String Manipulation: Practice Questions (Computer Science A 1.15)

These are original Marlbridge practice questions, not past exam questions. All code is invented for these questions and was compiled and run before publication. The rubrics are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. No calculator is allowed on the exam, so work everything by hand. Code fragments are assumed to be inside a main method. Write the index under each character when a question gives you a string: the first character is at index 0.

Question 1 (multiple choice · foundation)

What is the value of "banana split".length()?

(A) 11
(B) 12
(C) 13
(D) 5

Answer and explanation

Answer: (B). banana has 6 characters, the space is 1, and split has 5: 6 + 1 + 5 = 12. Spaces count.

(A) leaves out the space.
(C) counts one character too many (the last index is 11).
(D) is the length of split alone.

Question 2 (multiple choice · foundation)

What is the value of "computer".substring(2, 5)?

(A) "mput"
(B) "omp"
(C) "mpu"
(D) "mpute"

Answer and explanation

Answer: (C). The letters are c0 o1 m2 p3 u4 t5 e6 r7. The substring starts at index 2 and ends at index 4 (to - 1): m, p, u.

(A) includes index 5, as if the to index were included.
(B) starts at index 1.
(D) goes on to index 6.

Question 3 (multiple choice · core)

What is printed by the following statement?

System.out.println(1 + 2 + "3" + 4 + 5);

(A) 3345
(B) 15
(C) 12345
(D) 3354

Answer and explanation

Answer: (A). The operators go left to right. 1 + 2 is the int 3. 3 + "3" is the String "33". Then "33" + 4 is "334", and "334" + 5 is "3345".

(B) adds everything as numbers.
(C) joins every item as text from the start, forgetting that 1 + 2 is evaluated first as addition.
(D) is not produced by any order of evaluation.

Question 4 (multiple choice · core)

What is printed by the following code segment?

String w = "dog";
w.substring(1);
w = w + "s";
System.out.println(w);

(A) og
(B) ogs
(C) dog
(D) dogs

Answer and explanation

Answer: (D). A String is immutable. w.substring(1); computes "og" but the result is not stored, so w still refers to "dog". Then w = w + "s"; joins "dog" and "s", giving "dogs".

(A) and (B) assume that substring changed w.
(C) forgets the assignment w = w + "s";.

Question 5 (multiple choice · core)

What is displayed by the following statement?

System.out.println("mississippi".indexOf("ss") + "mississippi".indexOf("pp"));

(A) 13
(B) 10
(C) 2
(D) 8

Answer and explanation

Answer: (B). The letters are m(0) i(1) s(2) s(3) i(4) s(5) s(6) i(7) p(8) p(9) i(10). The first "ss" begins at index 2 and the first "pp" begins at index 8. The sum is 2 + 8 = 10.

(A) uses index 5, the second occurrence of "ss"; indexOf returns the first occurrence.
(C) is the first call alone.
(D) is the second call alone.

Question 6 (multiple choice · core)

Which statement about "pear".compareTo("peach") is true?

(A) It returns a value less than 0
(B) It returns a value greater than 0
(C) It returns 0
(D) It cannot be compared, because the strings have different lengths

Answer and explanation

Answer: (B). The first three letters, p, e and a, are the same. The fourth letters differ: r comes after c in the alphabet, so "pear" is greater than "peach" and the result is above 0.

(A) would be correct for the reverse comparison.
(C) is for identical strings.
(D) is wrong: strings of different lengths are compared normally.

Question 7 (multiple choice · core)

For String s = "hello"; (length 5), which call throws a StringIndexOutOfBoundsException?

(A) s.substring(3, 6)
(B) s.substring(5)
(C) s.substring(2, 5)
(D) s.substring(0, 5)

Answer and explanation

Answer: (A). The to index 6 is greater than the length 5, which is outside the allowed range.

(B) is legal: starting at the length gives the empty string.
(C) takes indices 2, 3 and 4: "llo".
(D) is the whole string.

Question 8 (multiple choice · core)

What is printed by the following statement?

System.out.println("Java".equals("java") + " " + "Java".equals("Java"));

(A) true true
(B) true false
(C) false false
(D) false true

Answer and explanation

Answer: (D). equals compares the characters exactly, including capital letters, so "Java" and "java" are not equal. The second call compares identical strings. The two boolean results are joined to a space as text.

(A) ignores the difference between J and j.
(B) reverses the two results.
(C) assumes that even identical strings are not equal.

Question 9 (extracting parts · core)

The variable full holds "Ada Lovelace".

(a) Write code using indexOf and substring that stores the first name in first and the last name in last.

