Array Traversal Patterns in DSA C# - Time & Space Complexity

Time Complexity: Array Traversal Patterns

O(n)

Understanding Time Complexity

We want to understand how long it takes to go through all items in an array.

How does the time grow when the array gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int SumArray(int[] arr) {
    int sum = 0;
    for (int i = 0; i < arr.Length; i++) {
        sum += arr[i];
    }
    return sum;
}
    

This code adds up all numbers in the array by visiting each element once.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Accessing each element of the array once.
• How many times: Exactly once for each element, so as many times as the array length.

How Execution Grows With Input

As the array gets bigger, the number of steps grows directly with its size.

Input Size (n)	Approx. Operations
10	10 steps
100	100 steps
1000	1000 steps

Pattern observation: Doubling the array size doubles the work done.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete grows in a straight line with the number of elements.

Common Mistake

[X] Wrong: "Since we only add numbers, the time is constant no matter the array size."

[OK] Correct: Even though adding is simple, we must do it for every element, so time grows with array length.

Interview Connect

Understanding how simple loops scale helps you explain and optimize code clearly in interviews.

Self-Check

"What if we nested another loop inside to compare each element with every other? How would the time complexity change?"

Practice

1. What does the following C# code do?
int[] arr = {1, 2, 3, 4};
for (int i = 0; i < arr.Length; i++) {
Console.Write(arr[i] + " ");
}

easy

A. Prints all elements of the array from start to end

B. Prints all elements of the array in reverse order

C. Prints only the first element of the array

D. Causes a runtime error due to wrong loop condition

Solution

1. Step 1: Understand the loop range

   The loop starts at 0 and runs while i is less than arr.Length (4), so i goes 0,1,2,3.

2. Step 2: Print elements in order

   Each iteration prints arr[i], so it prints 1 2 3 4 in order.

3. Final Answer:

   Prints all elements of the array from start to end -> Option A

4. Quick Check:

   Loop from 0 to length-1 prints all elements [OK]

Hint: Loop from 0 to length-1 prints all elements in order [OK]

Common Mistakes:

• Thinking loop prints in reverse
• Confusing loop condition with <= instead of <
• Assuming only first element prints

2. Which of the following is the correct syntax to traverse an array arr in C# using a foreach loop?

easy

A. foreach (int i in arr) { Console.Write(i); }

B. foreach (arr in int i) { Console.Write(i); }

C. foreach (int arr in i) { Console.Write(i); }

D. foreach (int i; arr) { Console.Write(i); }

Solution

1. Step 1: Recall foreach syntax

   The correct syntax is: foreach (type variable in collection) { ... }

2. Step 2: Match syntax to options

   foreach (int i in arr) { Console.Write(i); } matches correct syntax: foreach (int i in arr) { Console.Write(i); }

3. Final Answer:

   foreach (int i in arr) { Console.Write(i); } -> Option A

4. Quick Check:

   foreach loops use 'variable in collection' syntax [OK]

Hint: Remember foreach uses 'variable in collection' syntax [OK]

Common Mistakes:

• Swapping variable and collection order
• Using semicolon instead of 'in'
• Incorrect variable declaration inside foreach

3. What is the output of the following code?

int[] arr = {10, 20, 30, 40};
for (int i = arr.Length - 1; i >= 0; i--) {
  Console.Write(arr[i] + " ");
}

medium

A. 40 30 20

B. 10 20 30 40

C. 40 30 20 10

D. Runtime error due to loop condition

Solution

1. Step 1: Analyze loop start and end

   Loop starts at arr.Length - 1 = 3 and runs while i >= 0, decrementing i each time.

2. Step 2: Trace printed elements

   Prints arr[3]=40, arr[2]=30, arr[1]=20, arr[0]=10 in that order.

3. Final Answer:

   40 30 20 10 -> Option C

4. Quick Check:

   Loop from end to start prints reversed array [OK]

Hint: Loop from length-1 down to 0 prints array in reverse [OK]

Common Mistakes:

• Confusing loop direction and printing forward
• Off-by-one errors in loop condition
• Assuming runtime error due to negative index

4. Find the error in this array traversal code:

int[] arr = {5, 10, 15};
for (int i = 0; i <= arr.Length; i++) {
  Console.Write(arr[i] + " ");
}

medium

A. Array declaration is incorrect

B. Variable i should start from 1, not 0

C. Console.Write should be Console.WriteLine

D. Loop condition should be i < arr.Length, not <= arr.Length

Solution

1. Step 1: Check loop condition

   Loop runs while i <= arr.Length, but arr indices go from 0 to arr.Length - 1.

2. Step 2: Identify out-of-range access

   When i == arr.Length (3), arr[3] is accessed causing runtime error (IndexOutOfRangeException).

3. Final Answer:

   Loop condition should be i < arr.Length, not <= arr.Length -> Option D

4. Quick Check:

   Array indices go 0 to length-1, use < not <= [OK]

Hint: Use i < arr.Length to avoid out-of-range errors [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Starting loop at 1 instead of 0
• Confusing Write and WriteLine methods

5. Given an integer array arr, which code snippet correctly prints only the even-indexed elements (0, 2, 4, ...) of the array?

hard

A. foreach (int i in arr) { if (i % 2 == 0) Console.Write(i + " "); }

B. for (int i = 0; i < arr.Length; i += 2) { Console.Write(arr[i] + " "); }

C. for (int i = 1; i < arr.Length; i += 2) { Console.Write(arr[i] + " "); }

D. for (int i = 0; i < arr.Length; i++) { if (arr[i] % 2 == 0) Console.Write(arr[i] + " "); }

Solution

1. Step 1: Understand requirement

   We want to print elements at even indexes: 0, 2, 4, ...

2. Step 2: Analyze options

   for (int i = 0; i < arr.Length; i += 2) { Console.Write(arr[i] + " "); } increments i by 2 starting at 0, so it accesses even indexes correctly.
   for (int i = 1; i < arr.Length; i += 2) { Console.Write(arr[i] + " "); } accesses odd indexes.
   foreach (int i in arr) { if (i % 2 == 0) Console.Write(i + " "); } checks element values for evenness, not indexes.
   for (int i = 0; i < arr.Length; i++) { if (arr[i] % 2 == 0) Console.Write(arr[i] + " "); } prints elements with even values, not indexes.

3. Final Answer:

   for (int i = 0; i < arr.Length; i += 2) { Console.Write(arr[i] + " "); } -> Option B

4. Quick Check:

   Increment index by 2 from 0 to get even indexes [OK]

Hint: Increment loop index by 2 starting at 0 for even indexes [OK]

Common Mistakes:

• Confusing even index with even element value
• Starting loop at 1 for even indexes
• Using foreach without index tracking