Break the number N in odd powers of 2
Given a positive integer N. The task is to print the array in decreasing order in which the numbers are odd powers of 2 and the sum of all the numbers in the array is N and the size of the array should be minimum if it is not possible to form the array then print -1.
Examples:
Input: 18
Output: 8 8 2
Explanation: Array with sum 18 and which consists
minimum numbers with odd powers of 2 are [8 8 2] .
Since 8+8+2=19 and 21=2, 23=8 where 1, 3 are odd numbers.Input: 7
Output: -1
Explanation: Since there is no array with sum equal to 7.
Approach: To solve the problem follow the below idea:
The idea is to create a binary array of size (32) and check the set bit position.If the bit position is odd, store the power of position else store power of (position-1).
Follow the below steps to solve the problem:
- Create a binary array of max size 32 which will store the number’s binary.
- If the number is odd then print -1.
- If the number is even traverse the binary array from the reverse side and check if the bit is 1 and the position of the bit is odd then take its power of 2 and store it in the answer array.
- If the bit is 1 and the position of the bit is even then take the 2 instances of 2^(position-1).
- If the bit is 0 then continue traversing the binary array.
- Print the answer array
Below is the Implementation of the above approach.
C++
// Code for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate power of 2 .
int power(int x, int y)
{
if (y == 0)
return 1;
else if (y % 2 == 0)
return power(x, y / 2) * power(x, y / 2);
else
return x * power(x, y / 2) * power(x, y / 2);
}
// Function performing Calculation
vector<int> solve(int N)
{
vector<int> binary(32);
vector<int> answer;
// Storing the binary of the number
// N in the binary array.
for (int i = 31; i >= 0; i--) {
// Checking if the bit is set
// or not
if ((N & (1 << i)) != 0)
binary[i] = 1;
}
// Storing power of 2 in the answer
// array
for (int i = 31; i >= 0; i--) {
// If the bit is set and is at odd
// position the store the power of
// 2 of i.
if (binary[i] == 1 && i & 1) {
answer.push_back(power(2, i));
}
// If the bit is set and is at
// even position
// then store 2 numbers with
// the power of 2 of i-1.
else if (binary[i] == 1 && !(i & 1)) {
answer.push_back(power(2, i - 1));
answer.push_back(power(2, i - 1));
}
else
continue;
}
return answer;
}
// Driver Function
int main()
{
int N = 18;
// Checking if the number is odd
if (N & 1) {
cout << -1 << endl;
return 0;
}
// Function call
vector<int> ans = solve(N);
for (int x : ans)
cout << x << " ";
return 0;
}
Java
// Java Code for the above approach
import java.io.*;
import java.util.*;
class GFG
{
// Function to calculate power of 2 .
public static int power(int x, int y)
{
if (y == 0)
return 1;
else if (y % 2 == 0)
return power(x, y / 2) * power(x, y / 2);
else
return x * power(x, y / 2) * power(x, y / 2);
}
// Function performing Calculation
public static ArrayList<Integer> solve(int N)
{
int binary[] = new int[32];
ArrayList<Integer> answer
= new ArrayList<Integer>();
// Storing the binary of the number
// N in the binary array.
for (int i = 31; i >= 0; i--) {
// Checking if the bit is set
// or not
if ((N & (1 << i)) != 0)
binary[i] = 1;
}
// Storing power of 2 in the answer
// array
for (int i = 31; i >= 0; i--) {
// If the bit is set and is at odd
// position the store the power of
// 2 of i.
if (binary[i] == 1 && (i & 1) != 0) {
answer.add(power(2, i));
}
// If the bit is set and is at
// even position
// then store 2 numbers with
// the power of 2 of i-1.
