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