Consider the same method again: public boolean isprefix(String s1, String s2) { int i = 0; if(s1.length s2.length) return false; while(i

Question

Consider the same method again:

public boolean isprefix(String s1, String s2) {

int i = 0;
    if(s1.length > s2.length) 
        return false;

while(i < s1.length) {
        if(s1[i] != s2[i])
            return false;
        i++;
    }
    return true;
}

Use the active operation approach and determine the exact number of times the active operation is executed in the worst case.  Express your answer in terms of n, the length of the string s1.

Hint:  simplify your final answer as much as possible, and do not put spaces in your answer.  Use juxtaposition for the multiplication operator, for example to write "nine times n" write "9n" not "9xn" or "9*n";  to write "four times (n+2)" write "4(n+2)", not "4x(n+2)" or "4*(n+2)".  Do not write your answer in Big-O notation.  Write the exact number of lines executed.

BlackTom AI · Accepted Answer

The question asks for the exact count of the active operation executions in the worst case, expressed in terms of n where n is the length of s1.

First, observe the control flow: if the length of s1 is greater than the length of s2, the function returns false immediately. In the worst case, this branch does not trigger right away, so we proceed to the while loop.

Inside the while loop, the key operation is the character comparison: if (s1[i] != s2[i]) return false;. This comparison is evaluated on every iteration of the loop until a mismatch is found or the end is reached.

The loop condition is while (i < s1.length), and i starts at 0 and is incremented by 1 on each iteration (i++). Therefore, the loop runs exactly i = 0, 1, 2, ..., n-1, which totals n iterations in the worst case (where s1 and s2 match character by character up to length n).

During each of those n iterations, the active operation (the comparison s1[i] != s2[i]) is performed once. Hence the number of times this active operation executes scales linearly with n and equals n in the worst case.

Other operations inside the function, such as evaluating the loop condition or performing i++ and array indexing, are constant-time with respect to n and do not affect the exact count that grows with n. Since the prompt asks for the exact number of times the primary active operation executes, we focus on the comparison per iteration.

Putting it together, the worst-case count of the active operation is n, matching the given answer.

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