Count Array Element Frequencies

Given an integer array values, count how many times each distinct value appears.
Return a two-dimensional integer array in which each row is [value, frequency]. Sort the rows by value in ascending order.

Function
countFrequencies(values: int[]) → int[][]

Examples
Example 1
values = [4,2,4,3,2,4]
return = [[2,2],[3,1],[4,3]]
The value 2 appears twice, 3 appears once, and 4 appears three times.
Example 2
values = [-1,-1,0,5,0]
return = [[-1,2],[0,2],[5,1]]
Negative and zero values are counted normally, and rows remain ordered by value.
Example 3
values = [7]
return = [[7,1]]
The only value appears once.

Constraints
1 <= values.length <= 10^5.
-10^9 <= values[i] <= 10^9.

The trick is a hash map from value to count. One pass over the array builds it in O(n). Then pull out the keys, sort them ascending, and build the rows. Total cost is O(n + k log k), where k is the number of distinct values. Don't use a fixed-size counting array. Values range from -10^9 to 10^9, so that blows up memory. Another pitfall is sorting the original array first and then scanning runs. It works, but it's slower to write cleanly and easier to get wrong. Watch negatives and zero, which the examples call out. Return type is a 2D int array, so size it to k rows. In a language with a sorted map, you can skip the explicit sort. If the Wells Fargo OA clock is tight and your hands freeze, StealthCoder is the hedge sitting quietly on your screen. This one should take about ten minutes.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.