Check if Number is Prime in DSA C++ - Build from Scratch

Check if Number is Prime

📖 Scenario: You are helping a friend who wants to know if a number is prime. A prime number is a number greater than 1 that has no divisors other than 1 and itself.We will write a simple program to check if a given number is prime.

🎯 Goal: Build a program that checks if a number is prime and prints "Prime" if it is, or "Not Prime" if it is not.

📋 What You'll Learn

Create an integer variable called number with the value 29

Create a boolean variable called isPrime and set it to true

Use a for loop with variable i from 2 to number - 1 to check divisibility

If number is divisible by i, set isPrime to false and break the loop

Print "Prime" if isPrime is true, otherwise print "Not Prime"

💡 Why This Matters

🌍 Real World

Checking if a number is prime is useful in cryptography, coding theory, and computer security.

💼 Career

Understanding prime numbers and loops is fundamental for programming jobs, especially in algorithm design and software development.

1

Create the number variable

Create an integer variable called number and set it to 29.

DSA C++

# Your code here

Hint

Use int number = 29; to create the variable.

2

Create the isPrime variable

Create a boolean variable called isPrime and set it to true.

DSA C++

int number = 29;
# Your code here

Hint

Use bool isPrime = true; to create the variable.

3

Check divisibility with a for loop

Use a for loop with variable i from 2 to number - 1. Inside the loop, if number % i == 0, set isPrime to false and break the loop.

DSA C++

int number = 29;
bool isPrime = true;
// Your code here

Hint

Use a for loop and check if number is divisible by i.

4

Print the result

Print "Prime" if isPrime is true, otherwise print "Not Prime".

DSA C++

int number = 29;
bool isPrime = true;
for (int i = 2; i < number; i++) {
    if (number % i == 0) {
        isPrime = false;
        break;
    }
}
// Your code here

Hint

Use std::cout to print the result based on isPrime.

Practice

1. What does it mean if a number is prime?

easy

A. It has exactly two distinct positive divisors: 1 and itself.

B. It is divisible by 2 only.

C. It is an even number greater than 2.

D. It has more than two divisors.

Solution

1. Step 1: Understand the definition of prime number

   A prime number is a number greater than 1 that has no divisors other than 1 and itself.

2. Step 2: Compare options with definition

   Only It has exactly two distinct positive divisors: 1 and itself. correctly states that a prime number has exactly two distinct positive divisors: 1 and itself.

3. Final Answer:

   It has exactly two distinct positive divisors: 1 and itself. -> Option A

4. Quick Check:

   Prime number = two divisors [OK]

Hint: Prime means divisible only by 1 and itself [OK]

Common Mistakes:

• Thinking all even numbers are prime
• Confusing prime with odd numbers
• Assuming 1 is prime

2. Which of the following C++ code snippets correctly checks if a number n is prime?

easy

A. for (int i = 2; i < n; i++) { if (n % i == 0) return false; } return true;

B. for (int i = 2; i <= n; i++) { if (n % i == 0) return false; } return true;

C. for (int i = 1; i < n; i++) { if (n % i == 0) return false; } return true;

D. for (int i = 2; i < n/2; i++) { if (n % i == 0) return true; } return false;

Solution

1. Step 1: Analyze loop range for prime check

   We check divisors from 2 up to but not including n to find any divisor.

2. Step 2: Check correctness of each option

   for (int i = 2; i < n; i++) { if (n % i == 0) return false; } return true; correctly checks divisors from 2 to n-1 and returns false if divisible, else true. for (int i = 2; i <= n; i++) { if (n % i == 0) return false; } return true; incorrectly includes n itself, causing false negatives. for (int i = 1; i < n; i++) { if (n % i == 0) return false; } return true; starts from 1, which divides all numbers. for (int i = 2; i < n/2; i++) { if (n % i == 0) return true; } return false; returns true on first divisor found, which is incorrect logic.

3. Final Answer:

   for (int i = 2; i < n; i++) { if (n % i == 0) return false; } return true; -> Option A

4. Quick Check:

   Loop from 2 to n-1 and check divisibility [OK]

Hint: Check divisors from 2 to n-1 only [OK]

Common Mistakes:

• Checking divisibility by 1 or n itself
• Looping beyond n
• Returning true too early

3. What is the output of the following C++ code?

int n = 11;
bool isPrime = true;
for (int i = 2; i <= n / 2; i++) {
    if (n % i == 0) {
        isPrime = false;
        break;
    }
}
std::cout << (isPrime ? "Prime" : "Not Prime") << std::endl;

medium

A. Runtime Error

B. Not Prime

C. Compilation Error

D. Prime

Solution

1. Step 1: Understand the loop and condition

   The loop checks divisors from 2 to n/2 (which is 5 for n=11). It tests if 11 is divisible by any number in this range.

2. Step 2: Check divisibility for each i

   11 % 2 != 0, 11 % 3 != 0, 11 % 4 != 0, 11 % 5 != 0. No divisor found, so isPrime remains true.

3. Final Answer:

   Prime -> Option D

4. Quick Check:

   11 has no divisors except 1 and itself [OK]

Hint: Check divisors only up to n/2 for efficiency [OK]

Common Mistakes:

• Looping up to n instead of n/2
• Not breaking loop on divisor found
• Misinterpreting output

4. Find the error in this C++ code that checks if n is prime:

bool isPrime(int n) {
    if (n <= 1) return false;
    for (int i = 2; i < n; i++) {
        if (n % i == 0) return true;
    }
    return false;
}

medium

A. Loop should start from 1, not 2.

B. Returns true when a divisor is found, should return false instead.

C. Function should return true for n <= 1.

D. No error, code is correct.

Solution

1. Step 1: Analyze return values on divisor found

   The code returns true immediately when a divisor is found, which means it claims the number is prime incorrectly.

2. Step 2: Correct logic for prime check

   When a divisor is found, the function should return false (not prime). If no divisor is found, return true.

3. Final Answer:

   Returns true when a divisor is found, should return false instead. -> Option B

4. Quick Check:

   Return false on divisor found [OK]

Hint: Return false when divisor found, true otherwise [OK]

Common Mistakes:

• Returning true on divisor found
• Loop starting at 1
• Not handling n <= 1 correctly

5. You want to optimize prime checking for large numbers by checking divisors only up to the square root of n. Which code snippet correctly implements this?

bool isPrime(int n) {
    if (n <= 1) return false;
    for (int i = 2; i <= sqrt(n); i++) {
        if (n % i == 0) return false;
    }
    return true;
}

hard

A. Incorrect: returns true inside the loop when divisor found.

B. Incorrect: loop should go up to n/2, not sqrt(n).

C. Correct: checks divisors up to sqrt(n) and returns false if divisible.

D. Incorrect: should start loop from 1 instead of 2.

Solution

1. Step 1: Understand optimization using square root

   Any divisor larger than sqrt(n) pairs with a smaller divisor, so checking up to sqrt(n) is enough.

2. Step 2: Verify code logic

   The loop runs from 2 to sqrt(n), returns false if divisor found, else true after loop ends. This is correct and efficient.

3. Final Answer:

   Correct: checks divisors up to sqrt(n) and returns false if divisible. -> Option C

4. Quick Check:

   Check divisors up to sqrt(n) for efficiency [OK]

Hint: Check divisors only up to sqrt(n) for speed [OK]

Common Mistakes:

• Looping up to n/2 instead of sqrt(n)
• Starting loop at 1
• Returning true inside loop on divisor found