Linear Search Algorithm in DSA Java - Build from Scratch

Linear Search Algorithm

📖 Scenario: You are helping a store clerk find a specific product ID in a list of product IDs on the shelf.

🎯 Goal: Build a simple program that uses linear search to find the position of a product ID in an array.

📋 What You'll Learn

Create an integer array called productIDs with the exact values: 101, 203, 405, 302, 101, 504

Create an integer variable called targetID and set it to 302

Use a for loop with variable i to iterate over productIDs

Inside the loop, use an if statement to check if productIDs[i] equals targetID

Print the index i when the target is found and stop the loop

If the target is not found after the loop, print -1

💡 Why This Matters

🌍 Real World

Stores and warehouses often need to find products quickly in a list or shelf. Linear search helps find items when the list is small or unsorted.

💼 Career

Understanding linear search is a foundation for learning more complex search algorithms used in software development and data processing.

1

Create the product ID array

Create an integer array called productIDs with these exact values: 101, 203, 405, 302, 101, 504

DSA Java

// Your code here

Hint

Use int[] productIDs = { ... }; to create the array with the exact numbers.

2

Set the target product ID

Create an integer variable called targetID and set it to 302

DSA Java

public class LinearSearch {
    public static void main(String[] args) {
        int[] productIDs = {101, 203, 405, 302, 101, 504};
        // Your code here
    }
}

Hint

Use int targetID = 302; to set the target product ID.

3

Implement the linear search loop

Use a for loop with variable i to iterate over productIDs. Inside the loop, use an if statement to check if productIDs[i] equals targetID. If yes, print i and use break to stop the loop. After the loop, if the target was not found, print -1.

DSA Java

public class LinearSearch {
    public static void main(String[] args) {
        int[] productIDs = {101, 203, 405, 302, 101, 504};
        int targetID = 302;
        // Your code here
    }
}

Hint

Use a for loop from 0 to productIDs.length. Check each element with if. Print index and break if found. Print -1 if not found.

4

Print the search result

Run the program to print the index of targetID in productIDs. The output should be the index number or -1 if not found.

DSA Java

public class LinearSearch {
    public static void main(String[] args) {
        int[] productIDs = {101, 203, 405, 302, 101, 504};
        int targetID = 302;
        boolean found = false;
        for (int i = 0; i < productIDs.length; i++) {
            if (productIDs[i] == targetID) {
                System.out.println(i);
                found = true;
                break;
            }
        }
        if (!found) {
            System.out.println(-1);
        }
    }
}
// Run the program to see the output

Hint

The program should print 3 because 302 is at index 3 in the array.

Practice

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

easy

A. Sort the list first then search using binary method

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

C. Use a hash table to find the target instantly

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

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 D

4. Quick Check:

   Linear search = sequential check [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 linear search uses extra memory

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

   Java arrays start at index 0 and go up to n-1, so loop must run from 0 to less than n.

2. Step 2: Check loop conditions

   for (int i = 0; i < n; i++) uses i = 0 to i < n, which covers all valid indices correctly.

3. Final Answer:

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

4. Quick Check:

   Array index 0 to n-1 = for loop i=0 to i<n [OK]

Hint: Array indices start at 0, loop till less than length [OK]

Common Mistakes:

• Using <= n causes index out of bounds error
• Starting loop at 1 misses first element
• Confusing array length with last index

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

int[] arr = {4, 8, 15, 16, 23, 42};
int target = 15;
int index = -1;
for (int i = 0; i < arr.length; i++) {
  if (arr[i] == target) {
    index = i;
    break;
  }
}
System.out.println(index);

medium

A. 0

B. -1

C. 2

D. 3

Solution

1. Step 1: Loop through array to find target 15

   Indices: 0=4, 1=8, 2=15. At i=2, arr[i] == 15 matches target.

2. Step 2: Assign index and break loop

   index is set to 2 and loop stops immediately.

3. Final Answer:

   2 -> Option C

4. Quick Check:

   Target 15 found at index 2 [OK]

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

Common Mistakes:

• Not breaking loop causes last index output
• Printing target value instead of index
• Confusing index with array value

4. Identify the error in the following linear search code snippet and choose the correct fix:

int[] arr = {1, 3, 5, 7};
int target = 5;
int index = -1;
for (int i = 1; i <= arr.length; i++) {
  if (arr[i] == target) {
    index = i;
    break;
  }
}
System.out.println(index);

medium

A. Change loop to: for (int i = 0; i <= arr.length; i++)

B. Change loop to: for (int i = 0; i < arr.length; i++)

C. Change loop to: for (int i = 1; i < arr.length; i++)

D. No change needed, code is correct

Solution

1. Step 1: Identify loop index error

   Loop starts at 1 and goes to i <= arr.length, which causes index out of bounds at i = arr.length.

2. Step 2: Correct loop to valid indices

   Loop should start at 0 and run while i < arr.length to cover all valid indices safely.

3. Final Answer:

   Change loop to: for (int i = 0; i < arr.length; i++) -> Option B

4. Quick Check:

   Loop indices 0 to length-1 avoid errors [OK]

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

Common Mistakes:

• Using <= arr.length causes runtime error
• Starting loop at 1 misses first element
• Ignoring array index bounds

5. You have an array arr = {2, 4, 7, 6, 8, 10}. You want to find the index of the first even number greater than 5 using linear search. Which code snippet correctly implements this?

hard

A. for (int i = 0; i < arr.length; i++) { if (arr[i] > 5 && arr[i] % 2 == 0) { System.out.println(i); break; } }

B. for (int i = 0; i < arr.length; i++) { if (arr[i] % 2 == 0) { System.out.println(i); break; } }

C. for (int i = 0; i < arr.length; i++) { if (arr[i] > 5) { System.out.println(i); break; } }

D. for (int i = 0; i < arr.length; i++) { if (arr[i] < 5 && arr[i] % 2 == 0) { System.out.println(i); break; } }

Solution

1. Step 1: Understand the condition to find

   We want the first element that is even and greater than 5.

2. Step 2: Check each option's condition

   for (int i = 0; i < arr.length; i++) { if (arr[i] > 5 && arr[i] % 2 == 0) { System.out.println(i); break; } } checks both conditions correctly; others miss one or use wrong condition.

3. Final Answer:

   for (int i = 0; i < arr.length; i++) { if (arr[i] > 5 && arr[i] % 2 == 0) { System.out.println(i); break; } } -> Option A

4. Quick Check:

   Check both conditions with && for correct index [OK]

Hint: Use && to combine conditions in linear search [OK]

Common Mistakes:

• Checking only one condition instead of both
• Using || instead of && causing wrong matches
• Not breaking loop after first match