Week 32026-06-04

Practice, Debugging, and Arrays

Loop tracing · Function review · Reading many values · Array basics

Part 1 — Week 2 Practice Review

We will review last week by doing small exercises first, then comparing against a clean answer.

Problem 1 — Factorial

Practice first

Trace factorial(5): write down result after each loop.
Fill the two blanks in the C function below.
Check the edge case: what should factorial(0) return?

factorial_practice.c

int factorial(int n) {    int result = ____;    for (int i = 1; i <= n; i++) {        result = result * ____;    }    return result;}

i	1	2	3	4	5
result	1	2	6	24	120

Reference answer

factorial.c

#include <stdio.h>int factorial(int n) {    int result = 1;    for (int i = 1; i <= n; i++) {        result *= i;    }    return result;}int main() {    int n;    scanf("%d", &n);    printf("%d\n", factorial(n));    return 0;}

factorial.py

def factorial(n):    result = 1    for i in range(1, n + 1):        result *= i    return resultn = int(input())print(factorial(n))

Problem 2 — Reverse Integer

Practice first

For n = 120030, fill digit, result, and n after each loop.
Then fill the missing update line in the function.
Run the same thinking for n = 0.

reverse_practice.c

int reverse_int(int n) {    int result = 0;    while (n > 0) {        int digit = n % 10;        result = __________________;        n = n / 10;    }    return result;}

Round	digit	result	n
1	0	0	12003
2	3	3	1200
3	0	30	120
4	0	300	12
5	2	3002	1
6	1	30021	0

Reference answer

reverse.c

#include <stdio.h>int reverse_int(int n) {    int result = 0;    while (n > 0) {        int digit = n % 10;        result = result * 10 + digit;        n /= 10;    }    return result;}int main() {    int n;    scanf("%d", &n);    printf("%d\n", reverse_int(n));    return 0;}

reverse.py

def reverse_int(n):    result = 0    while n > 0:        digit = n % 10        result = result * 10 + digit        n //= 10    return resultn = int(input())print(reverse_int(n))

Problem 3 — Primes in a Range

Practice first

For input 14 20, mark each number as prime or not prime.
For input 8 10, decide why the output is NONE.
Fill the two TODO parts in the C code.

YES

primes_range_practice.c

int is_prime(int n) {    for (int i = 2; i * i <= n; i++) {        if (__________) {            return 0;        }    }    return 1;}int count = 0;for (int x = L; x <= R; x++) {    if (is_prime(x)) {        printf("%d\n", x);        __________;    }}if (count == 0) {    printf("NONE\n");}

Reference answer

primes_range.c

#include <stdio.h>int is_prime(int n) {    if (n < 2) {        return 0;    }    for (int i = 2; i * i <= n; i++) {        if (n % i == 0) {            return 0;        }    }    return 1;}int main() {    int L, R;    scanf("%d %d", &L, &R);    int count = 0;    for (int x = L; x <= R; x++) {        if (is_prime(x)) {            printf("%d\n", x);            count++;        }    }    if (count == 0) {        printf("NONE\n");    }    return 0;}

primes_range.py

def is_prime(n):    if n < 2:        return False    i = 2    while i * i <= n:        if n % i == 0:            return False        i += 1    return TrueL, R = map(int, input().split())count = 0for x in range(L, R + 1):    if is_prime(x):        print(x)        count += 1if count == 0:    print("NONE")

Part 2 — Slow Down and Debug

This week is intentionally slower. Before learning more syntax, we need to make loops, conditions, functions, and input/output feel reliable.

1. The workflow for every problem

When a program does not work, the problem is usually not the C language itself. It is usually one of these: the input was misunderstood, the loop range is wrong, a variable was not initialized, or the output format does not match.

1. Read

What is given?

2. Trace

Try one sample by hand.

3. Code

Write the smallest loop.

4. Test

Check edge cases.

Rule: Do not write a full solution first. Write one loop, test it, then wrap it into a function only after the loop is correct.

2. Trace before you code

Tracing means writing down how variables change after each loop iteration. This catches many mistakes before the compiler does.

factorial_trace.c

int n = 5;int result = 1;for (int i = 1; i <= n; i++) {    result = result * i;}

i	Before update	After update
1	result = 1	result = 1
2	result = 1	result = 2
3	result = 2	result = 6
4	result = 6	result = 24
5	result = 24	result = 120

3. Common beginner bugs

Bug	Symptom	Fix
int sum;	The answer changes randomly.	int sum = 0;
i < n	The last value is missing.	i <= n
scanf("%d", n)	The program may crash.	scanf("%d", &n)
if (x = 0)	The condition is not comparing.	if (x == 0)

Part 3 — Function Review

1. What should go into a function?

A function should do one useful job. In most homework problems, main should handle input/output, and helper functions should handle computation.

function_shape.c

return_type function_name(parameter_list) {    // compute something    return answer;}

read inputcompute answerprint outputtest edge cases

2. return is not printf

Many beginners print inside every function. That makes the function harder to reuse. Prefer returning the result, then printing it in main.

