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bigoh6

Language/Type: C++ algorithm analysis big-oh
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Author: Marty Stepp (on 2016/06/16)

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; i++) {
    int k = 4000;
    for (int j = 1; j <= k; j++) {
        sum++;
    }
}
cout << sum << endl;
answer:
// b)
int sum = 0;
for (int i = 1; i <= N; i++) {
    sum += 2;
}
for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N * N; j++) {
        sum++;
    }
}
cout << sum << endl;
answer:
// c)
Stack<int> stack;
for (int i = 1; i <= N; i++) {
    stack.push(i);
}
Set<int> set;
while (!stack.isEmpty()) {
    int k = stack.pop();
    set.add(k);
}
cout << "done!" << endl;
answer:
// d)
HashMap<int, int> map1;
for (int i = 1; i <= N; i++) {
    map1.put(i, i);
}

Map<int, int> map2;
for (int i = 1; i <= N; i++) {
    int k = map1.get(i);
    map2.put(k, k);
    map2.put(n + k, n + k);
}
cout << "done!" << endl;
answer:
// e)
Vector<int> v;
for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) {
        v.insert(0, i);
    }
    v.clear();
}
while (!v.isEmpty()) {
    v.remove(0);
}
cout << "done!" << endl;
answer:

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