text: Minimize operations to reduce N to 2 by repeatedly reducing by 3 or dividing by 5 | Function to find the minimum number of operations to reduce N to 2 by dividing N by 5 or decrementing by 3 ; Initialize the dp array ; Initialize the array dp [ ] ; For N = 2 number of operations needed is zero ; Iterating over the range [ 1 , N ] ; If it 's not possible to  create current N ; Multiply with 5 ; Adding the value 3 ; Checking if not possible to make the number as 2 ; Return the minimum number of operations ; Driver Code
code: def minimumOperations ( N ) : NEW_LINE INDENT dp = [ 0 for i in range ( N + 1 ) ] NEW_LINE for i in range ( N + 1 ) : NEW_LINE INDENT dp [ i ] = 1000000000 NEW_LINE DEDENT dp [ 2 ] = 0 NEW_LINE for i in range ( 2 , N + 1 , 1 ) : NEW_LINE INDENT if ( dp [ i ] == 1000000000 ) : NEW_LINE INDENT continue NEW_LINE DEDENT if ( i * 5 <= N ) : NEW_LINE INDENT dp [ i * 5 ] = min ( dp [ i * 5 ] , dp [ i ] + 1 ) NEW_LINE DEDENT if ( i + 3 <= N ) : NEW_LINE INDENT dp [ i + 3 ] = min ( dp [ i + 3 ] , dp [ i ] + 1 ) NEW_LINE DEDENT DEDENT if ( dp [ N ] == 1000000000 ) : NEW_LINE INDENT return - 1 NEW_LINE DEDENT return dp [ N ] NEW_LINE DEDENT if __name__ == ' _ _ main _ _ ' : NEW_LINE INDENT N = 25 NEW_LINE print ( minimumOperations ( N ) ) NEW_LINE DEDENT