Test: Asymptotic Worst Case Time & Space Complexity- 1

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);

  }

}

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.

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 =  

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

We can get it by taking an example like n = 54321. After 2 iterations, rev would be 12 and n would be 543.

The bubble sort is at its best if the input data is sorted. i.e. If the input data is sorted in the same order as expected output. This can be achieved by using one boolean variable. The boolean variable is used to check whether the values are swapped at least once in the inner loop. Consider the following code snippet:
[sourcecode language="C"]

int main() {
int arr[] = {10, 20, 30, 40, 50}, i, j, isSwapped;
int n = sizeof(arr)/ sizeof(*arr);
isSwapped = 1; for(i = 0; i < n - 1 && isSwapped; ++i)
{
isSwapped = 0;
for(j = 0; j < n - i - 1; ++j)
if (arr[j] > arr[j + 1])
{
swap(&arr[j], &arr[j + 1]);
isSwapped = 1;
} }
for(i = 0; i < n; ++i)

printf("%d ", arr[i]); return 0;
}
[/sourcecode]
Please observe that in the above code, the outer loop runs only once.

Applying binary search to calculate the position of the data to be inserted doesn't reduce the time complexity of insertion sort. This is because insertion of a data at an appropriate position involves two steps: 1. Calculate the position. 2. Shift the data from the position calculated in step #1 one step right to create a gap where the data will be inserted. Using binary search reduces the time complexity in step #1 from O(N) to O(logN). But, the time complexity in step #2 still remains O(N). So, overall complexity remains O(N^2).

The number of comparisons that a comparison sort algorithm requires increases in proportion to Nlog(N), where N is the number of elements to sort. This bound is asymptotically tight: Given a list of distinct numbers (we can assume this because this is a worst-case analysis), there are N factorial permutations exactly one of which is the list in sorted order. The sort algorithm must gain enough information from the comparisons to identify the correct permutations. If the algorithm always completes after at most f(N) steps, it cannot distinguish more than 2^f(N) cases because the keys are distinct and each comparison has only two possible outcomes. Therefore, 2^f(N) >= N! or equivalently f(N) >= log(N!). Since log(N!) is Omega(NlogN), the answer is NlogN. For more details, read here

The time complexity is given by: T(N) = T(N/3) + T(2N/3) + N Solving the above recurrence relation gives, T(N) = N(logN base 3/2)

In the first look, the time complexity seems to be O(n^2) due to two loops. But, please note that the variable j is not initialized for each value of variable i. So, the inner loop runs at most n times. Please observe the difference between the function given in question and the below function:
[sourcecode language="C"]
void fun(int n, int arr[])
{
int i = 0, j = 0;
for(; i < n; ++i)
{
j = 0;
while(j < n && arr[i] < arr[j]) j++;
} }
[/sourcecode]

The time complexity of first for loop is O(n). The time complexity of second for loop is O(n/2), equivalent to O(n) in asymptotic analysis. The time complexity of third for loop is O(logn). The fourth for loop doesn't terminate.

Order of growth of [Tex]log n![/Tex] and [Tex]nlog n[/Tex] is same for large values of [Tex]n[/Tex], i.e., [Tex] heta (log n!) = heta (nlog n)[/Tex]. So time complexity of fun() is [Tex] heta (nlog n)[/Tex]. The expression [Tex] heta (log n!) = heta (nlog n)[/Tex] can be easily derived from following Stirling's approximation (or Stirling's formula). [Tex] log n! = nlog n - n +O(log(n)) [/Tex]

In asymptotic analysis we consider growth of algorithm in terms of input size. An algorithm X is said to be asymptotically better than Y if X takes smaller time than y for all input sizes n larger than a value n0 where n0 > 0.

Floyd–Warshall algorithm uses three nested loops to calculate all pair shortest path. So, time complexity is O(n^3). Read here for more details.