In the following code, what is the algorithm's complexity? minIndex = 0
CurrentIndex = 1
While currentIndex < len(lyst) :
If lyst[currentIndex] < lyst[minIndex]:
MinIndex = currentIndex
CurrentIndex += 1
Return minIndex
A) O( n 2)
B) O( n )
C) O(log2 n )
D) O2
Correct Answer:
Verified
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