Question

(Java) For the following code segment, match the different values of the empty line with the resulting Big O complexity of the algorithm.

int n = 
int k = 0;
for (int i=0; i <= n; i++){
    int j = i;
    while (j > 0) {
        // ___MISSING CODE___
        k++;
    }
}

j = 0;

j--;

j /= 2;

j = j * 2;

Options for each are: O(n^3)            O(n * log n)            an infinite loop results            O(1)            O(n^2)            O(n)      

Answer 1

int n = 
int k = 0;
for (int i=0; i <= n; i++){
    int j = i;
    while (j > 0) {
        j = 0;
        k++;
    }
}
Time complexity = O(n)


int n =
int k = 0;
for (int i=0; i <= n; i++){
    int j = i;
    while (j > 0) {
        j--;
        k++;
    }
}
Time complexity = O(n^2)


int n =
int k = 0;
for (int i=0; i <= n; i++){
    int j = i;
    while (j > 0) {
        j /= 2;
        k++;
    }
}
Time complexity = O(n * log(n))


int n =
int k = 0;
for (int i=0; i <= n; i++){
    int j = i;
    while (j > 0) {
        j = j * 2;
        k++;
    }
}
Answer: an infinite loop results