Check if Number is Prime in DSA C++ - Execution Trace

Concept Flow - Check if Number is Prime

Start with number n

↓

Check if n <= 1?

Yes→Not Prime, Stop

No↓

Set divisor = 2

↓

Is divisor * divisor <= n?

No→Prime, Stop

Yes↓

Check if n % divisor == 0?

Yes→Not Prime, Stop

No↓

Increment divisor by 1

↩Back to divisor check

Start with the number, check divisors from 2 up to sqrt(n). If any divides n, it's not prime; else it is prime.

Execution Sample

DSA C++

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

This code checks if a number n is prime by testing divisors from 2 up to sqrt(n).

Execution Table

Step	Operation	Divisor (i)	Condition (i*i <= n)	Check n % i == 0	Result	Reason/Action
1	Check if n <= 1	-	-	-	False	n=11 > 1, continue
2	Set divisor i=2	2	2*2=4 <= 11 True	11 % 2 == 0? False	Continue	Increment i
3	Check divisor i=3	3	3*3=9 <= 11 True	11 % 3 == 0? False	Continue	Increment i
4	Check divisor i=4	4	4*4=16 <= 11 False	-	Prime	No divisor found, 11 is prime
5	End	-	-	-	-	Stop checking

💡 Loop ends when divisor squared exceeds n; no divisor found means n is prime.

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 4	Final
n	11	11	11	11	11
i (divisor)	-	2	3	4	4
Condition i*i <= n	-	True	True	False	False
n % i == 0	-	False	False	-	-
Result	-	Continue	Continue	Prime	Prime

Key Moments - 3 Insights

Why do we check divisors only up to i*i <= n instead of i <= n?

Because if n has a divisor larger than sqrt(n), it must have a smaller paired divisor already checked. See execution_table rows 3 and 4 where i=4 stops the loop.

What happens if n is less than or equal to 1?

The function immediately returns false as numbers <=1 are not prime. This is shown in execution_table row 1.

Why do we return false immediately when n % i == 0?

Because finding any divisor means n is not prime. This is checked in each loop iteration, and if true, the function stops early.

Visual Quiz - 3 Questions

Look at the execution_table, what is the divisor i when the loop stops?

A 4

B 3

C 2

D 5

💡 Hint

Check the 'Divisor (i)' and 'Condition (i*i <= n)' columns in rows 3 and 4.

At which step does the code decide that 11 is prime?

A Step 2

B Step 3

C Step 4

D Step 5

💡 Hint

Look at the 'Result' and 'Reason/Action' columns in the execution_table.

If n was 9 instead of 11, at which divisor i would the function return false?

A 2

B 3

C 4

D 5

💡 Hint

Recall that 9 % 3 == 0, so check when divisor i equals 3.

Concept Snapshot

Check if number n is prime:
- If n <= 1, not prime
- Test divisors i from 2 to sqrt(n)
- If n % i == 0, not prime
- If no divisor found, n is prime
- Stop checking when i*i > n

Full Transcript

This concept checks if a number is prime by testing divisors from 2 up to the square root of the number. If any divisor divides the number evenly, it is not prime. Otherwise, it is prime. The process stops early if a divisor is found or when the divisor squared exceeds the number.

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