Consider this pseudocode for a shared counter increment:

medium📝 Analysis Q13 of 15

LLD - Advanced LLD Concepts

lock.acquire()
counter = counter + 1
lock.release()
print(counter)

If two threads run this code simultaneously starting with counter = 0, what is the possible output?

DAny number greater than 2

Step-by-Step Solution

Solution:

Step 1: Understand lock usage in code
The lock ensures only one thread increments the counter at a time, preventing race conditions.
Step 2: Calculate final counter value
Two threads each increment once, so counter goes from 0 to 2 safely.
Final Answer:
2 -> Option C
Quick Check:
Lock ensures increments are safe, so counter = 2 [OK]

Quick Trick: Locks prevent lost updates, so increments add up [OK]

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