Hard to reuse

not_reusable.c

void square(int x) {    printf("%d\n", x * x);}

Reusable

reusable.c

int square(int x) {    return x * x;}

Rule: If the function name sounds like a calculation, return a value. If the function name sounds like an action, such as print_line, then void may be appropriate.

3. Review examples

review_functions.c

int is_even(int n) {    return n % 2 == 0;}int max2(int a, int b) {    if (a > b) {        return a;    }    return b;}int sum_to_n(int n) {    int sum = 0;    for (int i = 1; i <= n; i++) {        sum += i;    }    return sum;}

Part 4 — Array Basics

1. Why do we need arrays?

A variable stores one value. An array stores many values of the same type, using one name and an index.

Without array

many_variables.c

int score1 = 80;int score2 = 91;int score3 = 76;int score4 = 88;int score5 = 95;

With array

array.c

int scores[5] = {80, 91, 76, 88, 95};

2. Memory model: contiguous slots

An array is not just a convenient syntax. In C, it is a continuous block of memory. The animation below shows how declaration, indexing, reading, writing, and out-of-bounds access work.

Animated memory model

How int a[5] lives in memory

Current C line

int a[5];

CPU does

reserve stack space

Stack memory

Declare the array

1/7

array size = 5 × sizeof(int) = 20 bytes

0x1000

0x1004

0x1008

0x100C

0x1010

0x1014

a[0]4B

a[1]4B

a[2]4B

a[3]4B

a[4]4B

outside4B

other

reserved for a

not reserved

The stack reserves 5 adjacent int slots for a.

Each int takes 4 bytes in this example. The values are not meaningful until we write to them.

Legend

write

read

address focus

out of bounds

Teacher script: Every a[i] becomes an address calculation. Reading uses the value stored at that address. Writing changes the value stored there. In C, the compiler does not automatically stop you from using a[5], so the programmer must keep indexes inside bounds.

3. Index starts at 0

For an array of length 5, the valid indexes are 0, 1, 2, 3, and 4. There is no index 5.

array_index.c

int scores[5] = {80, 91, 76, 88, 95};printf("%d\n", scores[0]);  // 80printf("%d\n", scores[4]);  // 95// scores[5] is wrong: valid indexes are 0 to 4.

Hint: If the loop reads or prints every element of an array with length n, the loop is usually for (int i = 0; i < n; i++), not i <= n.

4. Read n values

The most common input pattern for arrays is: first read n, then read n values into the array.

read_array.c

#include <stdio.h>int main() {    int n;    int a[100];    scanf("%d", &n);    for (int i = 0; i < n; i++) {        scanf("%d", &a[i]);    }    for (int i = 0; i < n; i++) {        printf("%d\n", a[i]);    }    return 0;}

python

n = int(input())a = list(map(int, input().split()))for x in a:    print(x)

5. Common array computations

Most beginner array problems are still loop problems. The only new idea is that the loop reads from a[i].

array_functions.c

int sum_array(int a[], int n) {    int sum = 0;    for (int i = 0; i < n; i++) {        sum += a[i];    }    return sum;}int max_array(int a[], int n) {    int best = a[0];    for (int i = 1; i < n; i++) {        if (a[i] > best) {            best = a[i];        }    }    return best;}int count_at_least(int a[], int n, int threshold) {    int count = 0;    for (int i = 0; i < n; i++) {        if (a[i] >= threshold) {            count++;        }    }    return count;}

Check: For functions that use arrays in C, pass both the array and its length: int a[] and int n.

Part 5 — Assignments

Requirements

3 required problems. Estimated time: 1.5 – 2.5 hours.
Each problem requires a C version (.c) and a Python version (.py).
Use loops yourself. In Python, do not use built-in helpers such as sum, max, min, or list.index for these problems.
For C array problems, assume the maximum array length is 100 and declare arrays like int a[100];.
These are array problems. Read the values into an array/list first, then pass the array/list to the required functions.
Output format must match the samples exactly.

1Reverse Array

You are given n integers. Store them in an array, then print them in reverse order.

This problem forces you to keep the earlier values. If you only print while reading, you cannot print the input backwards.

