Anagram

Problem Solving

# Understand the problem
# I have 2 strings
# I need to determine if these strings have the same number of the same characters. robed bored

# Input - Output
# "robed", "bored" -> true
# "cats", "dogs" -> false

# Five Ideas
# 1. Break into individual characters and compare chars to eachother. 
# 2. Check if characters in first string exist in second string. Remove chars in second string if found. if not found, is not anagram.
# 3. Sort characters in alphabetical order
# 4. Is there a way to strict match strings without caring about order?
# 5. Maps? Maybe we can convert this to a map which doesn't care about order.

# A: Understand the problem
# B: Come up with 5+ Ideas
# C: Compare and Contrast Pros/Cons
# D*: Decide which idea to try next

## Identify Edge Cases
# 1. Double Character Issues ccab vs bacc

## Breaking Down The Problem

# Input Output
# "robed", "bored" -> true

# will not work
%{c1: "r", c2: "o", c3: "b"} == %{c1: "b", c2: "o", c3: "r"}

# duplicate key issues using map
%{a: nil, b: nil, a: nil} == %{b: nil, a: nil, a: nil}
# fix the issue by tracking how many of a character there are.
# "baa", "aba"
%{a: 2, b: 1} == %{b: 1, a: 2}

# sorting

# I: "cab" "bac"
# O: "abc" "abc"

Solution Ideas

# Sorting

# "cab" "bac"
# "abc" "abc"

# "cab" "bac"
# ["c", "a", "b"] ["b", "a", "c"]

# "caba" "bacc"
["a", "a", "b", "c"] == ["a", "b", "c", "c"]

# Removing


"cab", "bac"

# if not found, is not anagram.

# ["c", "a", "b"] ["b", "a", "c"]

# c

# ["b", "a"]

# a

["b"]
# b

[]

# when finished looking through first string.

# if list is empty, then is anagram.

# Example Solving Without Enum!

# check length
# split
# remove all characters
["a", "b", "c"] -- ["b", "a", "c"]
# if list is empty, is anagram.

input1 = "c"
input2 = "cc"

same_length = String.length(input1) == String.length(input2)

split_string1 = String.split(input1, "")
split_string2 = String.split(input2, "")

IO.inspect(split_string1)
IO.inspect(split_string2)

result1 = split_string1 -- split_string2
# result2 = split_string2 -- split_string1

result1 == [] and same_length

Graphemes vs Split vs Codepoints

String.split("abc", "", trim: true)
String.split("a,b,c", ",")
String.graphemes("é")
String.codepoints("é")