Describe the worst case running time of the following pseudocode functions in Big-Oh notation in terms of the variable n.
O(n²). O(n³ log n) O(n log n), O(n), O(n² log n), O(n⁵), O(2ⁿ), O(n³), O(log n), O(1), O(n⁴), O(nⁿ)
e) void silly(int n, int x, int y) {
if (x < y) {
for (int i = 0; i < n; ++i)
for (int j = 0; j < n * i; ++j)
System.out.println("y = " + y);
} else {
System.out.println("x = " + x);
}
} __________________
f) void silly(int n) {
j = 0;
while (j < n) {
for (int i = 0; i < n; ++i) {
System.out.println("j = " + j);
}
j = j + 5;
}
} __________________
g) void silly(int n) {
for (int i = 0; i < n * n; ++i) {
for (int j = 0; j < n; ++j) {
for (int k = 0; k < i; ++k)
System.out.println("k = " + k);
for (int m = 0; m < 100; ++m)
System.out.println("m = " + m);
}
}
} __________________
h) void happy(int n) {
for (int i = n*n; i > 0; i--) {
for (int k = 0; k < n; ++k)
System.out.println("k = " + k);
for (int j = 0; j < i; ++j)
System.out.println("j = " + j);
for (int m = 0; m < 5000; ++m)
System.out.println("m = " + m);
}
} __________________
i) Consider the following function:
int mystery(int n) {
int answer;
if (n > 0) {
answer =(mystery(n-2)+3*mystery(n/2) + 5);
return answer;
}
else
return 1;
}
