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Check if N can be represented as sum of distinct powers of 3

  • Last Updated : 29 Jul, 2021

Given a positive integer N, the task is to check whether the given number N can be represented as the sum of the distinct powers of 3. If found to be true, then print “Yes”. Otherwise, “No”.

Examples:

Input: N = 28
Output: Yes
Explanation:
The number N(= 28) can be represented (1 + 7) = (30 + 33), which is a perfect power 2.

Input: N = 6
Output: No

Approach: The simplest approach to solve the given problem is to generate all possible permutations of all distinct powers of 3 and if there exists any such combination whose sum is a perfect power of 3. As 315 > 107 so there are only 16 distinct powers i.e., [0, 15].



Below is the implementation of the above approach:

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to try all permutations
// of distinct powers
bool PermuteAndFind(vector<long> power, int idx,
                    long SumSoFar, int target)
{
    // Base Case
    if (idx == power.size()) {
 
        // If the distinct powers
        // sum is obtained
        if (SumSoFar == target)
            return true;
 
        // Otherwise
        return false;
    }
 
    // If current element not selected
    // in power[]
    bool select
        = PermuteAndFind(power, idx + 1, SumSoFar, target);
 
    // If current element selected in
    // power[]
    bool notselect = PermuteAndFind(
        power, idx + 1, SumSoFar + power[idx], target);
 
    // Return 1 if any permutation
    // found
    return (select || notselect);
}
 
// Function to check the N can be
// represented as the sum of the
// distinct powers of 3
void DistinctPowersOf3(int N)
{
    // Stores the all distincts powers
    // of three to [0, 15]
    vector<long> power(16);
    power[0] = 1;
    for (int i = 1; i < 16; i++)
        power[i] = 3 * power[i - 1];
 
    // Function Call
    bool found = PermuteAndFind(power, 0, 0L, N);
 
    // print
    if (found == true) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}
 
// Driven Code
int main()
{
    int N = 91;
    DistinctPowersOf3(N);
    return 0;
}

Java




// Java program for the above approach
class GFG
{
   
    // Function to try all permutations
    // of distinct powers
    public static boolean PermuteAndFind(int[] power, int idx,
                                         int SumSoFar, int target)
    {
        // Base Case
        if (idx == power.length)
        {
 
            // If the distinct powers
            // sum is obtained
            if (SumSoFar == target)
                return true;
 
            // Otherwise
            return false;
        }
 
        // If current element not selected
        // in power[]
        boolean select = PermuteAndFind(power, idx + 1, SumSoFar, target);
 
        // If current element selected in
        // power[]
        boolean notselect = PermuteAndFind(power, idx + 1, SumSoFar + power[idx], target);
 
        // Return 1 if any permutation
        // found
        return (select || notselect);
    }
 
    // Function to check the N can be
    // represented as the sum of the
    // distinct powers of 3
    public static void DistinctPowersOf3(int N)
    {
       
        // Stores the all distincts powers
        // of three to [0, 15]
        int[] power = new int[16];
        power[0] = 1;
        for (int i = 1; i < 16; i++)
            power[i] = 3 * power[i - 1];
 
        // Function Call
        boolean found = PermuteAndFind(power, 0, 0, N);
 
        // print
        if (found == true) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
 
    // Driven Code
    public static void main(String args[]) {
        int N = 91;
        DistinctPowersOf3(N);
    }
}
 
// This code is contributed by _saurabh_jaiswal.

Python3




# Python3 program for the above approach
 
# Function to try all permutations
# of distinct powers
def PermuteAndFind(power, idx, SumSoFar, target):
     
    # Base Case
    if (idx == len(power)):
         
        # If the distinct powers
        # sum is obtained
        if (SumSoFar == target):
            return True
             
        # Otherwise
        return False
 
    # If current element not selected
    # in power[]
    select = PermuteAndFind(power, idx + 1,
                            SumSoFar, target)
 
    # If current element selected in
    # power[]
    notselect = PermuteAndFind(power, idx + 1,
                                 SumSoFar + power[idx],
                                 target)
 
    # Return 1 if any permutation
    # found
    return(select or notselect)
 
# Function to check the N can be
# represented as the sum of the
# distinct powers of 3
def DistinctPowersOf3(N):
     
    # Stores the all distincts powers
    # of three to[0, 15]
    power = [0 for x in range(16)]
    power[0] = 1
     
    for i in range(1, 16):
        power[i] = 3 * power[i - 1]
 
    # Function Call
    found = PermuteAndFind(power, 0, 0, N)
 
    # Print
    if (found == True):
        print("Yes")
    else:
        print("No")
 
# Driver Code
N = 91
 
DistinctPowersOf3(N)
 
# This code is contributed by amreshkumar3

Javascript




<script>
// Javascript program for the above approach
 
// Function to try all permutations
// of distinct powers
function PermuteAndFind(power, idx, SumSoFar, target) {
  // Base Case
  if (idx == power.length) {
    // If the distinct powers
    // sum is obtained
    if (SumSoFar == target) return true;
 
