The worst case of QuickSort occurs when the picked pivot is always one of the corner elements in sorted array. In worst case, QuickSort recursively calls one subproblem with size 0 and other subproblem with size (n-1). So recurrence is T(n) = T(n-1) + T(0) + O(n) The above expression can be rewritten as T(n) = T(n-1) + O(n) 1 void exchange(int *a, int *b) { int temp; temp = *a; *a = *b; *b = temp; } int partition(int arr[], int si, int ei) { int x = arr[ei]; int i = (si - 1); int j; for (j = si; j <= ei - 1; j++) { if(arr[j] <= x) { i++; exchange(&arr[i], &arr[j]); } } exchange (&arr[i + 1], &arr[ei]); return (i + 1); } /* Implementation of Quick Sort arr[] --> Array to be sorted si --> Starting index ei --> Ending index */ void quickSort(int arr[], int si, int ei) { int pi; /* Partitioning index */ if(si < ei) { pi = partition(arr, si, ei); quickSort(arr, si, pi - 1); quickSort(arr, pi + 1, ei); } } [/sourcecode]

If we use median as a pivot element, then the recurrence for all cases becomes T(n) = 2T(n/2) + O(n) The above recurrence can be solved using Master Method. It falls in case 2 of master method.

See http://www.geeksforgeeks.org/nearly-sorted-algorithm/ for explanation and implementation.

See http://www.geeksforgeeks.org/lower-bound-on-comparison-based-sorting-algorithms/ for point A. See http://www.geeksforgeeks.org/stability-in-sorting-algorithms/ for B. C is true, count sort is an Integer Sorting algorithm.

For a input integer n, the innermost statement of fun() is executed following times. n + n/2 + n/4 + ... 1 So time complexity T(n) can be written as T(n) = O(n + n/2 + n/4 + ... 1) = O(n) The value of count is also n + n/2 + n/4 + .. + 1

The time complexity can be calculated by counting number of times the expression "count = count + 1;" is executed. The expression is executed 0 + 1 + 2 + 3 + 4 + .... + (n-1) times. Time complexity = Theta(0 + 1 + 2 + 3 + .. + n-1) = Theta (n*(n-1)/2) = Theta(n^2)

Following are the steps to follow to solve Tower of Hanoi problem recursively.

Let the three pegs be A, B and C. The goal is to move n pegs from A to C.
To move n discs from peg A to peg C:
    move n-1 discs from A to B. This leaves disc n alone on peg A
    move disc n from A to C
    move n?1 discs from B to C so they sit on disc n

The recurrence function T(n) for time complexity of the above recursive solution can be written as following. T(n) = 2T(n-1) + 1

The worst case time complexity is always greater than or same as the average case time complexity.

The order of growth of option c is n^2.5 which is higher than n^2.

  f1(n) = 2^n
  f2(n) = n^(3/2)
  f3(n) = nLogn
  f4(n) = n^(Logn)

Except f3, all other are exponential. So f3 is definitely first in output. Among remaining, n^(3/2) is next. One way to compare f1 and f4 is to take Log of both functions. Order of growth of Log(f1(n)) is Θ(n) and order of growth of Log(f4(n)) is Θ(Logn * Logn). Since Θ(n) has higher growth than Θ(Logn * Logn), f1(n) grows faster than f4(n). Following is another way to compare f1 and f4. Let us compare f4 and f1. Let us take few values to compare

n = 32, f1 = 2^32, f4 = 32^5 = 2^25
n = 64, f1 = 2^64, f4 = 64^6 = 2^36
...............
............... 

Also see http://www.wolframalpha.com/input/?i=2^n+vs+n^%28log+n%29 Thanks to fella26 for suggesting the above explanation.