An array Names[1:20] stores the names of 20 students in a class. Some names may appear more than once if students share the same name. (a) Write pseudocode ...

(a) The CountOccurrences function iterates through the entire array, comparing each element to the target name and incrementing a counter for each match: [3]

FUNCTION CountOccurrences(Target : STRING) RETURNS INTEGER
    DECLARE Count : INTEGER
    DECLARE i : INTEGER
    Count ← 0
    FOR i ← 1 TO 20
        IF Names[i] = Target THEN
            Count ← Count + 1
        ENDIF
    NEXT i
    RETURN Count
ENDFUNCTION

[1 mark for loop, 1 mark for comparison and counting, 1 mark for return]

(b) To find duplicates without reporting the same name twice, the algorithm checks earlier positions in the array. If the same name appears at a lower index, it has already been reported: [4]

DECLARE i : INTEGER
DECLARE j : INTEGER
DECLARE AlreadyReported : BOOLEAN

FOR i ← 1 TO 20
    IF CountOccurrences(Names[i]) > 1 THEN
        AlreadyReported ← FALSE
        FOR j ← 1 TO i - 1
            IF Names[j] = Names[i] THEN
                AlreadyReported ← TRUE
            ENDIF
        NEXT j
        IF AlreadyReported = FALSE THEN
            OUTPUT Names[i], " appears ", CountOccurrences(Names[i]), " times"
        ENDIF
    ENDIF
NEXT i

The inner loop from 1 to i-1 checks whether Names[i] appeared earlier. If it did, this occurrence is a later duplicate and skipping it prevents the same name from being output more than once. [1 mark for checking count > 1, 1 mark for preventing duplicate output, 1 mark for checking earlier positions, 1 mark for correct output]

(c) CountOccurrences("Aisha") returns 3, because "Aisha" appears at positions 3, 8, and 15 in the array, giving a count of three matches. [1]