# CodeStepByStep

## bigoh2

Language/Type: C++ algorithm analysis big-oh

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++; } cout << sum << endl;``` 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++; } cout << sum << endl;``` 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++; } cout << sum << endl;``` answer: ```// d) set set; for (int i = 1; i <= N * 2; i++) { set.insert(i); } for (int k : set) { cout << k << endl; } cout << "done!" << endl;``` answer: ```// e) vector vec; for (int i = 1; i <= N + 100; i++) { vec.push_back(i); } stack stack; while (!vec.empty()) { stack.push(vec[vec.size() - 1]); vec.erase(vec.begin() + vec.size() - 1); } while (!stack.empty()) { stack.pop(); } cout << "done!" << endl;``` answer: