Linear Search Algorithm in DSA C - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Linear Search on Array

What is the output of this C code that performs a linear search for the value 7 in the array?

DSA C

int arr[] = {3, 5, 7, 9, 11};
int n = 5;
int target = 7;
int i;
int index = -1;
for(i = 0; i < n; i++) {
    if(arr[i] == target) {
        index = i;
        break;
    }
}
printf("%d", index);

A 3

B -1

C 1

D 2

💡 Hint

Remember that array indexing in C starts at 0.

The target value 7 is at index 2 in the array. The loop stops when it finds 7 and prints 2.

🧠 Conceptual

intermediate

Linear Search Worst Case Scenario

In linear search, what is the worst case time complexity when searching for an element in an array of size n?

A O(n)

B O(1)

C O(log n)

D O(n^2)

💡 Hint

Think about how many elements you might check if the target is not in the array.

In the worst case, linear search checks every element once, so the time complexity is O(n).

🔧 Debug

advanced

Find the Bug in Linear Search Implementation

What error will this C code produce when searching for 4 in the array?

DSA C

int arr[] = {1, 2, 3, 5, 6};
int n = 5;
int target = 4;
int i;
int index = -1;
for(i = 0; i < n; i++) {
    if(arr[i] == target) {
        index = i;
        break;
    }
}
printf("%d", index);

A Runs and prints -1

B Runs and prints 4

C Compilation error

D Runtime error (out of bounds access)

💡 Hint

Check the loop condition carefully for array bounds.

The original loop condition was incorrect (i <= n). It should be i < n to avoid out of bounds access. With the corrected loop, the code runs and prints -1 since 4 is not in the array.

🚀 Application

advanced

Linear Search on Linked List

Given a singly linked list, which approach correctly implements linear search to find a target value?

A Jump to middle node and compare, then decide to go left or right

B Traverse nodes from head, compare each node's data with target, stop if found or reach null

C Sort the list first, then use binary search

D Use a hash map to store nodes and search in O(1)

💡 Hint

Think about how linked lists are structured and accessed.

Linear search on linked list means checking each node one by one from the start until the target is found or list ends.

❓ Predict Output

expert

Output of Linear Search with Multiple Occurrences

What is the output of this C code searching for 3 in the array with multiple 3s?

DSA C

int arr[] = {1, 3, 5, 3, 7};
int n = 5;
int target = 3;
int i;
int index = -1;
for(i = 0; i < n; i++) {
    if(arr[i] == target) {
        index = i;
        break;
    }
}
printf("%d", index);

A 3

B 2

C 1

D -1

💡 Hint

Linear search stops at the first match.

The first occurrence of 3 is at index 1, so the loop breaks and prints 1.

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