Why Linear Search Algorithm in DSA C++?

The Big Idea

What if you could find anything in a messy list without sorting it first?

The Scenario

Imagine you have a messy drawer full of different colored socks, and you want to find a red sock. You start picking up each sock one by one, checking its color until you find the red one.

The Problem

Doing this by hand is slow and tiring, especially if the drawer is very full. You might lose track or miss some socks, making the search frustrating and error-prone.

The Solution

The linear search algorithm works just like checking each sock one by one, but it does it quickly and without mistakes. It goes through each item in a list until it finds the one you want or finishes checking all.

Before vs After

✗ Before

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

✓ After

int linearSearch(const std::vector<int>& data, int target) {
  for (size_t index = 0; index < data.size(); ++index) {
    if (data[index] == target) return static_cast<int>(index);
  }
  return -1;
}

What It Enables

Linear search lets you quickly find any item in a list without needing it to be sorted or organized.

Real Life Example

Finding a friend's phone number in a small, unsorted contact list by checking each entry until you find the right one.

Key Takeaways

Linear search checks each item one by one.

It works on any list, sorted or not.

Simple but can be slow for big lists.

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