Question 6 sm .Is 2^(n+1) = O(2^n )? Is 2^(2n) = O(2^n ) ?  Suppose that on a particular computer, an Algorithm A, with inputs of size n, runs in 10n^2 steps, and another Algorithm B runs in 60n lg n steps [Note: lg x = log2(x)]. For what values of positive integers n is Algorithm A better than Algorithm B? Full explain this question and text typing work only      We should answer our question within 2 hours takes more time then we will reduce Rating Dont ignore this line

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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.