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bigoh2

Language/Type: Java algorithm analysis big-oh
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Give a tight bound of the nearest runtime complexity class for each of the following code fragments in Big-Oh notation, in terms of the variable N. In other words, write the code's growth rate as N grows. Write a simple expression that gives only a power of N using a caret ^ character for exponentiation, such as O(N^2) to represent O(N2) or O(log N) to represent O(log2 N). Do not write an exact calculation of the runtime such as O(2N3 + 4N + 14). 

// a)
int sum = 0;
for (int i = 1; i <= N - 2; i++) {
    for (int j = 1; j <= i + 4; j++) {
        sum++;
    }   
    sum++;
}

for (int i = 1; i <= 100; i++) {
    sum++;
}
System.out.println(sum);
answer: 
// b)
int sum = 0;
for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N * N; j++) {
        sum++;
    }
    for (int j = 1; j <= 100; j++) {
        sum++;
    }
    for (int j = 1; j <= N; j++) {
        sum++;
    }
    sum++;
}
System.out.println(sum);
answer: 
// c)
int sum = 0;
for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= 100; j++) {
        sum++;
    }
}

for (int k = 1; k <= 10000; k++) {
    sum++;
}
System.out.println(sum);
answer: 
// d)
TreeSet<Integer> set = new TreeSet<Integer>();
for (int i = 1; i <= N * 2; i++) {
    set.add(i);
}

for (int k : set) {
    System.out.println(k);
}
System.out.println("done!");
answer: 
// e)
ArrayList<Integer> vec = new ArrayList<Integer>();
for (int i = 1; i <= N + 100; i++) {
    vec.add(i);
}
Stack<Integer> stack = new Stack<Integer>();
while (!vec.isEmpty()) {
    stack.push(vec.get(vec.size() - 1));
    vec.remove(vec.size() - 1);
}
while (!stack.isEmpty()) {
    stack.pop();
}
System.out.println("done!");
answer:

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