    // Otherwise
    return false;
  }
 
  // If current element not selected
  // in power[]
  let select = PermuteAndFind(power, idx + 1, SumSoFar, target);
 
  // If current element selected in
  // power[]
  let notselect = PermuteAndFind(power, idx + 1, SumSoFar + power[idx], target);
 
  // Return 1 if any permutation
  // found
  return select || notselect;
}
 
// Function to check the N can be
// represented as the sum of the
// distinct powers of 3
function DistinctPowersOf3(N) {
  // Stores the all distincts powers
  // of three to [0, 15]
  let power = new Array(16);
  power[0] = 1;
  for (let i = 1; i < 16; i++) power[i] = 3 * power[i - 1];
 
  // Function Call
  let found = PermuteAndFind(power, 0, 0, N);
 
  // print
  if (found == true) {
    document.write("Yes");
  } else {
    document.write("No");
  }
}
 
// Driven Code
 
let N = 91;
DistinctPowersOf3(N);
 
</script>
Output
Yes

Time Complexity: O(2N)
Auxiliary Space: O(2N)

Another Approach: The above approach can also be optimized by observing the fact that N in ternary form (Base 3). Each digit is 3i, and the ternary number(N) is the sum of them.

To have distinct powers of 3, to sum up to N, in ternary form the ith digit can be either 0,1 or 2(In Binary, it is 0 and 1). Thus, there can be three cases for each digit at ith position: 

  • Digit can be 0 i.e. there is No 3i number in N.
  • Digit can be 1 i.e, there is One 3i number in N.
  • Digit cannot be 2 because then there are 2 of 3i, therefore, not distinct.

Follow the below steps to solve the problem: 

  • Iterate until N becomes 0 and perform the following steps:
    • If the value of N%3 is 2, then print “No”.
    • Otherwise, divide N by 3.
  • After completing the above steps, if the value of N is 0 then print “Yes”. Otherwise, “No”.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check whether the given
// N can be represented as the sum of
// the distinct powers of 3
void DistinctPowersOf3(int N)
{
    // Iterate until N is non-zero
    while (N > 0) {
 
        // Termination Condition
        if (N % 3 == 2) {
            cout << "No";
            return;
        }
 
        // Right shift ternary bits
        // by 1 for the next digit
        N /= 3;
    }
 
    // If N can be expressed as the
    // sum of perfect powers of 3
    cout << "Yes";
}
 
// Driver Code
int main()
{
    int N = 91;
 
    DistinctPowersOf3(N);
 
    return 0;
}

Java




// Java program for the above approach
class GFG
{
   
// Function to check whether the given
// N can be represented as the sum of
// the distinct powers of 3
public static void DistinctPowersOf3(int N)
{
   
    // Iterate until N is non-zero
    while (N > 0) {
 
        // Termination Condition
        if (N % 3 == 2) {
            System.out.println("No");
            return;
        }
 
        // Right shift ternary bits
        // by 1 for the next digit
        N /= 3;
    }
 
    // If N can be expressed as the
    // sum of perfect powers of 3
    System.out.println("Yes");
}
 
// Driver Code
public static void main(String args[])
{
    int N = 91;
 
    DistinctPowersOf3(N);
}
}
 
// This code is contributed by _saurabh_jaiswal.

Python3




# Python3 program for the above approach
 
# Function to check whether the given
# N can be represented as the sum of
# the distinct powers of 3
def DistinctPowersOf3(N):
   
    # Iterate until N is non-zero
    while (N > 0):
 
        # Termination Condition
        if (N % 3 == 2):
            cout << "No"
            return
 
        # Right shift ternary bits
        # by 1 for the next digit
        N //= 3
 
    # If N can be expressed as the
    # sum of perfect powers of 3
    print ("Yes")
 
# Driver Code
if __name__ == '__main__':
    N = 91
 
    DistinctPowersOf3(N)
 
# This code is contributed by mohit kumar 29.

Javascript




<script>
// Javascript program for the above approach
 
// Function to check whether the given
// N can be represented as the sum of
// the distinct powers of 3
function DistinctPowersOf3(N)
{
 
    // Iterate until N is non-zero
    while (N > 0) {
  
        // Termination Condition
        if (N % 3 == 2)
        {
            document.write("No");
            return;
        }
  
        // Right shift ternary bits
        // by 1 for the next digit
        N = Math.floor(N/ 3);
    }
  
    // If N can be expressed as the
    // sum of perfect powers of 3
    document.write("Yes");
}
 
// Driver Code
let N = 91;
DistinctPowersOf3(N);
 
// This code is contributed by patel2127
</script>
Output
Yes

Time Complexity: O(log3N)
Auxiliary Space: O(1)

 

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