else if (binary[i] == 1 && (i & 1) == 0) {
answer.add(power(2, i - 1));
answer.add(power(2, i - 1));
}
else
continue;
}
return answer;
}
// Driver Code
public static void main(String[] args)
{
int N = 18;
// Checking if the number is odd
if ((N & 1) != 0) {
System.out.println(-1);
return;
}
// Function call
ArrayList<Integer> ans = solve(N);
for (Integer x : ans)
System.out.print(x + " ");
}
}
// This code is contributed by Rohit Pradhan
Python3
# Python code for the above approach
import math
# Function to calculate power of 2 .
def power( x, y):
if (y == 0):
return 1;
elif (y % 2 == 0):
return power(x, math.floor(y / 2)) * power(x, math.floor(y / 2));
else:
return x * power(x, math.floor(y / 2)) * power(x, math.floor(y / 2));
# Function performing Calculation
def solve( N):
binary = [0]*(32);
answer =[];
# Storing the binary of the number
# N in the binary array.
for i in range(31,0,-1) :
# Checking if the bit is set
# or not
if ((N & (1 << i)) != 0):
binary[i] = 1;
# Storing power of 2 in the answer
# array
for i in range(31,0,-1):
# If the bit is set and is at odd
# position the store the power of
# 2 of i.
if binary[i] == 1 and i & 1:
answer.append(power(2, i));
# If the bit is set and is at
# even position
# then store 2 numbers with
# the power of 2 of i-1.
elif binary[i] == 1 and (i & 1)==0:
answer.append(power(2, i - 1));
answer.append(power(2, i - 1));
else:
continue;
return answer;
# Driver Function
N = 18;
# Checking if the number is odd
if (N & 1):
print(-1)
# Function call
ans = solve(N);
for x in ans:
print(x,end=" ");
# This code is contributed by Potta Lokesh
C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to calculate power of 2 .
public static int power(int x, int y)
{
if (y == 0)
return 1;
else if (y % 2 == 0)
return power(x, y / 2) * power(x, y / 2);
else
return x * power(x, y / 2) * power(x, y / 2);
}
// Function performing Calculation
public static List<int> solve(int N)
{
int[] binary = new int[32];
List<int> answer
= new List<int>();
// Storing the binary of the number
// N in the binary array.
for (int i = 31; i >= 0; i--) {
// Checking if the bit is set
// or not
if ((N & (1 << i)) != 0)
binary[i] = 1;
}
// Storing power of 2 in the answer
// array
for (int i = 31; i >= 0; i--) {
// If the bit is set and is at odd
// position the store the power of
// 2 of i.
if (binary[i] == 1 && (i & 1) != 0) {
answer.Add(power(2, i));
}
// If the bit is set and is at
// even position
// then store 2 numbers with
// the power of 2 of i-1.
else if (binary[i] == 1 && (i & 1) == 0) {
answer.Add(power(2, i - 1));
answer.Add(power(2, i - 1));
}
else
continue;
}
return answer;
}
// Driver Code
public static void Main()
{
int N = 18;
// Checking if the number is odd
if ((N & 1) != 0) {
Console.Write(-1);
return;
}
// Function call
List<int> ans = solve(N);
foreach (int x in ans)
Console.Write(x + " ");
}
}
// This code is contributed by sanjoy_62.
Javascript
<script>
// Code for the above approach
// Function to calculate power of 2 .
const power = (x, y) => {
if (y == 0)
return 1;
else if (y % 2 == 0)
return power(x, parseInt(y / 2)) * power(x, parseInt(y / 2));
else
return x * power(x, parseInt(y / 2)) * power(x, parseInt(y / 2));
}
// Function performing Calculation
const solve = (N) => {
let binary = new Array(32).fill(0);
let answer = [];
// Storing the binary of the number
// N in the binary array.
for (let i = 31; i >= 0; i--) {
// Checking if the bit is set
// or not
if ((N & (1 << i)) != 0)
binary[i] = 1;
}
// Storing power of 2 in the answer
// array
for (let i = 31; i >= 0; i--) {
// If the bit is set and is at odd
// position the store the power of
// 2 of i.
if (binary[i] == 1 && i & 1) {
answer.push(power(2, i));
}
// If the bit is set and is at
// even position
// then store 2 numbers with
// the power of 2 of i-1.
else if (binary[i] == 1 && !(i & 1)) {
answer.push(power(2, i - 1));
answer.push(power(2, i - 1));
}
else
continue;
}
return answer;
}
// Driver Function
let N = 18;
// Checking if the number is odd
if (N & 1) {
document.write("-1<br/>");
}
// Function call
let ans = solve(N);
for (let x in ans)
document.write(`${ans[x]} `);
// This code is contributed by rakeshsahni
</script>
Output
8 8 2
Time Complexity: O(N*log N) where in the worst case N is 32.
Auxiliary Space: O(N) where N is 32.
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