Algorithm Analysis - Computer Science

The run time can best be analyzed by breaking the code down by line. The line iterating the value of sum is performed in constant time, or O(1). This constant time operation is performed n times in the inner "for" loop, and n times in the outer "for" loop. Thus, the run time is .

The run time can best be analyzed by breaking the code down by line. The line iterating the value of sum is performed in constant time, or O(1). This constant time operation is performed times in the inner "for" loop, and times in the outer "for" loop. Thus, the overall run time is .

The run-time can best be analyzed by breaking the code down by line. The line iterating the value of sum is performed in constant time, or O(1). The "for" loop that controls this constant-time operation is initialized at i = 1, and i is iterated by multiplying by 2 until it is equal to or greater than n. Thus, this loop will run times, and the run-time will be O(log(n))

The run time of this code is best analyzed by plugging in numbers. Take the value of bar(2) for example. bar(2) leads to two calls to bar(1), which lead to four calls to bar(0). A call to bar(4) leads to two calls to bar(3), four calls to bar(2), eight calls to bar(1), and sixteen calls to bar(0). Thus, it can be shown that the number of calls is proportional to 2^n, and that that run time is O(2^n).

The function is O(n2) because it has a nested loop of size 2, with no extras thrown in that could possibly add on a log(n). A good rule of thumb is this: if there are nested loops, then the BigO is usually O(nm), with m being the levels of loops in the longest part.

At first glance we might be tempted to pick O( ) because there are 2 for loops. But, upon closer inspection we can see that the first loop will yield a O( n ) running time but the second loop does not. The second loop has only an O( log(n) ) running time because "j" doubles each iteration and does not increase linearly. That will yield O( log(n) ) since O( log(n) ) is a much faster running time. So the final result is O( n log(n) ).

Code sample #1 relies on i in the second loop where int j = i. Since the code relies on i in the second loop, the order goes from O(N) to O(N2)

Code sample #2 has two separate loops that do not rely on each other. The first for loop loops through the array arr and the second for loop loops through the array arr2. Since the two loops are exclusive, the order is O(N)

O(NlogN) is O(N) * O(logN) which is greater than O(N) alone.

O(N2) is O(N*N) which is greater than O(N).

O(N3) is O(N*N*N) which is greater than O(N).

O(N) is the smallest and therefore is the quickest to compile. Therefore, O(N) is the correct answer.

The two code snippets have the same efficiency. Both operate in O(N) time. ArrayLists use arrays as their underlying data structure so access to the data is also the same.

Code snippet A has a running time of O(N2). Code snippet B has a running time of O(N). While the two code snippets may look the same, the second for loop in code snippet A sets j=i. Since j is relying on i, it's multiplying the first for loop's running time by the second for loop's running time. This gives us O(N*N) or just O(N2).

Start by counting the number of times the first loop will cycle. Notice that the loop control variable (namely, i) is increased by 2 for each looping. This means that you will loop for i = 0, 2, 4, 6, 8. It will not loop for 10, for then, i will equal a. That will terminate the loop. Thus, you have five loopings thus far. Note, also that this means that b is decremented 5 times. Thus, at the end of this loop, b now equals 10.

For the second loop, you must be careful, for it has two loops. The outer loop will run from i = 10 to i = 0. This is a total of 11 loopings. (This is important to note because it runs not only for 10 through 1 but also for the final number, 0. The loop within that loop will run from j = 0 to j = 6 (i.e. one less than c, which is 7). This is a total of 7 loopings. Therefore, these nested loops will run a total of or 77 times. The total calls to System.out.print will be , or 82.

Let us count for each loop, paying attention to variables as needed. The first loop will run a total of 54 times. Thus it will produce 54 total executions. Notice also how it initializes the array. This will be used in the second loop. Each element is equal to 3i + 1. Thus, we will have:

a\[0\] = 1; a\[1\] = 4; a\[2\] = 7; a\[3\] = 10; a\[4\] = 13; a\[5\] = 16; etc.

Now, loop two will use a[5] in controlling its execution. Notice that it runs from i = 0 to i = 15. For each execution, it reduces the value of a[i] by one. Thus, be careful. It will eventually run for a[5], making it to be 15. This will then be used for the loop control. This means that it actually runs only from i = 0 to i = 14. Thus, the loop runs a total of 15times.

The last loop runs from i = 9 to i = 1. (Note: This is because of the decrement in the second loop.) This is a total of 9 executions.

Thus, the total count of executions is:

The statements executed in the code snippet are in order:

1. int i = 0
2. i < arr.length
3. i++ (which is equivalent to i = i + 1)
4. i == 2
5. i++

These statements are all executions that happen in order each time the for loop is run. Sometimes there may only be 4 executions if the if statement is false.

Performing a bitwise excludive OR constitutes in taking both binary values and evaluating as follows: either one or the other (1) never both (0). This is the truth table for a bitwise XOR operation:

Taking both a and b and performing the operation bit by bit we get the following result:

Yes, there is a runtime exception in the code snippet. The int wait_time is defined twice which will give a runtime exception. This can be fixed by not declaring int before the second assignment of the variable wait_time.