Computational Thinking · Quiz 2 · Jan 2026

This is a reference card showing the schemas and sample rows of Scores, Words, Library, Olympics and Shopping Bills. It supplies context for later pseudocode questions rather than asking for a computed result. Use the field names and row meanings when tracing the later algorithms; the displayed ellipses indicate omitted rows, not a complete dataset. **Answer: A.** A: **Correct:** The source selects the acknowledgement that useful dataset information is supplied. B: **Incorrect:** This describes its reference role too, but is not the source's selected administrative acknowledgement; these are not competing computational answers.

process returns the last element of each inner list, giving Z, then Y, then X. The statement outList=[Z]+outList prepends each new result rather than appending it. Successive outList values are [Z], [Y,Z], and [X,Y,Z]. Hence the final order is ['X','Y','Z']. **Answer: B.** A: **Incorrect:** Would result from appending each value rather than prepending it. B: **Correct:** Correctly reverses the sequence of last elements by repeated prepending. C: **Incorrect:** Does not match any step of the prepend trace. D: **Incorrect:** Does not match the order produced by repeated prepending.

checkPattern keeps the preceding character in prev and visits each following character as curr. It returns True when prev is a vowel and curr is a consonant, so it detects an adjacent vowel-consonant pair anywhere in a word. The outer cnt increases once per qualifying word, because the procedure returns immediately after the first match. Words shorter than two letters cannot qualify. **Answer: C.** A: **Incorrect:** The second member must be a consonant, not a vowel. B: **Incorrect:** Reverses the required order of the two character types. C: **Correct:** Precisely describes an adjacent vowel followed by a consonant. D: **Incorrect:** The search is over every adjacent pair, not just the two ends.

The first pass stores each city's minimum Mathematics score: replace D[city] only when a smaller score is found. The second pass finds the minimum among these city minima. When a smaller value is found, B becomes that value and A resets to 1; when a city equals the current B, A increases. Thus B is the global lowest score and A counts the cities containing at least one student with that score. The initial B=100 assumes marks are at most 100, as in this dataset. **Answer: C.** A: **Incorrect:** The comparisons track minima, and A is a count rather than a list of cities. B: **Incorrect:** The comparisons find the lowest, not highest, marks. C: **Correct:** Correctly identifies the minimum score and the number of cities attaining it. D: **Incorrect:** A can be one; neither claimed property is ensured.

doSomething builds a list of first occurrences, using a dictionary to prevent duplicate words. The outer code collects words until a full stop, removes duplicates within that sentence, appends the resulting list as one element of sList, and resets wList. Therefore sList is a list of per-sentence unique-word lists, preserving first occurrence order within each. It assumes sentences end with the specified full-stop marker. **Answer: D.** A: **Incorrect:** sList contains nested lists rather than one flat word list. B: **Incorrect:** Duplicates are removed per sentence, not globally into one flat list. C: **Incorrect:** The per-sentence words are deduplicated by doSomething. D: **Correct:** Matches both the sentence grouping and within-sentence deduplication.

Each bill's item prices are summed into billTotal. globalSum/billCount computes the average bill value. Separately, CustomerTotal adds all bill totals for each customer. The final loop increments Val once for each customer whose accumulated spending exceeds the average single-bill value. These are different aggregation levels: customer totals are compared with a bill-level average. **Answer: A.** A: **Correct:** Exactly matches the final loop over customer names. B: **Incorrect:** The final loop visits customers, not individual bills. C: **Incorrect:** Val increments by 1, so it counts people rather than summing their expenditure. D: **Incorrect:** CustomerTotal sums bills; it does not keep a customer's maximum bill.

Invert the registration relation: for every student, visit every course in that student's list and append that student's name to the course's roster. Initialise a new roster once, then append for later students. A and C differ only in which branch handles the first occurrence. For example Math receives Rahul and Amar, while Physics receives Rahul and Sita. **Answer: A, C.** A: **Correct:** Initialises a course once and accumulates every registered student. B: **Incorrect:** Overwrites the roster on every visit, losing earlier students. C: **Correct:** Uses the same correct logic with the new-course case first. D: **Incorrect:** Treats Registration's student-name keys as course names and its course lists as student lists, reversing their actual meanings.

Deleting the first instance requires remembering whether a deletion has already happened, or using the stated remove operation that deletes only the first occurrence. A skips the first X and then preserves everything else. C likewise skips the first X, retaining subsequent copies. D checks existence and calls remove once. These operations preserve the relative order of all retained values, hence preserve sorted order. **Answer: A, C, D.** A: **Correct:** The found flag ensures exactly one X is skipped. B: **Incorrect:** Drops every X rather than only the first. C: **Correct:** The first X sets found without appending; subsequent X values are retained. D: **Correct:** The stipulated remove deletes one first occurrence and the membership test makes absence safe.

The starting dictionary already has alphaDict['t']=2. The word 'reluctant' contains t twice: at positions 6 and 9 when counted from 1. Each occurrence increments the existing t value by one. Therefore the final value is 2+2=4. The procedure adds to initial counts rather than resetting them for the new word. **Answer: 4.**

Trace each value using its updated D[k] for the final threshold test. A:10 adds 10, halves to 5, then subtracts 5, net +5. B:25 subtracts 25, updates to 30, then adds 10, net -15. C:30 adds 30, halves to 15, then adds 10, net +40. D:40 adds 40, halves to 20, then adds 10, net +50. E:18 subtracts 18, updates to 23, then adds 10, net -8. Total=5-15+40+50-8=72. Each key is processed once. **Answer: 72.**

For a new country the algorithm creates an empty player list. It increments count only when the current player is not yet in that country's list, then adds that player. Repeated medals or years for the same country-player pair do not increment it again. The same name under a different country would be a distinct pair. Thus count measures unique country-player pairs. **Answer: B.** A: **Incorrect:** Repeated rows for an existing pair do not increment count. B: **Correct:** Counts one first occurrence of each country-player pair. C: **Incorrect:** A country can contribute several different players, so this is not merely a country count. D: **Incorrect:** Repeated medals for one player-country pair are deduplicated.

Removing the membership test means every row appends its player name to the country's list. A player appearing in several event or year rows is therefore appended several times. The dictionary still groups by country, but its lists no longer deduplicate players. **Answer: B.** A: **Incorrect:** Uniqueness depended on the removed membership check. B: **Correct:** Every matching row contributes a name, including repeated player names. C: **Incorrect:** Appending extends the list rather than replacing it with a single name. D: **Incorrect:** The stored values remain lists of names, not numeric counts.