Exercise I

  • Express the number of steps for a given code segment as a function of the input size in the worst-case scenario.
  • Appreciate that counting the exact number of RAM instructions requires too many details.

Consider the following code snippet:

int sum = 0;
String s = keyboard.nextLine(); // keyboard is an initialized Scanner
String v = "aeiou";
for (int  i = 0; i < v.length(); i++) {
  if (s.indexOf(v.charAt(i)) >= 0) {
    sum++;
  }
}

Exercise Count the number of steps taken by the above program. Write the count as a function T(N)T(N) where NN is the size of the "input." You need to determine what quantity is a suitable choice for the "size of input" here.

Solution

Note that while indexOf may seem like a constant operation, it is actually a linear search; it tells you to search through an array (in this case, a word) until you find the proper position. So, we are effectively running a linear search for each vowel (i.e., a constant number of times)

The running time of indexOf on string s is a function of the length of the string s. The program's running time can then be expressed as a function of NN where NN is the length of s. 

int sum = 0;                            // 1
String s = keyboard.nextLine();         // 1 + 1
String v = "aeiou";                     // 1
for (int  i = 0; i < v.length(); i++) { // 1 + 6 + 5
  if (s.indexOf(v.charAt(i)) >= 0) {    // (1 + (4N + 3) + 1) * 5
    sum++;                              // 1 * 5
  }
}

T(N)=20N+56 T(N) = 20N + 56

We don't know the exact running time of indexOf function. However, we do know it performs a linear search. Therefore, we can consider it 4N+34N + 3 by assuming its running time is the same as the (worst-case) running time of indexOf we analyzed earlier.

Case in point: It requires detailed knowledge of the program's constituents to count the exact number of RAM instructions a program takes to execute.