Input format

text

First line: one integer n.Second line: n integers.Constraint: 1 ≤ n ≤ 100

Output format

text

One line containing the numbers in reverse order, separated by single spaces.

Sample 1

Input

text

510 20 30 40 50

Output

text

50 40 30 20 10

Sample 2

Input

text

43 3 8 1

Output

text

1 8 3 3

Sample 3

Input

text

Output

text

Required structure

reverse_array.c

#include <stdio.h>void print_reverse(int a[], int n) {    // TODO: print a[n-1], a[n-2], ..., a[0]    // Use one space between numbers and one newline at the end.}int main() {    int n;    int a[100];    scanf("%d", &n);    for (int i = 0; i < n; i++) {        scanf("%d", &a[i]);    }    print_reverse(a, n);    return 0;}

reverse_array.py

def print_reverse(a):    # TODO: print the list in reverse order    passn = int(input())a = list(map(int, input().split()))print_reverse(a)

Hint: The last valid index is n - 1. A reverse loop usually starts at i = n - 1 and continues while i >= 0.

2Scores Below Average

You are given n student scores. Store all scores in an array, compute the average, then print every score below the average in the original order.

This requires a second pass over the same data after the average is known. That is why the array is necessary.

Input format

text

First line: one integer n.Second line: n integers, the scores.Constraint: 1 ≤ n ≤ 100, 0 ≤ score ≤ 100

Output format

text

Print two lines:avg=AVERAGEbelow=SCORESIf no score is below average, print below=NONE.

Sample 1

Input

text

580 91 76 88 95

Output

text

avg=86.00below=80 76

Sample 2

Input

text

360 60 90

Output

text

avg=70.00below=60 60

Sample 3

Input

text

475 75 75 75

Output

text

avg=75.00below=NONE

Required structure

below_average.c

#include <stdio.h>int sum_array(int a[], int n) {    // TODO}void print_below_average(int a[], int n, double avg) {    // TODO: print "below=" followed by scores below avg.    // If there are none, print "below=NONE".}int main() {    int n;    int scores[100];    scanf("%d", &n);    for (int i = 0; i < n; i++) {        scanf("%d", &scores[i]);    }    int total = sum_array(scores, n);    double avg = (double) total / n;    printf("avg=%.2lf\n", avg);    print_below_average(scores, n, avg);    return 0;}

below_average.py

def sum_array(a):    # TODO    passdef print_below_average(a, avg):    # TODO: print "below=" followed by scores below avg.    # If there are none, print "below=NONE".    passn = int(input())scores = list(map(int, input().split()))total = sum_array(scores)avg = total / nprint(f"avg={avg:.2f}")print_below_average(scores, avg)

Hint: Use one loop to compute the sum. After avg is known, use a second loop to decide which stored scores are below avg. Keep a counter so you know whether to print NONE.

3Left Rotate Array

You are given an array and an integer k. Rotate the array left by k positions, then print the result.

Left rotation by 2 changes 10 20 30 40 50 into 30 40 50 10 20.

Input format

text

First line: one integer n.Second line: n integers.Third line: one non-negative integer k.Constraint: 1 ≤ n ≤ 100, 0 ≤ k ≤ 10000

Output format

text

One line containing the rotated array, separated by single spaces.

Sample 1

Input

text

510 20 30 40 502

Output

text

30 40 50 10 20

Sample 2

Input

text

41 2 3 40

Output

text

1 2 3 4

Sample 3

Input

text

41 2 3 45

Output

text

2 3 4 1

Sample 4

Input

text

Output

text

Required structure

left_rotate.c

#include <stdio.h>void left_rotate(int a[], int n, int k) {    // TODO: rotate a left by k positions}void print_array(int a[], int n) {    // TODO: print the array on one line}int main() {    int n, k;    int a[100];    scanf("%d", &n);    for (int i = 0; i < n; i++) {        scanf("%d", &a[i]);    }    scanf("%d", &k);    left_rotate(a, n, k);    print_array(a, n);    return 0;}

left_rotate.py

def left_rotate(a, k):    # TODO: return a new list rotated left by k positions    passdef print_array(a):    # TODO: print the list on one line    passn = int(input())a = list(map(int, input().split()))k = int(input())a = left_rotate(a, k)print_array(a)

Hint: First reduce k with k = k % n. One simple method: create a temporary array temp, set temp[i] = a[(i + k) % n], then copy temp back into a.

Optional challenge

After finishing the required problems, try rotating the array to the right by k positions instead of left. Think about how the index formula changes.

— Week 3 End —