Linear Search Algorithm in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Linear Search for Existing Element

What is the output of the following JavaScript code that performs a linear search for the number 7 in the array?

DSA Javascript

function linearSearch(arr, target) {
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] === target) {
      return i;
    }
  }
  return -1;
}

const numbers = [3, 5, 7, 9, 11];
console.log(linearSearch(numbers, 7));

A 1

B 3

C -1

D 2

💡 Hint

Remember that array indexes start at 0 in JavaScript.

The function checks each element in order. The number 7 is at index 2 in the array, so the function returns 2.

❓ Predict Output

intermediate

Output When Element is Not Found

What does the following code print when searching for 10 in the array?

DSA Javascript

function linearSearch(arr, target) {
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] === target) {
      return i;
    }
  }
  return -1;
}

const numbers = [1, 2, 3, 4, 5];
console.log(linearSearch(numbers, 10));

A -1

B 5

C 0

D undefined

💡 Hint

If the target is not in the array, what does the function return?

Since 10 is not in the array, the loop finishes without finding it, so the function returns -1.

🧠 Conceptual

advanced

Time Complexity of Linear Search

What is the worst-case time complexity of the linear search algorithm when searching in an array of size n?

A O(n log n)

B O(n)

C O(1)

D O(log n)

💡 Hint

Think about how many elements the algorithm might check in the worst case.

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

🔧 Debug

advanced

Identify the Bug in Linear Search Implementation

What is the output of the following code when searching for 4 in the array?

DSA Javascript

function linearSearch(arr, target) {
  for (let i = 0; i <= arr.length; i++) {
    if (arr[i] === target) {
      return i;
    }
  }
  return -1;
}

const numbers = [1, 2, 3, 4, 5];
console.log(linearSearch(numbers, 4));

A TypeError: Cannot read property 'undefined' of undefined

B -1

C Returns 3

D SyntaxError

💡 Hint

Check the loop condition and array indexing carefully.

The code returns 3 because 4 is found at index 3 before the out-of-bounds access. However, the loop uses i <= arr.length, which is a bug that accesses arr[arr.length] (undefined) if the target is not found early.

🚀 Application

expert

Number of Comparisons in Linear Search

If you perform a linear search for the target value 8 in the array [2, 4, 6, 8, 10], how many comparisons does the algorithm make before finding the target?

A 4

B 3

C 5

D 2

💡 Hint

Count how many elements the algorithm checks until it finds 8.

The algorithm checks elements at indexes 0, 1, 2, and 3 before finding 8, so it makes 4 comparisons.

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 steps

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

   Binary search divides list, hash uses extra memory, linear is simple sequential check.

3. Final Answer:

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

4. Quick Check:

   Linear search = sequential check [OK]

Hint: Linear search checks items one by one [OK]

Common Mistakes:

• Confusing linear search with binary search
• Thinking linear search requires sorted list
• Assuming linear search uses extra memory

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

easy

A. for (let i = 0; i <= arr.length; i++)

B. for (let i = 0; i < arr.length; i++)

C. for (let i = 1; i < arr.length; i++)

D. for (let i = 1; i <= arr.length; i++)

Solution

1. Step 1: Understand array indexing in JavaScript

   Arrays start at index 0 and go up to length - 1.

2. Step 2: Choose loop condition to avoid out-of-bound error

   Loop must run while i is less than arr.length, not equal or greater.

3. Final Answer:

   for (let i = 0; i < arr.length; i++) -> Option B

4. Quick Check:

   Loop runs from 0 to length-1 [OK]

Hint: Use i < arr.length to avoid index errors [OK]

Common Mistakes:

• Using <= instead of < causing out-of-bound errors
• Starting loop from 1 instead of 0
• Confusing array length with last index

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

const arr = [4, 8, 15, 16, 23, 42];
function linearSearch(array, target) {
  for (let i = 0; i < array.length; i++) {
    if (array[i] === target) {
      return i;
    }
  }
  return -1;
}
console.log(linearSearch(arr, 15));

medium

A. 2

B. 3

C. -1

D. 15

Solution

1. Step 1: Trace the loop to find target 15

   Indices: 0=4, 1=8, 2=15. At i=2, array[i] === 15 is true.

2. Step 2: Return index when target found

   Function returns 2 immediately when target matches.

3. Final Answer:

   2 -> Option A

4. Quick Check:

   Index of 15 in array = 2 [OK]

Hint: Return index when target matches during loop [OK]

Common Mistakes:

• Returning value instead of index
• Returning -1 even if target exists
• Off-by-one errors in index

4. Find the error in this linear search function and choose the correct fix:

function linearSearch(arr, target) {
  for (let i = 0; i <= arr.length; i++) {
    if (arr[i] == target) {
      return i;
    }
  }
  return -1;
}

medium

A. Initialize i to 1 instead of 0

B. Change == to === in if condition

C. Change loop condition to i < arr.length

D. Add a break statement after return

Solution

1. Step 1: Identify loop condition error

   Loop runs while i <= arr.length, which causes out-of-bound access at i = arr.length.

2. Step 2: Fix loop condition to prevent error

   Change condition to i < arr.length to stay within valid indices.

3. Final Answer:

   Change loop condition to i < arr.length -> Option C

4. Quick Check:

   Loop must stay within array bounds [OK]

Hint: Use i < arr.length to avoid index errors [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Thinking == vs === causes runtime error here
• Adding break after return (unnecessary)

5. You want to find the first occurrence of the number 7 in the array [3, 5, 7, 7, 9] using linear search. Which code snippet correctly returns the index of the first 7?

hard

A. for (let i = arr.length - 1; i >= 0; i--) { if (arr[i] === 7) return i; } return -1;

B. for (let i = 0; i < arr.length; i++) { if (arr[i] == 7) return arr[i]; } return -1;

C. for (let i = 1; i <= arr.length; i++) { if (arr[i] === 7) return i; } return -1;

D. for (let i = 0; i < arr.length; i++) { if (arr[i] === 7) return i; } return -1;

Solution

1. Step 1: Understand requirement to find first occurrence

   Start from index 0 and return index when first 7 is found.

2. Step 2: Analyze each option for correctness

   for (let i = 0; i < arr.length; i++) { if (arr[i] === 7) return i; } return -1; loops forward, returns index correctly. for (let i = arr.length - 1; i >= 0; i--) { if (arr[i] === 7) return i; } return -1; loops backward (last occurrence). for (let i = 1; i <= arr.length; i++) { if (arr[i] === 7) return i; } return -1; has wrong loop bounds. for (let i = 0; i < arr.length; i++) { if (arr[i] == 7) return arr[i]; } return -1; returns value, not index.

3. Final Answer:

   for (let i = 0; i < arr.length; i++) { if (arr[i] === 7) return i; } return -1; -> Option D

4. Quick Check:

   Forward loop returns first match index [OK]

Hint: Loop forward and return index on first match [OK]

Common Mistakes:

• Looping backward returns last occurrence, not first
• Returning value instead of index
• Using wrong loop bounds causing errors