Writing a clean/cleaner solution to "Valid Anagram" in Elixir - functional-programming

Question

Trying to level up my Elixir understanding by doing algo/leetCode style problems using Elixir.
As I'm a relatively new programmer (around a year in) and was trained on traditionally OOP languages like Ruby and JS, it's still somewhat hard for me to wrap my head around doing algo questions in a functional paradigm, though I felt I understood the Udemy course I took on Elixir/Phoenix.
I wrote a solution to the LeetCode "valid anagram" problem using Elixir and Repl and wanted to see if people had any ideas for improving/understanding the problem or if there was a best approach way of thinking for this problem.
For an answer, I'd take a code review, a book recommendation or even just suggestions of what I could do differently.
Thank you for your time and hope this (my first question on this site) is clear.
###
Given two strings s and t , write a function to determine if t is an anagram of s.
Example 1:
Input: s = "anagram", t = "nagaram"
Output: true
Example 2:
Input: s = "rat", t = "car"
Output: false
Note:
You may assume the string contains only lowercase alphabets.
###
defmodule Algos do
def is_anagram(str1, str2) do
case String.length(str1) == String.length(str2) do
false ->
IO.puts(false)
true ->
both_trackers(str1, str2)
|> check_trackers
|> IO.puts
end
end
def both_trackers(str1, str2) do
t1 = make_tracker(str1)
t2 = make_tracker(str2)
{t1, t2}
end
def check_trackers({t1, t2}) do
Map.keys(t1)
|> Enum.reduce_while(true, fn x, acc ->
if t1[x] == t2[x], do: {:cont, acc}, else: {:halt, false}
end)
end
def make_tracker(str) do
tracker = String.split(str, "", trim: true)
|> Enum.reduce(%{},
fn x,acc -> Map.merge(acc,
case !!acc[x] do
false ->
%{x => 1}
true ->
%{x => acc[x] + 1}
end
)
end
)
tracker
end
end
Algos.is_anagram("sloop ", "pools")

Answer 1

New elixir has Enum.frequencies, which generates a histogram from an enumerable, which basically solves this problem out of the box:
defmodule Algos do
def anagram?(a, b) do
Enum.frequencies(to_charlist(a)) == Enum.frequencies(to_charlist(b))
end
end
Algos.anagram?("a gentleman", "elegant man") # => true
Algos.anagram?("alice", "bob") # => false