Question

For the following parts, try to get the best Big-O estimate that you can and briefly justify your answers.

Part a)

int i, j;

int n = 100;

for (i = 1; i <= n; i++) {

for (j = 3*i; j <= n; j++) {

printf("programming is fun\n");

}

}

Part b)

int i, j;

int n = 1000000;

for (i = 1; i <= n; i++) {

for (j = 1; j <= 10000; j++) {

printf("%d %d\n", i, j);

}

}

Part c)

int i = 0;

int n = 10;

int j;

while (i < n) {

i++;

j = i;

while (i < n) {

printf("hello %d\n", i);

i++;

}

i = j;

}

Part d)

int i = 0;

int n = 10;

int j;

Page 5 of 6

while (i < n) {

i++;

j = i;

while (i < n) {

printf("hello %d\n", i);

i++;

break;

}

i = j;

}

Answer 1

Part a)

int i, j;

int n = 100;

for (i = 1; i <= n; i++) {

for (j = 3*i; j <= n; j++) {

printf("programming is fun\n");

}

}
================================================================
Answer: O(n²)
Explanation: As we can see there are two for loop one nested on another. Lets say if n = 3, the program will run for 3*3 => 9 times. So the idea if the upper for loop will run for n time and then inner runs each n times for every iteration of upper for loop
n + n + n + n + n .. ..n times => n*n
=================================================================

Part b)

int i, j;

int n = 1000000;

for (i = 1; i <= n; i++) {

for (j = 1; j <= 10000; j++) {

printf("%d %d\n", i, j);

}

}
================================================================
Answer: O(n²)
Explanation: As we can see there are two for loop one nested on another. Lets say if n = 3, the program will run for 3*3 => 9 times. So the idea if the upper for loop will run for n time and then inner runs each n times for every iteration of upper for loop
n + n + n + n + n .. ..n times => n*n
If ith n = 1000000 and for jth we have 1000 , Then TOTAL TIMES = 1000000 * 1000 = 10⁹
=================================================================

Part c)

int i = 0;

int n = 10;

int j;

while (i < n) {

i++;

j = i;

while (i < n) {

printf("hello %d\n", i);

i++;

}

i = j;

}

================================================================
Answer: O(n²)
Explanation: As we can see there are two while loop one nested on another. Lets say if n = 3, the program will run for 3*3 => 9 times. So the idea if the upper while loop will run for n time and then inner runs each n times for every iteration of upper for loop
n + n + n + n + n .. ..n times => n*n
=================================================================

Part d)

int i = 0;

int n = 10;

int j;

Page 5 of 6

while (i < n) {

i++;

j = i;

while (i < n) {

printf("hello %d\n", i);

i++;

break;

}

i = j;

}
================================================================
Answer: O(n)
Explanation: As we can see there are two for loop one nested on another.So the idea if the upper while loop will run for n time and then inner runs once once as there is break statement
1 + 1 + 1 + 1 + 1 + ...n times = n
=================================================================