Linear Search Algorithm in DSA C# - Execution Trace

Concept Flow - Linear Search Algorithm

Start at index 0

↓

Check if element at index == target?

Yes→Return index

No↓

Move to next index

↓

Is index < array length?

Yes→Repeat check

No↓

Return -1 (not found)

Start from the first element, check each one against the target. If found, return index; if not, move to next until end.

Execution Sample

DSA C#

int LinearSearch(int[] arr, int target) {
  for (int i = 0; i < arr.Length; i++) {
    if (arr[i] == target) return i;
  }
  return -1;
}

Searches array elements one by one to find the target value and returns its index or -1 if not found.

Execution Table

Step	Operation	Index (i)	Element Checked	Target	Match?	Action	Result
1	Start loop	0	5	7	No	Move to next index	Continue searching
2	Check element	1	3	7	No	Move to next index	Continue searching
3	Check element	2	7	7	Yes	Return index	Found at index 2

💡 Element found at index 2, loop exits early.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3 (Final)
i	undefined	0	1	2
arr[i]	undefined	5	3	7
Result	undefined	searching	searching	2 (found)

Key Moments - 3 Insights

Why does the search stop immediately when a match is found?

Because the algorithm returns the index as soon as arr[i] == target is true (see execution_table step 3), no need to check further.

What happens if the target is not in the array?

The loop completes all indices without finding a match, then returns -1 (not shown in this trace but implied by exit condition).

Why do we check i < arr.Length in the loop?

To avoid accessing outside the array bounds, ensuring each element is checked safely (see concept_flow step 'Is index < array length?').

Visual Quiz - 3 Questions

Look at the execution_table, what is the value of 'i' when the target is found?

A 2

B 1

C 0

D 3

💡 Hint

Check the 'Index (i)' column at the step where 'Match?' is Yes.

At which step does the algorithm decide to return the index?

A Step 1

B Step 3

C Step 2

D After loop ends

💡 Hint

Look for the step where 'Action' is 'Return index' in the execution_table.

If the target was 10 (not in array), what would happen in the execution_table?

A The loop would return index 0

B The loop would stop at index 2

C The loop would return -1 after checking all elements

D The loop would return index 3

💡 Hint

Recall the key_moments answer about what happens if target is not found.

Concept Snapshot

Linear Search:
- Check each element from start to end
- If element == target, return index immediately
- If not found, return -1
- Time: O(n), Space: O(1)
- Simple, no sorting needed

Full Transcript

Linear Search checks each element in the array one by one from the start. It compares the current element to the target. If they match, it returns the current index immediately. If no element matches after checking all, it returns -1. This method is simple and works on any array but can be slow for large arrays because it may check every element.

Practice

1. What does the linear search algorithm do when searching for a value in a list?

easy

A. It checks each element one by one until it finds the value or reaches the end.

B. It divides the list into halves to find the value quickly.

C. It sorts the list before searching for the value.

D. It uses a hash table to find the value instantly.

Solution

1. Step 1: Understand linear search process

   Linear search checks each element in order from start to end.

2. Step 2: Compare with other search methods

   Other methods like binary search divide the list, but linear search does not.

3. Final Answer:

   It checks each element one by one until it finds the value or reaches the end. -> Option A

4. Quick Check:

   Linear search = check elements one by one [OK]

Hint: Linear search checks elements one by one [OK]

Common Mistakes:

• Confusing linear search with binary search
• Thinking linear search sorts the list first
• Assuming linear search uses extra data structures

2. Which of the following is the correct syntax to start a linear search loop in C# for an array named arr?

easy

A. for (int i = 0; i < arr.Length; i++)

B. for (int i = 1; i <= arr.Length; i++)

C. foreach (int i in arr.Length)

D. while (i < arr.Length) i++"

Solution

1. Step 1: Identify correct for loop syntax for arrays

   In C#, arrays use zero-based indexing, so loop starts at 0 and runs while i < arr.Length.

2. Step 2: Check each option

   for (int i = 0; i < arr.Length; i++) uses correct syntax: for (int i = 0; i < arr.Length; i++). Others have errors like starting at 1 or wrong foreach syntax.

3. Final Answer:

   for (int i = 0; i < arr.Length; i++) -> Option A

4. Quick Check:

   Loop from 0 to Length-1 = for (int i = 0; i < arr.Length; i++) [OK]

Hint: Array loops start at 0 and go to Length-1 [OK]

Common Mistakes:

• Starting loop at 1 instead of 0
• Using <= instead of < in loop condition
• Incorrect foreach syntax for arrays

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

int[] arr = {4, 7, 1, 3, 9};
int target = 3;
int index = -1;
for (int i = 0; i < arr.Length; i++) {
  if (arr[i] == target) {
    index = i;
    break;
  }
}
Console.WriteLine(index);

medium

A. 4

B. 1

C. -1

D. 3

Solution

1. Step 1: Trace the loop to find target 3

   Check elements: arr[0]=4 (no), arr[1]=7 (no), arr[2]=1 (no), arr[3]=3 (yes), so index = 3 and break.

2. Step 2: Print the index found

   After loop, index = 3, so Console.WriteLine prints 3.

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Target 3 found at index 3 [OK]

Hint: Stop loop when target found, print index [OK]

Common Mistakes:

• Forgetting to break loop after finding target
• Printing value instead of index
• Confusing index with element value

4. The following code is intended to perform a linear search but has a bug. What is the bug?

int[] arr = {2, 5, 8, 12};
int target = 8;
int index = -1;
for (int i = 0; i <= arr.Length; i++) {
  if (arr[i] == target) {
    index = i;
    break;
  }
}
Console.WriteLine(index);

medium

A. The index variable should be initialized to 0.

B. The target value is not in the array.

C. The loop condition should be i < arr.Length, not i <= arr.Length.

D. The break statement is missing.

Solution

1. Step 1: Analyze loop condition

   The loop runs while i <= arr.Length, but valid indices are 0 to arr.Length - 1. Using <= causes out-of-range error.

2. Step 2: Correct the loop condition

   Change condition to i < arr.Length to avoid accessing invalid index.

3. Final Answer:

   The loop condition should be i < arr.Length, not i <= arr.Length. -> Option C

4. Quick Check:

   Loop condition must be less than Length [OK]

Hint: Use i < arr.Length to avoid index errors [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Assuming target not found is a bug
• Changing index initialization unnecessarily

5. You have an array int[] arr = {5, 3, 7, 3, 9}. You want to find the index of the second occurrence of the value 3 using linear search. Which approach is correct?

hard

A. Sort the array first, then find the value 3.

B. Use a counter to track occurrences and return index when counter reaches 2.

C. Return the first index where value 3 is found.

D. Use binary search to find the second occurrence.

Solution

1. Step 1: Understand the problem of finding second occurrence

   Linear search must count how many times target appears and return index at second time.

2. Step 2: Evaluate options

   Sort the array first, then find the value 3. sorts the array, changing indices. Return the first index where value 3 is found. returns only the first occurrence. Use binary search to find the second occurrence. uses binary search, unsuitable for unsorted array and specific occurrences. Use a counter to track occurrences and return index when counter reaches 2. correctly uses a counter to track occurrences and return the index of the second match.

3. Final Answer:

   Use a counter to track occurrences and return index when counter reaches 2. -> Option B

4. Quick Check:

   Count occurrences to find second index [OK]

Hint: Count matches, return index at second match [OK]

Common Mistakes:

• Returning first occurrence instead of second
• Sorting array which changes original order
• Using binary search on unsorted array