Still O(n) time complexity as args.Min() has to iterate through the whole string in worst case similar to a for loop

Are u serious? I`m gonna practice more...

Another opportunity to practice with LINQ (◕‿◕)

There are 54 data points but only 3 unique values. How can a histogram have more than 3 bins?

[2, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 2] should equal [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 2]

pass

Python:

• New test framework should be used

• Corresponding imports are missing in test cases

oh

Check {0,0,9,9}, Bro :)

okay don't need to complain about the solution actualy obviously that's a basic question so it doesn't matter how you solve it

You're making a histogram from a set of data. It's a bar graph of "bins" (ranges) of numbers vs. their frequency in the set. The return value is the heights of the columns in the histogram.

For the data [1, 1, 0, 1, 3, 2, 6], binWidth = 1 in the example, there is one 0, three 1s, one 2, one 3, no 4s, no 5s and one 6, so the output will be [1, 3, 1, 1, 0, 0, 1] -- the frequencies of each element for all the possible bins. Each bin just contains a single number.

If binWidth = 2, each bin covers two numbers. The first bin matches 0 and 1, the second 2 and 3, and so forth.

No matter how many times I read the description, I don't understand what is being asked.

Good question. Not necessary, not sure why I had them included.

The expected result was wrong in the tests. Fixed.