one-dimensional (1D)

The use of arrays (or equivalent) when solving problems: one-dimensional (1D)

2.2.3 Additional Programming Techniques

Additional Programming Techniques — Note 6 of 12

Arrays as fixed length or static structures

What you need to know

A one-dimensional (1D) array is a data structure that stores a fixed-size collection of elements of the same data type under a single name. Each element is accessed using an index (position number), starting from 0. Arrays are static, meaning their size cannot change once they are created.

What to know

Arrays store multiple items of the same data type under one name.
Elements are accessed using an index, starting from 0 (zero-indexed).
![1D array indexes from 0, and a 2D array where scores[1][2] is week 2, Wednesday](/knowledge/unit-2-knowledge-images/arrays-1d-2d.svg)
Arrays are static (fixed size): their length is set when created and cannot change.
Declaring: array names[5] creates an array with 5 elements (index 0 to 4).
Assigning: names[0] = "Alice" puts "Alice" at position 0.
Arrays are useful for storing lists of data, e.g. scores, names, or colours.
You can loop through an array using a FOR loop: for count = 0 to theTeam.length - 1.
Python lists are used to mimic arrays. Lists can change size, but in exam answers treat an array as a fixed size.
len(scores) gives the number of items, so for i in range(len(scores)): visits every index. This is safer than hard-coding range(3), which breaks if the array changes size.
FOUND FLAG: set found = False before the loop, set it to True inside the loop when a match is found, then report the result AFTER the loop has finished.

Exam technique

Tip:A FOR loop iterating through an array should use the loop counter as the INDEX: theTeam[count], NOT theTeam[i] if the counter is called count.
Tip:Off-by-one errors are common. If the array has 6 elements (index 0-5), the loop should go 0 to 5. Python's range(6) gives 0-5.
Tip:theTeam.length returns the number of elements. To loop from 0, use for i = 0 to theTeam.length - 1.
Tip:scores = [3, 6, 6, 9, 2, 8] → scores[2] = 6 (third element, 0-indexed). Remember arrays are ZERO-INDEXED.
Mistake:Putting else: print("Not found") inside the loop prints 'Not found' once for every item that doesn't match.

Example:Linear search fills: for count = 0 to theTeam.length - 1; if theTeam[count] == studentName then return True. Use the COUNTER as the array INDEX.

Code examples

1D array operations

python

# Create and access a 1D array (list in Python)
scores = [10, 20, 30, 40, 50]

print(scores[0])   # 10 (first element)
print(scores[2])   # 30 (third element)
scores[1] = 25     # Update second element

# Loop through the array
for i in range(len(scores)):
    print("Index " + str(i) + ": " + str(scores[i]))

Array basics: define, access, loop, change

python

scores = [4, 7, 8]     # define an array

print(scores[1])        # 7: index 1 is the SECOND element

for i in range(len(scores)):   # index 0, 1, 2
    print(scores[i])           # print each item

scores[0] = 10          # replace the first element (index 0) with 10

Total and average of an array

python

scores = [4, 7, 8]
total = 0                      # running total

for i in range(len(scores)):   # loop through each index
    total += scores[i]         # add each score to the total

average = total / len(scores)
print("Total:", total)         # Total: 19
print("Average:", average)     # Average: 6.333...

Check if a value appears in an array

python

scores = [12, 7, 5, 10, 3, 2, 7]
target = 7       # the value we are searching for
found = False    # flag: nothing found yet

for i in range(len(scores)):
    if scores[i] == target:
        found = True   # set the flag when a match is found

if found:              # checked AFTER the loop
    print("Found")
else:
    print("Not found")

Count how many times a value appears

python

scores = [12, 5, 10, 6, 7, 12]
target = 12
count = 0        # how many times the target appears

for i in range(len(scores)):
    if scores[i] == target:
        count += 1   # increase the count on each match

print("Appears", count, "times")   # Appears 2 times

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# Create and access a 1D array (list in Python) scores = [10, 20, 30, 40, 50] print(scores[0]) # 10 (first element) print(scores[2]) # 30 (third element) scores[1] = 25 # Update second element # Loop through the array for i in range(len(scores)): print("Index " + str(i) + ": " + str(scores[i]))

scores = [4, 7, 8] # define an array print(scores[1]) # 7: index 1 is the SECOND element for i in range(len(scores)): # index 0, 1, 2 print(scores[i]) # print each item scores[0] = 10 # replace the first element (index 0) with 10

scores = [4, 7, 8] total = 0 # running total for i in range(len(scores)): # loop through each index total += scores[i] # add each score to the total average = total / len(scores) print("Total:", total) # Total: 19 print("Average:", average) # Average: 6.333...

scores = [12, 7, 5, 10, 3, 2, 7] target = 7 # the value we are searching for found = False # flag: nothing found yet for i in range(len(scores)): if scores[i] == target: found = True # set the flag when a match is found if found: # checked AFTER the loop print("Found") else: print("Not found")

scores = [12, 5, 10, 6, 7, 12] target = 12 count = 0 # how many times the target appears for i in range(len(scores)): if scores[i] == target: count += 1 # increase the count on each match print("Appears", count, "times") # Appears 2 times