Avoiding zeros in ArrayReshape

list = {1, 2, 3, 4, 5, 6, 7, 8};
Partition[list, UpTo[3]]

{{1, 2, 3}, {4, 5, 6}, {7, 8}}

It seems that using ArrayReshape and modifying, if needed, the last entry to remove trailing zeros is faster than using Partition or using Inactive @ Nothing padding:

ClearAll[f1, f2, f3, f4]
f1 = Partition[#, UpTo[#2]] &; (* from Okkes's answer *)
f2 = Partition[#, #2, #2, 1, {}] &;
f3 = Activate @ ArrayReshape[#, 
  {Ceiling[Length[#]/#2], #2}, Inactive @ Nothing] &; (* from MarcoB's answer *)
f4 = Module[{p = ArrayReshape[#, {Floor[Length[#]/#2], #2}], m = Mod[Length@#, #2]},
      If[m == 0, p, Join @@ {p, {#[[-m ;;]]}}]] &;

Timings:

n = 999999;
k = 199;
list = Range[n];

First[RepeatedTiming[r1 = f1[list, k]]]

0.078

First[RepeatedTiming[r2 = f2[list, k]]]

0.075

First[RepeatedTiming[r3 = f3[list, k]]]

0.094

First[RepeatedTiming[r4 = f4[list, k]]]

0.012

r1 == r2 == r3 == r4

True

When n is divisible by k (so that ArrayReshape produces a rectangular array with no paddings):

k = 99;

First[RepeatedTiming[r1 = f1[list, k]]]

0.083

First[RepeatedTiming[r2 = f2[list, k]]]

0.075

First[RepeatedTiming[r3 = f3[list, k]]]

0.0072

First[RepeatedTiming[r4 = f4[list, k]]]

0.0071

r1 == r2 == r3 == r4

True

Using Partition with UpTo is clearly the most expedient answer here, as shown above. However, this got me wondering how one could force ArrayReshape into service anyway, just for fun. For instance, we could twist the use of the padding parameter in ArrayReshape to force it to pad with Nothing (or the vanishing function ##&[]), as in either of the following, which is then removed when evaluated:

ArrayReshape[Range[8], {3, 3}, Hold@Nothing] // ReleaseHold
ArrayReshape[Range[8], {3, 3}, Inactive@Nothing] // Activate

{{1, 2, 3}, {4, 5, 6}, {7, 8}}

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After posting the answer, I refreshed the page to see that this was brought up in comments as well.