True , Lamport Bakery Algorithm satisfies deadlock free and starvation . We know that in Bakery Algorithm Each process is given a ticket number say (i). Now all those process which have ticket number less than i will enter and exit CS ,but for process with ticket number >i cant enter in CS because process with ticket =i is not yet enter .
And if you think process with ticket > i can enter CS, then its not possible because of the below statement
(the concept of bakery algorithm)
∀j:j!=i:Num[j] = 0∨(NUM[j], IDj)≥(NUM[i] ,IDi)
This question has been asked indirectly In gate2016 OS . However you can refer to below video where they have explained the approach for it