(b) Write a statement that displays the last name preceded by the first initial and a full stop and a space: A. Lovelace.

Worked solution

String full = "Ada Lovelace";
int sp = full.indexOf(" ");
String first = full.substring(0, sp);
String last = full.substring(sp + 1);
System.out.println(first.substring(0, 1) + ". " + last);

(c) The characters are A0 d1 a2 and the space at index 3, so full.indexOf(" ") is 3. Then first is full.substring(0, 3), which is "Ada"; last is full.substring(4), which is "Lovelace". The first initial is first.substring(0, 1), which is "A". The output is A. Lovelace.

Suggested Marlbridge rubric (5 marks):

Mark	Awarded for
1	`indexOf(" ")` used to find the space (value 3)
1	`first` taken as `substring(0, sp)`
1	`last` taken as `substring(sp + 1)` (starts after the space)
1	Initial taken with `substring(0, 1)`
1	Pieces joined to give `A. Lovelace`

Question 10 (tracing strings · stretch)

Trace the following code segment and state the output.

String s = "marlbridge";
String a = s.substring(0, 4);
String b = s.substring(4);
s = b + a;
int k = s.indexOf("ma");
String c = s.substring(k, k + 2) + s.length();
System.out.println(s + " " + a + " " + k + " " + c);

Worked solution

The letters of marlbridge are m0 a1 r2 l3 b4 r5 i6 d7 g8 e9.

After statement	s	a	b	k	c
`String s = "marlbridge";`	`marlbridge`	-	-	-	-
`String a = s.substring(0, 4);`	`marlbridge`	`marl`	-	-	-
`String b = s.substring(4);`	`marlbridge`	`marl`	`bridge`	-	-
`s = b + a;`	`bridgemarl`	`marl`	`bridge`	-	-
`int k = s.indexOf("ma");`	`bridgemarl`	`marl`	`bridge`	6	-
`String c = s.substring(k, k + 2) + s.length();`	`bridgemarl`	`marl`	`bridge`	6	`ma10`

In bridgemarl the letters are b0 r1 i2 d3 g4 e5 m6 a7 r8 l9, so "ma" is at index 6, and the length is 10. s.substring(6, 8) is "ma", and joining the int 10 gives "ma10". The output is bridgemarl marl 6 ma10.

Suggested Marlbridge rubric (5 marks):

Mark	Awarded for
1	`a` is `marl` and `b` is `bridge`
1	`s` becomes `bridgemarl` (a new string; `a` and `b` are unchanged)
1	`k` is 6
1	`c` is `ma10` (substring then the length joined as text)
1	Output line exactly `bridgemarl marl 6 ma10`

Question 11 (a method with a precondition · stretch)

A class method removes the first and last characters of a string.

public class Trim {
    /**
     * Returns s without its first and last characters.
     * Precondition: s.length() >= 2
     * Postcondition: the value returned has length s.length() - 2
     */
    public static String middle(String s) {
        return s.substring(1, s.length() - 1);
    }

    public static void main(String[] args) {
        System.out.println(middle("adventure"));
        System.out.println("[" + middle("ab") + "]");
    }
}

dventur
[]

(a) Explain why middle("adventure") returns dventur.

(b) The output for middle("ab") is the empty string. Explain why that is legal.

Worked solution

(a) The string has 9 characters (a0 d1 v2 e3 n4 t5 u6 r7 e8), so s.length() - 1 is 8. substring(1, 8) takes indices 1 to 7: d, v, e, n, t, u, r, which is dventur.

(b) For "ab" the call is substring(1, 1). Equal indices give the empty string, which is a valid result with length 0 = 2 - 2.

(c) For "a" the call is substring(1, 0): the from index is greater than the to index, so a StringIndexOutOfBoundsException is thrown. The precondition s.length() >= 2 is broken, and the method is not expected to check it, so the exception is the caller’s responsibility.

Suggested Marlbridge rubric (5 marks):

Mark	Awarded for
1	(a) length 9 and the range of indices 1 to 7
1	(a) result `dventur`
1	(b) `substring(1, 1)` gives an empty string
1	(c) `substring(1, 0)` throws a `StringIndexOutOfBoundsException`
1	(c) links this to the broken precondition, with the caller responsible

How did you do?

Questions 1, 2, 5 and 7: write the string with an index under every character, then apply the method’s rule.
Questions 3 and 8: re-read “Joining strings with + and +=” and “Comparing strings” in the study guide.
Questions 4 and 10: re-read “Strings cannot be changed”. Every method returns a new value.
Question 6: find the first position where the two strings differ, then compare those letters.
Questions 9 and 11: practise extracting parts of three other strings.

Then try the checklist.