Continue statement in C++ - Time & Space Complexity

Time Complexity: Continue statement

O(n)

Understanding Time Complexity

We want to see how using the continue statement affects how long a program takes to run.

Specifically, does skipping some steps change the overall work done as input grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

for (int i = 0; i < n; i++) {
    if (i % 2 == 0) {
        continue;
    }
    // Some constant time work here
    int x = i * 2;
}
    

This code loops through numbers from 0 to n-1, skipping even numbers using continue.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop runs from 0 to n-1.
• How many times: It runs n times, but the work inside after continue runs only for odd i values.

How Execution Grows With Input

As n grows, the loop runs n times, but the work after continue happens only about half the time.

Input Size (n)	Approx. Operations
10	10 loop checks, 5 work executions
100	100 loop checks, 50 work executions
1000	1000 loop checks, 500 work executions

Pattern observation: The total loop runs grow linearly with n, and the work inside grows roughly half as fast.

Final Time Complexity

Time Complexity: O(n)

This means the program's running time grows in a straight line as the input size increases, even with the continue skipping some steps.

Common Mistake

[X] Wrong: "Using continue makes the loop run faster and changes the time complexity to less than linear."

[OK] Correct: The loop still runs n times; skipping some steps only reduces work by a constant factor, not the overall growth pattern.

Interview Connect

Understanding how control statements like continue affect time helps you explain your code clearly and reason about efficiency in real projects.

Self-Check

"What if the continue condition depended on a nested loop? How would that change the time complexity?"

Practice

1. What does the continue statement do inside a loop in C++?

easy

A. It stops the entire loop immediately.

B. It skips the rest of the current loop iteration and moves to the next iteration.

C. It exits the current function.

D. It restarts the program from the beginning.

Solution

1. Step 1: Understand the role of continue in loops

   The continue statement tells the program to skip the remaining code in the current loop cycle.

2. Step 2: Identify what happens after continue

   After skipping, the loop moves to the next iteration immediately without executing the rest of the loop body.

3. Final Answer:

   It skips the rest of the current loop iteration and moves to the next iteration. -> Option B

4. Quick Check:

   Continue skips current iteration [OK]

Hint: Continue skips current loop step, not whole loop [OK]

Common Mistakes:

• Thinking continue stops the loop completely
• Confusing continue with break
• Assuming continue exits the function

2. Which of the following is the correct syntax to use continue inside a for loop in C++?

easy

A. continue;

B. continue()

C. continue{}

D. continue[]

Solution

1. Step 1: Recall the syntax of continue in C++

   The continue statement is a keyword and used alone followed by a semicolon.

2. Step 2: Check each option's syntax

   Only continue; is valid. The others use parentheses or braces which are incorrect.

3. Final Answer:

   continue; -> Option A

4. Quick Check:

   Continue ends with semicolon [OK]

Hint: Continue is a keyword, ends with semicolon [OK]

Common Mistakes:

• Adding parentheses after continue
• Using braces or brackets with continue
• Omitting the semicolon

3. What will be the output of the following C++ code?

for (int i = 1; i <= 5; i++) {
    if (i == 3) continue;
    std::cout << i << " ";
}

medium

A. 1 2 3 4 5

B. 1 2 3 4

C. 1 2 4 5

D. 2 3 4 5

Solution

1. Step 1: Analyze the loop and condition

   The loop runs from 1 to 5. When i == 3, the continue skips printing.

2. Step 2: Determine printed values

   Numbers 1 and 2 print normally. At 3, continue skips printing. Then 4 and 5 print.

3. Final Answer:

   1 2 4 5 -> Option C

4. Quick Check:

   Skip 3, print others [OK]

Hint: Continue skips printing 3, others print [OK]

Common Mistakes:

• Including 3 in output
• Stopping loop at 3
• Printing numbers after 3 incorrectly

4. Identify the error in the following C++ code snippet:

for (int i = 0; i < 5; i++) {
    if (i == 2)
        continue
    std::cout << i << " ";
}

medium

A. No error in the code.

B. Incorrect loop condition.

C. Wrong variable type for i.

D. Missing semicolon after continue statement.

Solution

1. Step 1: Check syntax of continue statement

   The continue statement must end with a semicolon.

2. Step 2: Identify missing semicolon

   In the code, continue is not followed by a semicolon, causing a syntax error.

3. Final Answer:

   Missing semicolon after continue statement. -> Option D

4. Quick Check:

   Continue needs semicolon [OK]

Hint: Always put semicolon after continue [OK]

Common Mistakes:

• Forgetting semicolon after continue
• Assuming continue needs parentheses
• Misreading loop condition as error

5. Consider the following code that sums numbers from 1 to 10 but skips multiples of 3 using continue. What is the final sum?

int sum = 0;
for (int i = 1; i <= 10; i++) {
    if (i % 3 == 0) continue;
    sum += i;
}
std::cout << sum;

hard

A. 52

B. 55

C. 45

D. 50

Solution

1. Step 1: Identify numbers skipped by continue

   Numbers divisible by 3 (3, 6, 9) are skipped and not added to sum.

2. Step 2: Calculate sum of numbers 1 to 10 excluding 3, 6, 9

   Sum 1+2+4+5+7+8+10 = 37 + 15 = 52

3. Final Answer:

   52 -> Option A

4. Quick Check:

   Sum excluding multiples of 3 = 52 [OK]

Hint: Skip multiples of 3, sum others [OK]

Common Mistakes:

• Including multiples of 3 in sum
• Adding all numbers 1 to 10
• Misunderstanding continue effect