The recurrence relation of ternary search is T(n)= T(n/3) + 4, How 4 is in recurrence relation, since in ternary search it's log to the base 3 N, so only 3 partitions should be there ?
Recurrence relation for ternary search
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Recurrence relation for ternary search is
T(n) = T(n/3) + O(1)or evenT(n) = T(2n/3) + O(1). The constant hidden in thisO(1)depends on concrete implementation and how analysis was conducted. It could be4or3, or some other value. Applying case 2 of Master theorem you still haveO(log n).