t.Errorf("Test %d: %s:  Expected to pass, but failed with: <ERROR> %s", i+1, instanceType, err.Error())
			continue
		}

		if len(result.Objects) != len(testCase.objectsAfterDelete) {
			t.Errorf("Test %d: %s: mismatch number of objects after delete, expected = %v, found = %v", i+1, instanceType, testCase.objectsAfterDelete, result.Objects)
			continue
		}

		for idx := range result.Objects {

                expectedBucketNumber - 1,
                maxNumberInBucket,
                noElementSplitFunction,
                canRunTogether,
            )
    }
}

/**
 * Determine the number of buckets for the first element in the list
 * when it needs to be split into several smaller pieces.
 *
 * The basic idea is:
 * 1. Make sure the rest elements has at least one bucket.

func (l *LastMinuteHistogram) Add(size int64, t time.Duration) {
	l[sizeToTag(size)].add(t)
}

// GetAvgData will return the average for each bucket from the last time minute.
// The number of objects is also included.
func (l *LastMinuteHistogram) GetAvgData() [sizeLastElemMarker]AccElem {
	var res [sizeLastElemMarker]AccElem
	for i, elem := range l[:] {
		res[i] = elem.getTotal()
	}
	return res

	}
	for i, tt := range tests {
		err := readCRLF(tt.reader)
		if err != tt.expectedErr {
			t.Errorf("Test %d: Expected %s, got %s this", i+1, tt.expectedErr, err)
		}
	}
}

// Tests parsing hex number into its uint64 decimal equivalent.
func TestParseHexUint(t *testing.T) {
	type testCase struct {
		in      string
		want    uint64
		wantErr string
	}
	tests := []testCase{

// NewDummyDataGen returns a ReadSeeker over the first `totalLength`
// bytes from the infinite stream consisting of repeated
// concatenations of `alphabets`.
//
// The skipOffset (generally = 0) can be used to skip a given number
// of bytes from the beginning of the infinite stream. This is useful
// to compare such streams of bytes that may be split up, because:
//
// Given the function:
//
//	f := func(r io.Reader) string {

	notify:
	  endpoint: "https://notify.endpoint" # notification endpoint to receive job status events
	  token: "Bearer xxxxx" # optional authentication token for the notification endpoint

	retry:
	  attempts: 10 # number of retries for the job before giving up
	  delay: "500ms" # least amount of delay between each retry
```

You can create and run multiple 'replication' jobs at a time there are no predefined limits set.

</p>

<p>
This rule generalizes the previous rule to buffered channels.
It allows a counting semaphore to be modeled by a buffered channel:
the number of items in the channel corresponds to the number of active uses,
the capacity of the channel corresponds to the maximum number of simultaneous uses,
sending an item acquires the semaphore, and receiving an item releases
the semaphore.
This is a common idiom for limiting concurrency.

   * its first position but with its final value. The {@code duplicates} map is modified.
   *
   * @param entries the original array of entries including duplicates
   * @param n the number of valid entries in {@code entries}
   * @param newN the expected number of entries once duplicates are removed
   * @param duplicates a map of canonical {@link Entry} objects for each duplicate key. This map

      out.write(bytes);
    }
  }

  /**
   * Writes all the bytes from the given {@code InputStream} to this sink. Does not close {@code
   * input}.
   *
   * @return the number of bytes written
   * @throws IOException if an I/O occurs while reading from {@code input} or writing to this sink
   */
  @CanIgnoreReturnValue
  public long writeFrom(InputStream input) throws IOException {

      out.write(bytes);
    }
  }

  /**
   * Writes all the bytes from the given {@code InputStream} to this sink. Does not close {@code
   * input}.
   *
   * @return the number of bytes written
   * @throws IOException if an I/O occurs while reading from {@code input} or writing to this sink
   */
  @CanIgnoreReturnValue
  public long writeFrom(InputStream input) throws IOException {