Performance deep dive

In this approach, we'll find out how to most efficiently determine if a string is an Isogram in Go.

The approaches page lists three idiomatic approaches to this exercise:

1. Using map.
2. Using strings.Count().
3. Using a bit field.

Benchmarks

To benchmark the approaches, we ran the following Benchmark code for each approach:

func BenchmarkIsIsogram(b *testing.B) {
	for i := 0; i < b.N; i++ {

		for _, c := range testCases {
			IsIsogram(c.input)
		}

	}
}

and received the following results:

map
BenchmarkIsIsogram-12    	  237565	      4960 ns/op	    1201 B/op	      13 allocs/op

strings.count
BenchmarkIsIsogram-12    	  896049	      1409 ns/op	      40 B/op	       3 allocs/op

bit field
BenchmarkIsIsogram-12    	 6214525	       185.2 ns/op	       0 B/op	       0 allocs/op

Generally, the fewer bytes allocated per op (B/op) the faster (i.e. the fewer ns/op) the implementation. More info on reading benchmarks can be found here.

The fastest is the Bitfeld approach, but it depends on all of the letters being ASCII. The fastest approach that supports Unicode is the strings.Count() approach.