# sorting by frequency of occurrence in an array

Is there an efficient way of doing this. I have an array

```a=[1,2,2,3,1,2]
```

I want to output the frequency of occurrence in an ascending order. Example

```[[3,1],[1,2],[2,3]]
```

Here is my code in ruby.

```b=a.group_by{|x| x}
out={}

b.each do |k,v|
out[k]=v.size
end

out.sort_by{|k,v| v}
```

```a = [1,2,2,3,1,2]
a.each_with_object(Hash.new(0)){ |m,h| h[m] += 1 }.sort_by{ |k,v| v }
#=> [[3, 1], [1, 2], [2, 3]]
```

Something like this:

```x = a.inject(Hash.new(0)) { |h, e| h[e] += 1 ; h }.to_a.sort{|a, b| a[1] <=> b[1]}
```

Are you trying to work out the algorithm or do you just want the job done? In the latter case, don't reinvent the wheel:

```require 'facets'
[1, 2, 2, 3, 1, 2].frequency.sort_by(&:last)
# => [[3, 1], [1, 2], [2, 3]]
```

use hashing, create a hash, traverse the array, for each number in array, update the count in hash. it will take linear time O(n) and space complexity will be equal to storing the hash O(n).

```def frequency(a)
a.group_by do |e|
e
end.map do |key, values|
[key, values.size]
end
end

a = [1, 2, 2, 3, 1, 2]
p frequency(a).sort_by(&:last)
```