Linear Search Algorithm in DSA C - Execution Trace

Concept Flow - Linear Search Algorithm

Start at index 0

↓

Check if element at index == target?

No→Move to next index

↓

Index < array length?

No→Target not found

↓

Return index

↓

End

Start from the first element, check each element one by one until the target is found or the list ends.

Execution Sample

DSA C

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

Searches for target in array by checking each element from start to end.

Execution Table

Step	Operation	Index (i)	Element Checked	Condition (arr[i] == target)	Action	Result
1	Start loop	0	5	5 == 3? False	Move to next index	Continue searching
2	Check element	1	3	3 == 3? True	Return index 1	Target found at index 1
3	End	-	-	-	-	Search ends

💡 Target found at index 1, loop stops

Variable Tracker

Variable	Start	After Step 1	After Step 2	Final
i	-	0	1	1
arr[i]	-	5	3	3
Result	-	-	1	1

Key Moments - 2 Insights

Why does the search stop immediately when the target is found?

Because the condition arr[i] == target becomes true at step 2, so the function returns the index immediately, stopping the loop as shown in the execution_table.

What happens if the target is not in the array?

The loop checks all elements until i reaches n, then returns -1 indicating target not found. This is not shown in this run but would be the exit condition if no match occurs.

Visual Quiz - 3 Questions

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

A 0

B 1

C 2

D -1

💡 Hint

Check the 'Index (i)' column at the step where 'Condition' is true.

At which step does the search stop?

A Step 1

B Step 2

C Step 3

D It never stops

💡 Hint

Look at the 'Action' column where the function returns the index.

If the target was 7, what would happen in the execution_table?

A The search would stop at step 2

B The search would find target at index 1

C The search would continue until the end and return -1

D The search would return index 0

💡 Hint

Think about what happens when no element matches the target.

Concept Snapshot

Linear Search Algorithm:
- Start from first element
- Check each element one by one
- If element == target, return index
- If end reached without match, return -1
- Simple but can be slow for large arrays

Full Transcript

Linear Search Algorithm checks each element in an array from the start to find a target value. It compares the target with each element one by one. If it finds the target, it returns the index immediately and stops. If it reaches the end without finding the target, it returns -1. This method is simple and easy to understand but can be slow if the array is large because it may check every element.

Practice

1. What is the main idea behind the Linear Search Algorithm?

easy

A. Check each element one by one until the target is found or list ends

B. Divide the list into halves repeatedly to find the target

C. Sort the list first then search using binary method

D. Use a hash table to find the target 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:

   Check each element one by one until the target is found or list ends -> 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 requires sorted list
• Assuming instant search with hash tables

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

easy

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

B. for (int i = 0; i <= n; i++)

C. for (int i = 1; i < n; i++)

D. for (int i = 1; i <= n; i++)

Solution

1. Step 1: Understand array indexing in C

   Arrays in C start at index 0 and go up to n-1 for size n.

2. Step 2: Choose correct loop condition

   The loop should run from 0 to less than n, so i < n is correct.

3. Final Answer:

   for (int i = 0; i < n; i++) -> Option A

4. Quick Check:

   Loop from 0 to n-1 = for(i=0; i<n; i++) [OK]

Hint: Array index starts at 0, loop till i < n [OK]

Common Mistakes:

• Using <= n causes out-of-bounds access
• Starting loop at 1 misses first element
• Confusing array size with last index

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

int arr[] = {4, 2, 7, 1, 9};
int n = 5;
int target = 7;
int index = -1;
for (int i = 0; i < n; i++) {
  if (arr[i] == target) {
    index = i;
    break;
  }
}
printf("%d", index);

medium

A. 3

B. 2

C. -1

D. 7

Solution

1. Step 1: Trace the loop to find target 7

   Check elements: arr[0]=4 (no), arr[1]=2 (no), arr[2]=7 (yes), stop loop.

2. Step 2: Identify index where target found

   Target 7 found at index 2, so index = 2.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Target 7 at arr[2] = 2 [OK]

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

Common Mistakes:

• Printing value instead of index
• Not breaking loop and printing last index
• Confusing index with element value

4. Find the error in this linear search code snippet in C:

int linearSearch(int arr[], int n, int target) {
  for (int i = 1; i <= n; i++) {
    if (arr[i] == target) {
      return i;
    }
  }
  return -1;
}

medium

A. The array parameter should be a pointer, not an array

B. The function should return the element, not the index

C. The return -1 should be inside the loop

D. Loop starts at 1 and goes to <= n, causing out-of-bounds access

Solution

1. Step 1: Check loop indexing for array access

   Array indices in C start at 0 and go to n-1, but loop starts at 1 and goes to n, causing out-of-bounds at arr[n].

2. Step 2: Identify correct loop bounds

   Loop should start at 0 and run while i < n to avoid invalid access.

3. Final Answer:

   Loop starts at 1 and goes to <= n, causing out-of-bounds access -> Option D

4. Quick Check:

   Array index out-of-bounds due to wrong loop limits [OK]

Hint: Array index starts at 0, loop i < n [OK]

Common Mistakes:

• Starting loop at 1 instead of 0
• Using <= n instead of < n in loop condition
• Returning element instead of index

5. You have an unsorted array arr with 10 elements where some elements repeat. You want to find the first occurrence index of the number 5 using linear search. Which approach is best to handle duplicates and return the correct index?

hard

A. Sort the array first, then use binary search to find 5

B. Traverse entire array, count all 5s, then return last index of 5

C. Traverse from start to end, return index immediately when 5 is found

D. Use a hash map to store counts, then find index of 5

Solution

1. Step 1: Understand requirement for first occurrence

   We want the first index where 5 appears, so searching from start is needed.

2. Step 2: Choose approach that stops at first match

   Linear search that returns immediately on first 5 found gives correct first occurrence index.

3. Final Answer:

   Traverse from start to end, return index immediately when 5 is found -> Option C

4. Quick Check:

   Stop at first match to get first occurrence [OK]

Hint: Return index immediately on first match [OK]

Common Mistakes:

• Counting all occurrences then returning last index
• Sorting changes original order, losing first occurrence
• Using hash map adds unnecessary complexity