combination if (sum != n): print("Impossible") # Possible solution is stored in A[] else: for i in range(0, k, 1): print(A[i], end = ' ')# Driver codeif __name__ == '__main__': n = 12 k = 6 FindAllElements(n, k)# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the above approach using System;class GfG { // Function to print k numbers // which are powers of two // and whose sum is equal to n public static void FindAllElements(int n, int k) { // Initialising the sum with k int sum = k; // Initialising an array A with k elements // and filling all elements with 1 int[] A = new int[k]; for(int i = 0; i < k; i++) A[i] = 1; for (int i = k - 1; i >= 0; --i) { // Iterating A[] from k-1 to 0 while (sum + A[i] <= n) { // Update sum and A[i] // till sum + A[i] is less than equal to n sum += A[i]; A[i] *= 2; } } // Impossible to find the combination if (sum != n) { Console.Write("Impossible"); } // Possible solution is stored in A[] else { for (int i = 0; i < k; ++i) Console.Write(A[i] + " "); } } // Driver code public static void Main(String []args) { int n = 12; int k = 6; FindAllElements(n, k); }}// This code contributed by Rajput-Ji |
PHP
<?php// PHP implementation of the above approach // Function to print k numbers which are // powers of two and whose sum is equal to n function FindAllElements($n, $k) { // Initialising the sum with k $sum = $k; // Initialising an array A with k elements // and filling all elements with 1 $A = array_fill(0, $k, 1) ; for ($i = $k - 1; $i >= 0; --$i) { // Iterating A[] from k-1 to 0 while ($sum + $A[$i] <= $n) { // Update sum and A[i] till // sum + A[i] is less than equal to n $sum += $A[$i]; $A[$i] *= 2; } } // Impossible to find the combination if ($sum != $n) { echo"Impossible"; } // Possible solution is stored in A[] else { for ($i = 0; $i < $k; ++$i) echo $A[$i], ' '; } } // Driver code $n = 12; $k = 6; FindAllElements($n, $k); // This code is contributed by Ryuga?> |
Javascript
<script>// Javascript implementation of the above approach // Function to print k numbers which are powers of two// and whose sum is equal to nfunction FindAllElements( n, k){ // Initialising the sum with k let sum = k; // Initialising an array A with k elements // and filling all elements with 1 let A = new Array(k).fill(1); for (let i = k - 1; i >= 0; --i) { // Iterating A[] from k-1 to 0 while (sum + A[i] <= n) { // Update sum and A[i] // till sum + A[i] is less than equal to n sum += A[i]; A[i] *= 2; } } // Impossible to find the combination if (sum != n) { document.write("Impossible"); } // Possible solution is stored in A[] else { for (let i = 0; i < k; ++i) document.write(A[i] + " "); }} // Driver Codelet n = 12;let k = 6; FindAllElements(n, k);</script> |
Output
1 1 1 1 4 4
Complexity Analysis:
- Time Complexity: O(k*log2(n-k))
- Auxiliary Space: O(k), since k extra space has been taken.
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