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bigoh6

Language/Type: C++ 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; 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.empty()) {
    int k = stack.top();
    stack.pop();
    set.insert(k);
}
cout << "done!" << endl;
answer: 
// d)
unordered_map<int, int> map1;
for (int i = 1; i <= N; i++) {
    map1[i] = i;
}

map<int, int> map2;
for (int i = 1; i <= N; i++) {
    int k = map1[i];
    map2[k] = k;
    map2[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(v.begin(), i);
    }
    v.clear();
}
while (!v.empty()) {
    v.erase(v.begin());
}
cout << "done!" << endl;
answer:

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