Linear Search Algorithm in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the index of the target element in the array or -1 if not found.

DSA C++

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

Drag options to blanks, or click blank then click option'

Ai

Btarget

Cn

Darr[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using the loop index i instead of the target value in the comparison.

Comparing arr[i] to itself instead of the target.

The code compares each element arr[i] with the target to find a match.

2fill in blank

medium

Complete the code to loop through the array correctly for linear search.

DSA C++

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

Drag options to blanks, or click blank then click option'

Ai

Btarget

Carr

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using the target value as the loop limit.

Using the loop index variable as the limit.

The loop should run from 0 to n - 1 to cover all elements.

3fill in blank

hard

Fix the error in the return statement to indicate the target was not found.

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];
}

Drag options to blanks, or click blank then click option'

A-1

B0

Ci

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Returning the loop index i after the loop ends.

Returning zero which could be a valid index.

Returning -1 indicates the target was not found in the array.

4fill in blank

hard

Fill both blanks to create a linear search that returns true if the target is found, false otherwise.

DSA C++

bool linearSearch(int arr[], int n, int target) {
    for (int i = 0; i < [1]; i++) {
        if (arr[i] == [2]) {
            return true;
        }
    }
    return false;
}

Drag options to blanks, or click blank then click option'

An

Btarget

Ci

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong variable for loop limit.

Comparing to the loop index instead of the target.

The loop runs up to n, and compares each element to target.

5fill in blank

hard

Fill all three blanks to create a linear search that returns the index or -1 if not found, using a while loop.

DSA C++

int linearSearch(int arr[], int n, int target) {
    int i = 0;
    while (i [1] n) {
        if (arr[i] == [2]) {
            return [3];
        }
        i++;
    }
    return -1;
}

Drag options to blanks, or click blank then click option'

A<

Btarget

Ci

D<=

Attempts:

3 left

💡 Hint

Common Mistakes

Using <= instead of < in the loop condition causing out-of-bounds access.

Returning the target value instead of the index.

The while loop runs while i < n, compares arr[i] to target, and returns i if found.

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