## Expert Answer

int i = 1;

while (i < n) {

printf(“Insert difficult work here!”); (This Line)

i = i +1;

}

The complexity is of O(n). The loop is running for n times as i starting with 1 and going up to n.

for(i=0; i<n; i++) {

for(j=0; j<n; i++) {

for(k=0; k<n; i++) {

if (i==j && j==k)

arr[i][j][k] = 1; (This Line)

}

}

}

This is again O(n) . For i = 1 the statement will be executed only once (i.e j = 1 && k = 1). So it means that for each value of i, the statement will be executed once and i is going from 1 to n so it is of order(n)