Linear Search Algorithm in DSA Java - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind the Linear Search Algorithm?

Linear Search checks each item in a list one by one until it finds the target or reaches the end.

beginner

What is the time complexity of Linear Search in the worst case?

The worst-case time complexity is O(n), where n is the number of items in the list.

beginner

In Linear Search, what happens if the target element is not in the list?

The algorithm checks all elements and returns -1 or a special value to show the target is not found.

beginner

Show the basic steps of Linear Search in order.

1. Start from the first element.
2. Compare it with the target.
3. If it matches, return the index.
4. Otherwise, move to the next element.
5. Repeat until found or end of list.

beginner

Why is Linear Search not efficient for large lists?

Because it checks every element one by one, it can take a long time if the list is large.

What does Linear Search do when it finds the target element?

A Deletes the element

B Continues searching the rest of the list

C Returns the index of the element

D Sorts the list

Linear Search stops and returns the index when it finds the target element.

What is the best case time complexity of Linear Search?

A O(1)

B O(n)

C O(log n)

D O(n^2)

If the target is the first element, Linear Search finds it immediately, so best case is O(1).

Which data structure is Linear Search best suited for?

A Hash table

B Sorted list

C Binary tree

D Unsorted list

Linear Search works on any list but is mainly used for unsorted lists.

If the list has 10 elements and the target is not present, how many comparisons does Linear Search make?

A 0

B 10

C 1

D 9

Linear Search compares each element until the end, so it makes 10 comparisons.

What value is commonly returned if the target is not found in Linear Search?

A -1

B 0

C 1

D null

Returning -1 is a common way to indicate the target was not found.

Explain how Linear Search works step-by-step on a list of numbers.

Think about checking each item one by one.

Describe when Linear Search is a good choice and when it is not.

Consider list size and sorting.

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