Linear Search Algorithm in DSA Java - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to start the linear search loop from the first index.

DSA Java

for (int i = [1]; i < arr.length; i++) {
    if (arr[i] == target) {
        return i;
    }
}

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

Aarr.length

B1

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting the loop at 1 skips the first element.

Using arr.length as start causes no iterations.

The linear search starts checking from the first element, which is at index 0.

2fill in blank

medium

Complete the code to return -1 if the target is not found after the loop.

DSA Java

for (int i = 0; i < arr.length; i++) {
    if (arr[i] == target) {
        return i;
    }
}
return [1];

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

A-1

Barr.length

Ctarget

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 can be mistaken as found at first index.

Returning target is incorrect as it's the value, not index.

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

3fill in blank

hard

Fix the error in the condition to correctly compare array element with target.

DSA Java

if (arr[i] [1] target) {
    return i;
}

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

A=

B!=

C==

D<

Attempts:

3 left

💡 Hint

Common Mistakes

Using '=' causes a syntax error or unexpected behavior.

Using '!=' checks for inequality, which is wrong here.

Use '==' to compare values for equality in Java, not '=' which is assignment.

4fill in blank

hard

Fill both blanks to create a linear search method header and return type.

DSA Java

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

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

AString

Bint

Cboolean

Dvoid

Attempts:

3 left

💡 Hint

Common Mistakes

Using void means no value is returned, causing errors.

Using boolean or String is incorrect for index return.

The method returns the index of the target, which is an integer.

5fill in blank

hard

Fill all three blanks to complete the linear search method with correct parameter names and return statement.

DSA Java

public static int linearSearch([1][] [2], int [3]) {
    for (int i = 0; i < [2].length; i++) {
        if ([2][i] == [3]) {
            return i;
        }
    }
    return -1;
}

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

Aint

Barr

Ctarget

Darray

Attempts:

3 left

💡 Hint

Common Mistakes

Using mismatched parameter names causes errors.

Using wrong data types for the array parameter.

The method takes an integer array named 'arr' and an integer 'target' to search for.

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