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bigoh7

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 - 999; i++) {
    for (int j = 1; j <= 0.0001 * N/2; j++) {
        sum++;
    }
}
answer: 
// b)
int sum = 0;
for (int i = 1; i <= 1000000; i++) {
    sum++;
}
int x = 999999;
for (int i = 1; i <= x; i++) {
    for (int j = 1; j <= 999; j++) {
        sum++;
    }
    for (int k = 1; k <= 999; k++) {
        sum++;
    }
}
cout << sum << endl;
answer: 
// c)
vector<int> vecC;
for (int i = 1; i <= N; i++) {
    vecC.insert(vecC.begin(), i);
}
set<int> setC;
for (int i = 0; i < vecC.size(); i++) {
    setC.insert(vecC[i]);
}
cout << "done!" << endl;
answer: 
// d)
map<int, int> mapD;
for (int i = 1; i <= N; i++) {
    mapD[i] = i/2;
}

set<int> setD;
for (int i = 1; i <= N; i++) {
    int value = mapD[i];
    setD.insert(value);
    mapD.erase(i);
}
cout << "done!" << endl;
answer: 
// e)
vector<int> vecE;
for (int i = 1; i <= N; i++) {
    vecE.push_back(i);
}
stack<int> stackE;
while (!vecE.empty()) {
    stackE.push(vecE[vecE.size() - 1]);
    vecE.erase(vecE.begin() + vecE.size() - 1);
}
cout << "done!" << endl;
answer:

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