Data Structures and Algorithms

Arranging the records in some logical order is referred to as sorting.

It is a linear data structure. A linear data structures elements are arranged in a list.

This field in ATM cell header is carried in the final octet of the header and provides an eight-bit cyclic redundancy check remainder that covers the entire header.

Using Master Theorem: a=3, b=4, f(n)=n^2. Since n^(log_4(3)) ≈ n^0.79 < n^2, we're in Case 3 where f(n) dominates. Therefore T(n) = O(n^2). The recursive calls at each level contribute polynomial work, but the n^2 term at the root determines overall complexity.

Using Master Theorem: a=3, b=4, f(n)=n log n. Since n^(log_4(3)) ≈ n^0.79 < n log n, we're in Case 3 where f(n) dominates by a polynomial factor. Therefore T(n) = O(n log n), which matches the claimed answer B.

Using Master Theorem: a=9, b=3, f(n)=n. Critical exponent: log_3(9)=2, so n^(log_b(a))=n^2. Since f(n)=n < n^2, we're in Case 1 where the recursive part dominates. Therefore T(n) = O(n^2). This matches the claimed answer A - this recurrence represents 9 recursive calls each on 1/3 of the input.

Using the Master Theorem: a=4, b=2, so n^(log_b a) = n^2. The function f(n) = n²√n = n^2.5 is polynomially larger than n^2 (by factor n^0.5). Since the regularity condition a·f(n/b) ≤ c·f(n) holds, case 3 applies: T(n) = Θ(f(n)) = Θ(n²√n). The n log n option would only apply if f(n) = O(n^2), which is false here.

The recursion tree has log n levels, each contributing n/log(n/2^i). Level 0: n/log n, Level 1: n/log(n/2), ..., Level log n: O(log n). Using the integral approximation, the sum of n/log(n/2^i) for i=0 to log n is O(n log log n). This is because we're essentially summing 1/log(x) which gives the log log factor.

At each level, the total work is n + n/2 + n/4 + n/8 + ... (sum of T(n/2), T(n/4), T(n/8) contributions) plus the constant n at the root. The sum n(1 + 1/2 + 1/4 + 1/8 + ...) converges to 2n = O(n). The O(n²) option would require quadratic growth at each level, and O(log n) ignores the linear work at each node.

Using the Master Theorem for recurrence relations of the form T(n) = aT(n/b) + f(n), here a = 3, b = 2, and f(n) = n log n. Comparing f(n) = n log n with n^(log_b a) = n^(log_2 3), since log_2 3 is approximately 1.585, n^1.585 grows faster than n log n. Thus, Case 1 applies, yielding O(n^(log 3)).

This recurrence adds log n at each of n steps. Summing log 1 + log 2 + ... + log n = log(n!) ≈ n log n - n + O(log n) by Stirling's approximation. Therefore T(n) = O(n log n). The O(n) option would require constant work per step, and O(n²) would require linearly growing work.