Minimum number of given powers of 2 required to represent a number
Given an integer x and an array arr[] each element of which is a power of 2. The task is to find the minimum number of integer powers of 2 from the array which when added give x. If it is not possible to represent x with the given array elements then print -1.
Examples:
Input: arr[] = {2, 4, 8, 2, 4}, x = 14
Output: 3
14 can be written as 8 + 4 + 2
Input: arr[] = {2, 4, 8, 2, 4}, x = 5
Output: -1
5 cannot be represented as the sum any elements from the given array.
Approach: For each power of 2 let’s calculate the number of elements in the given array with the value equals this. Let’s call it cnt. It is obvious that we can obtain the value x greedily (because all fewer values of elements are divisors of all greater values of elements).
Now let’s iterate over all powers of 2 from 30 to 0. Let’s deg be the current degree. We can take min(x / 2deg, cntdeg) elements with the value equals 2deg. Let it be cur. Add cur to the answer and subtract 2deg * cur from x. Repeat the process until the x can no longer be reduced. If after iterating over all powers, x is still non-zero then print -1. Otherwise, print the answer.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the minimum number// of given integer powers of 2 required// to represent a number as sum of these powersint power_of_two(int n, int a[], int x){ // To store the count of powers of two vector<int> cnt(31); for (int i = 0; i < n; ++i) { // __builtin_ctz(a[i]) returns the count // of trailing 0s in a[i] ++cnt[__builtin_ctz(a[i])]; } int ans = 0; for (int i = 30; i >= 0 && x > 0; --i) { // If current power is available // in the array and can be used int need = min(x >> i, cnt[i]); // Update the answer ans += need; // Reduce the number x -= (1 << i) * need; } // If the original number is not reduced to 0 // It cannot be represented as the sum // of the given powers of 2 if (x > 0) ans = -1; return ans;}// Driver codeint main(){ int arr[] = { 2, 2, 4, 4, 8 }, x = 6; int n = sizeof(arr) / sizeof(arr[0]); cout << power_of_two(n, arr, x); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG{ // __builtin_ctz(a[i]) returns the count// of trailing 0s in a[i]static int __builtin_ctz(int a){ int count = 0; for(int i = 0; i < 40; i++) if(((a >> i) & 1) == 0) { count++; } else break; return count;}// Function to return the minimum number// of given integer powers of 2 required// to represent a number as sum of these powersstatic int power_of_two(int n, int a[], int x){ // To store the count of powers of two Vector<Integer> cnt = new Vector<Integer>(); for (int i = 0; i < 31; ++i) cnt.add(0); for (int i = 0; i < n; ++i) { // __builtin_ctz(a[i]) returns the count // of trailing 0s in a[i] cnt.set(__builtin_ctz(a[i]), (cnt.get(__builtin_ctz(a[i]))==null) ? 1 : cnt.get(__builtin_ctz(a[i]))+1); } int ans = 0; for (int i = 30; i >= 0 && x > 0; --i) { // If current power is available // in the array and can be used int need = Math.min(x >> i, cnt.get(i)); // Update the answer ans += need; // Reduce the number x -= (1 << i) * need; } // If the original number is not reduced to 0 // It cannot be represented as the sum // of the given powers of 2 if (x > 0) ans = -1; return ans;}// Driver codepublic static void main(String args[]){ int arr[] = { 2, 2, 4, 4, 8 }, x = 6; int n = arr.length; System.out.println(power_of_two(n, arr, x));}}// This code is contributed by Arnab Kundu |
python
# Python3 implementation of the approach# Function to return the minimum number# of given eger powers of 2 required# to represent a number as sum of these powersdef power_of_two( n, a, x): # To store the count of powers of two cnt=[0 for i in range(31)] for i in range(n): # __builtin_ctz(a[i]) returns the count # of trailing 0s in a[i] count = 0 xx = a[i] while ((xx & 1) == 0): xx = xx >> 1 count += 1 cnt[count]+=1 ans = 0 for i in range(30,-1,-1): if x<=0: continue # If current power is available # in the array and can be used need = min(x >> i, cnt[i]) # Update the answer ans += need # Reduce the number x -= (1 << i) * need # If the original number is not reduced to 0 # It cannot be represented as the sum # of the given powers of 2 if (x > 0): ans = -1 return ans# Driver codearr=[2, 2, 4, 4, 8 ]x = 6n = len(arr)print(power_of_two(n, arr, x))# This code is contributed by mohit kumar 29 |
C#
// C# implementation of the approachusing System;class GFG{ // __builtin_ctz(a[i]) returns the count// of trailing 0s in a[i]static int __builtin_ctz(int a){ int count = 0; for(int i = 0; i < 40; i++) if(((a >> i) & 1) == 0) { count++; } else break; return count;}// Function to return the minimum number// of given integer powers of 2 required// to represent a number as sum of these powersstatic int power_of_two(int n, int []a, int x){ // To store the count of powers of two int[] cnt = new int[32]; for (int i = 0; i < n; ++i) { // __builtin_ctz(a[i]) returns the count // of trailing 0s in a[i] cnt[__builtin_ctz(a[i])] = cnt[__builtin_ctz(a[i])] == 0?1 : cnt[__builtin_ctz(a[i])] + 1; } int ans = 0; for (int i = 30; i >= 0 && x > 0; --i) { // If current power is available // in the array and can be used int need = Math.Min(x >> i, cnt[i]); // Update the answer ans += need; // Reduce the number x -= (1 << i) * need; } // If the original number is not reduced to 0 // It cannot be represented as the sum // of the given powers of 2 if (x > 0) ans = -1; return ans;}// Driver codestatic void Main(){ int []arr = { 2, 2, 4, 4, 8 }; int x = 6; int n = arr.Length; Console.WriteLine(power_of_two(n, arr, x));}}// This code is contributed by mits |
Javascript
<script>// JavaScript implementation of the approach// Function to return the minimum number// of given integer powers of 2 required// to represent a number as sum of these powersfunction power_of_two( n, a, x){ // To store the count of powers of two let cnt = []; for(let i = 0;i<31;i++) cnt.push(0); for (let i = 0; i < n; ++i) { // __builtin_ctz(a[i]) returns the count // of trailing 0s in a[i] let count = 0; let xx = a[i]; while ((xx & 1) == 0){ xx = xx >> 1 count += 1 } cnt[count]+=1; } let ans = 0; for (let i = 30; i >= 0 && x > 0; --i) { // If current power is available // in the array and can be used let need = Math.min(x >> i, cnt[i]); // Update the answer ans += need; // Reduce the number x -= (1 << i) * need; } // If the original number is not reduced to 0 // It cannot be represented as the sum // of the given powers of 2 if (x > 0) ans = -1; return ans;}// Driver codelet arr = [ 2, 2, 4, 4, 8 ], x = 6;let n = arr.length;document.write( power_of_two(n, arr, x));</script> |
Output
2
Time Complexity: O(N)
Auxiliary Space: O(32)
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