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 the following 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 index = -1;
for (int i = 0; i < n; i++) {
    if (arr[i] == target) {
        index = i;
        break;
    }
}
std::cout << index << std::endl;

A -1

B 2

C 3

D 0

💡 Hint

Remember that array indexing in C++ starts at 0.

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

❓ Predict Output

intermediate

Linear Search Result When Target Not Found

What will be the output of this C++ code when searching for 10 in the array?

DSA C++

int arr[] = {1, 2, 3, 4, 5};
int n = 5;
int target = 10;
int index = -1;
for (int i = 0; i < n; i++) {
    if (arr[i] == target) {
        index = i;
        break;
    }
}
std::cout << index << std::endl;

A -1

B 4

C 0

D 5

💡 Hint

If the target is not found, the index remains -1.

Since 10 is not in the array, the loop never updates index from -1, so -1 is printed.

🔧 Debug

advanced

Find the Bug in Linear Search Implementation

This code is supposed to perform a linear search but does not work correctly. What is the output and why?

DSA C++

int arr[] = {2, 4, 6, 8};
int n = 4;
int target = 10;
int index = -1;
for (int i = 1; i <= n; i++) {
    if (arr[i] == target) {
        index = i;
        break;
    }
}
std::cout << index << std::endl;

A 2

B -1

C 3

D Runtime error

💡 Hint

Check the loop boundaries and array indexing carefully.

The loop accesses arr[4] when i == 4, which is out of bounds causing undefined behavior or runtime error.

🧠 Conceptual

advanced

Time Complexity of Linear Search

What is the worst-case time complexity of linear search on an array of size n?

A O(n)

B O(log n)

C O(n log n)

D O(1)

💡 Hint

Consider how many elements you might check if the target is not present.

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

🚀 Application

expert

Linear Search on Linked List

Given a singly linked list with nodes containing values [10, 20, 30, 40], what is the output of searching for 30 using linear search?

DSA C++

struct Node {
    int data;
    Node* next;
};

Node* head = new Node{10, nullptr};
head->next = new Node{20, nullptr};
head->next->next = new Node{30, nullptr};
head->next->next->next = new Node{40, nullptr};

int target = 30;
int index = 0;
Node* current = head;
while (current != nullptr) {
    if (current->data == target) {
        break;
    }
    current = current->next;
    index++;
}
if (current == nullptr) index = -1;
std::cout << index << std::endl;

A 1

B 3

C 2

D -1

💡 Hint

Count nodes starting from 0 until you find the target value.

The value 30 is in the third node, which is index 2 starting from 0.

Practice

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

easy

A. Use a hash table to find the target instantly

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

C. Sort the list first and then search using binary steps

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

Solution

1. Step 1: Understand linear search process

   Linear search checks each element in order, starting from the first, to find the target.

2. Step 2: Compare with other search methods

   Other methods like binary search divide the list, but linear search simply checks one by one.

3. Final Answer:

   Check each element one by one until the target is found or list ends -> Option B

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 data
• Assuming linear search is faster than binary search always

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 = 1; i <= n; i++)

C. for (int i = 0; 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, so loop should run from 0 to less than n.

2. Step 2: Check loop conditions

   Loop with i < n ensures last index accessed is n-1, which is valid.

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Using <= n causes out-of-bound access
• Starting loop at 1 misses first element
• Using wrong loop condition causing runtime errors

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;
  }
}
std::cout << index;

medium

A. 2

B. 3

C. -1

D. 4

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), so index = 2 and break.

2. Step 2: Output the index

   After break, index is 2, so output is 2.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Target found at index 2 [OK]

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

Common Mistakes:

• Forgetting to break loop after finding target
• Printing target value instead of index
• Confusing index with position (1-based)

4. Identify the error in the following linear search code snippet:

int arr[] = {5, 3, 8, 6};
int n = 4;
int target = 6;
int index = -1;
for (int i = 1; i <= n; i++) {
  if (arr[i] == target) {
    index = i;
    break;
  }
}
std::cout << index;

medium

A. Both A and B are errors

B. Loop condition should be i < n, not i <= n

C. Loop starts at 1 instead of 0, missing first element

D. No error, code works fine

Solution

1. Step 1: Check loop start index

   Loop starts at 1, so arr[0] is never checked, missing first element.

2. Step 2: Check loop end condition

   Loop runs while i <= n, so i can be 4, but arr[4] is out of bounds (valid indices 0-3).

3. Final Answer:

   Both A and B are errors -> Option A

4. Quick Check:

   Loop must start at 0 and run i < n [OK]

Hint: Loop from 0 to less than n to avoid errors [OK]

Common Mistakes:

• Starting loop at 1 misses first element
• Using <= n causes out-of-bound access
• Ignoring array index limits

5. You have an unsorted array arr = {3, 5, 2, 9, 5, 2}. You want to find the first occurrence index of the number 5 using linear search. What will be the output and why?

medium

A. Error, because array has duplicates

B. 4, because linear search returns the last matching index

C. -1, because 5 is not found

D. 1, because linear search returns the first matching index

Solution

1. Step 1: Understand linear search behavior with duplicates

   Linear search returns the first index where target is found, ignoring later duplicates.

2. Step 2: Find first occurrence of 5 in array

   arr[0]=3 (no), arr[1]=5 (yes), so index = 1 is returned immediately.

3. Final Answer:

   1, because linear search returns the first matching index -> Option D

4. Quick Check:

   First occurrence index = 1 [OK]

Hint: Linear search stops at first match, returns its index [OK]

Common Mistakes:

• Returning last occurrence instead of first
• Assuming duplicates cause errors
• Returning -1 incorrectly when target exists