Linear Search Algorithm in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of linear search when element is present

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

DSA Java

public class Main {
    public static int linearSearch(int[] arr, int target) {
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == target) {
                return i;
            }
        }
        return -1;
    }

    public static void main(String[] args) {
        int[] numbers = {3, 5, 7, 9, 11};
        int index = linearSearch(numbers, 7);
        System.out.println(index);
    }
}

A 2

B 3

C -1

D 1

💡 Hint

Remember that array indices start at 0 in Java.

The element 7 is at index 2 in the array {3, 5, 7, 9, 11}. The linear search returns the index where the target is found.

❓ Predict Output

intermediate

Output when element is not found in linear search

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

DSA Java

public class Main {
    public static int linearSearch(int[] arr, int target) {
        for (int i = 0; i < arr.length; i++) {
            if (arr[i] == target) {
                return i;
            }
        }
        return -1;
    }

    public static void main(String[] args) {
        int[] numbers = {1, 2, 3, 4, 5};
        int index = linearSearch(numbers, 10);
        System.out.println(index);
    }
}

A 5

B 0

C -1

D 10

💡 Hint

If the element is not found, what does the method return?

The method returns -1 when the target is not found in the array.

🔧 Debug

advanced

Identify the error in linear search implementation

What error will the following Java code produce when compiled or run?

DSA Java

public class Main {
    public static int linearSearch(int[] arr, int target) {
        int i = 0;
        while (i <= arr.length) {
            if (arr[i] == target) {
                return i;
            }
            i++;
        }
        return -1;
    }

    public static void main(String[] args) {
        int[] numbers = {1, 3, 5};
        System.out.println(linearSearch(numbers, 99));
    }
}

A SyntaxError

B ArrayIndexOutOfBoundsException

C Returns -1

D Prints 2

💡 Hint

Check the loop condition and array indexing carefully.

The loop condition uses i <= arr.length, which causes the index to go out of bounds when i equals arr.length.

🧠 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(log n)

B O(1)

C O(n log n)

D O(n)

💡 Hint

Think about how many elements might be checked in the worst case.

Linear search checks each element one by one, so in the worst case it checks all n elements, making it O(n).

🚀 Application

expert

Result of linear search on a linked list

Given a singly linked list with nodes containing values [4, 8, 15, 16, 23, 42], what is the output of a linear search for the value 16? The search returns the zero-based position of the node or -1 if not found.

A 3

B 4

C 2

D -1

💡 Hint

Count nodes starting from zero at the head.

The value 16 is at the fourth node, which is index 3 starting from zero.

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