Provide the Big‐O notation for the following lines of Java code, make sure you include all the necessary steps to reach your answer to receive any credit:
a. for (i = 10; i < n – 8; i++)
{
for( j = 1; j < m; j *= 2)
x = i + j;
}
b. for(k = 0; k <= m; k += 2)
{
l = 0;
while(l < n/2)
{ y = l * 4 – x; l++; }
}
c. k = n;
while (k > 1)
{
for(j = 4; j < n; j++) ;
for(i = 1; i <= n; i++)
{
System.out.println(i+k);
z = i * k;
}
k /= 2;
}
a. Number of values for i are n-8-10 = n-18 Number of values for j are log(m) as j is incrementing twice every time. So, Total time complexity = O(nlog(m)) b. Number of values for k are m/2 Number of values for l are n So, Total time complexity = O(mn) c. Number of values for k are log(n) as k is decrementing by 2 every time. Number of values for i are n Number of values for j are n-4 As two for loop as are not nested (They are parallel) So, Total time complexity = O(log(n)(n+n-4)) = O(nlog(n))
Provide the Big‐O notation for the following lines of Java code, make sure you include all...
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