$1^{\text{st}}$ Pass: $37 \ 52 \ 12 \ 11 \ 25 \ 92$
$2^{\text{nd}}$ Pass: $37 \ 12 \ 11 \ 25 \ 52 \ 92$
$3^{\text{rd}}$ Pass: $12 \ 11 \ 25 \ 37 \ 52 \ 92$
$4^{\text{th}}$ Pass: $11 \ 12 \ 25 \ 37 \ 52 \ 92$
$5^{\text{th}}$ Pass: $11 \ 12 \ 25 \ 37 \ 52 \ 92$
Why $5^{\text{th}}$ pass? In $4^{\text{th}}$ pass, the array is already sorted, but the algorithm does not know if it is completed. The algorithm needs one whole pass without any swap to know it is sorted.