Find optimal weights which can be used to weigh all the weights in the range [1, X]

Given an integer X, the task is to find an optimal set of weights {w1, w2, w3, …, wn} such that we can weigh/determine all the weights from 1 to X using a two-sided weighing balance pan. Note that all the weights must be unique and n should be as minimum as possible.

Examples:

Input: X = 7
Output: 1 3 9

Weights Left Side Right Side
1 1 1
2 2 + 1 3
3 3 3
4 4 1 + 3
5 5 + 1 + 3 9
6 6 + 3 9
7 7 + 3 1 + 9

Input: X = 20
Output: 1 3 9 27

Approach:

  1. One optimal approach is to use weights which are a powers of 3 i.e. {1, 3, 9, 27, 81, 243, …}
  2. It can be proved through induction that using {1, 3, 9, 27, 81, …, pow(3, n)}, we can balance all the weights from 1 to (pow(3, n + 1) – 1) / 2.
  3. So, find the values of n such that all the values from 1 to X can be balanced and print the results.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the optimal weights
void findWeights(int X)
{
    int sum = 0;
  
    // Number of weights required
    int power = 0;
  
    // Finding the value of required powers of 3
    while (sum < X) {
        sum = pow(3, power + 1) - 1;
        sum /= 2;
        power++;
    }
  
    // Optimal Weights are powers of 3
    int ans = 1;
    for (int i = 1; i <= power; i++) {
        cout << ans << " ";
        ans = ans * 3;
    }
}
  
// Driver code
int main()
{
    int X = 2;
  
    findWeights(X);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*; 
  
public class GFG 
      
    // Function to find the optimal weights 
    static void findWeights(int X) 
    
        int sum = 0
  
        // Number of weights required 
        int power = 0
        int number = 3;
          
        // Finding the value of required powers of 3 
        while (sum < X)
        
            sum = number - 1
            sum /= 2
            power++;
            number *= 3;
        
  
        // Optimal Weights are powers of 3 
        int ans = 1
        for (int i = 1; i <= power; i++)
        
            System.out.print(ans + " "); 
            ans = ans * 3
        
    
  
      
    // Driver code 
    public static void main (String[] args) 
    
        int X = 2
  
        findWeights(X); 
  
    
  
// This code is contributed by Sam007. 

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Python3

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# Python3 implementation of the approach
  
# Function to find the optimal weights
def findWeights(X):
  
    sum = 0
  
    # Number of weights required
    power = 0
  
    # Finding the value of required powers of 3
    while (sum < X):
        sum = pow(3, power + 1) - 1
        sum //= 2
        power += 1
  
    # Optimal Weights are powers of 3
    ans = 1
    for i in range(1, power + 1):
        print(ans, end = " ")
        ans = ans * 3
  
# Driver code
X = 2
  
findWeights(X)
  
# This code is contributed by Mohit Kumar

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
      
    // Function to find the optimal weights 
    static void findWeights(int X) 
    
        int sum = 0; 
  
        // Number of weights required 
        int power = 0; 
        int number = 3;
          
        // Finding the value of required powers of 3 
        while (sum < X)
        
            sum = number - 1; 
            sum /= 2; 
            power++;
            number *= 3;
        
  
        // Optimal Weights are powers of 3 
        int ans = 1; 
        for (int i = 1; i <= power; i++)
        
            Console.Write(ans + " "); 
            ans = ans * 3; 
        
    
  
    // Driver code 
    static public void Main ()
    {
        int X = 2; 
        findWeights(X); 
    
  
// This code is contributed by ajit.

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Output:

1 3


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Improved By : krikti, mohit kumar 29, jit_t



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