What is the complexity of the following recurrence: T(n) = 4T(n/2) + n³, T(1) = 1 T(n) = O(nlogn) T(n) = 2(n² logn) T(n) = O(n³) T(n) = O(n²)

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A: Given Function: T(n)=3.Tn2+c.n2,n>1

Q: Solve the recurrence: T(n) = T(n/2) + 4n T(1) = 1

A: The solution to the given problem is below.

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A: In step 2, I have provided  ANSWER  with  BRIEF  EXPLANATION-------------------

Q: T(n)=2T(n/2)+1 where T(1)=1 and n

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A: Time complexity = O(n^2)

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Q: Build a recursion tree for the following recurrence equation and then solve for T(n)

A:

Q: 7. Prove or disprove: f(n) + g(n) = 0 (min(f(n), g(n)))

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Q: e the following recurrences exactly: (a) T(1) = 8, and for all n ≥ 2, T(n) = 3T(n − 1) + 1

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Q: What is the complexity of the following recurrence: T(n) = 4T(n/2) + n³, T(1) = 1 T(n) = 0(nlogn)…

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Q: T(1) = 1 T(n) = 1 + T(n-1) for n>1

A: We need to draw recursion tree and solve.

Counting sort

The counting sort algorithm sorts an array’s contents by counting the repetition of each element that occurs in the array. A counting sort algorithm is a stable algorithm that works the best when the range of elements in the array is small, say for example from 0 to 9. The count which is the number of repetitions of an element in the array is mapped to an index of an auxiliary array, which is then saved in an auxiliary array. The counting sort algorithm has a stable time complexity of the order of O(size of array + size of auxiliary array). Let’s discuss the algorithm in detail as follows.