Linear Search Algorithm in DSA Rust - Time & Space Complexity

Time Complexity: Linear Search Algorithm

O(n)

Understanding Time Complexity

We want to understand how the time taken by linear search changes as the list size grows.

Specifically, how many steps does it take to find an item or decide it's not there?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn linear_search(arr: &[i32], target: i32) -> Option {
    for (index, &value) in arr.iter().enumerate() {
        if value == target {
            return Some(index);
        }
    }
    None
}
    

This code looks through each item in the list one by one to find the target value.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Checking each element in the array one by one.
• How many times: Up to n times, where n is the number of elements in the array.

How Execution Grows With Input

As the list gets bigger, the number of checks grows roughly the same as the list size.

Input Size (n)	Approx. Operations
10	Up to 10 checks
100	Up to 100 checks
1000	Up to 1000 checks

Pattern observation: The number of steps grows directly with the size of the list.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the target grows linearly with the number of elements.

Common Mistake

[X] Wrong: "Linear search always takes the same time no matter the list size."

[OK] Correct: The search time depends on where the target is or if it is missing, so bigger lists usually take more time.

Interview Connect

Understanding linear search time helps you compare it with faster methods like binary search and shows you how to reason about simple algorithms.

Self-Check

"What if the array was sorted and we used binary search instead? How would the time complexity change?"

Practice

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

easy

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

B. Use a hash table to find the target instantly

C. Sort the list first then search using binary steps

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 means checking items one by one [OK]

Common Mistakes:

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

2. Which of the following is the correct Rust syntax to start a linear search loop over a vector arr?

easy

A. while i < arr.len() { }

B. for i in arr { }

C. for i in 0..arr.len() { }

D. loop i in arr.len() { }

Solution

1. Step 1: Identify correct Rust for loop syntax

   Rust uses for i in 0..arr.len() to loop over indices.

2. Step 2: Check other options for syntax errors

   for i in arr iterates over values, not indices; while needs initialization; loop i in is invalid.

3. Final Answer:

   for i in 0..arr.len() { } -> Option C

4. Quick Check:

   Rust for loop over range = for i in 0..len() [OK]

Hint: Use for i in 0..arr.len() to loop indices [OK]

Common Mistakes:

• Using for i in arr to get indices instead of values
• Writing invalid loop syntax like loop i in ...
• Forgetting to initialize loop variable in while loops

3. What will be the output of this Rust code snippet?

let arr = vec![3, 5, 7, 9];
let target = 7;
let mut index = -1;
for i in 0..arr.len() {
    if arr[i] == target {
        index = i as i32;
        break;
    }
}
println!("{}", index);

medium

A. 0

B. 2

C. 3

D. -1

Solution

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

   Loop checks arr[0]=3, arr[1]=5, arr[2]=7 matches target, sets index=2 and breaks.

2. Step 2: Print the index value

   After break, index is 2, so printed output is 2.

3. Final Answer:

   2 -> Option B

4. Quick Check:

   Target 7 found at index 2 [OK]

Hint: Stop loop at first match, print that index [OK]

Common Mistakes:

• Not breaking loop after finding target
• Printing index before loop ends
• Confusing index type casting

4. Identify the error in this Rust linear search code and choose the fix:

let arr = vec![1, 2, 3];
let target = 4;
let mut found = false;
for i in 0..arr.len() {
    if arr[i] = target {
        found = true;
        break;
    }
}
println!("{}", found);

medium

A. Use '==' for comparison instead of '='

B. Change 'found' to integer type

C. Remove the break statement

D. Use while loop instead of for loop

Solution

1. Step 1: Check the if condition syntax

   The code uses '=' which is assignment, not comparison. It should be '==' to compare values.

2. Step 2: Verify other options

   Changing 'found' type or removing break won't fix syntax error. Using while loop is valid but not necessary.

3. Final Answer:

   Use '==' for comparison instead of '=' -> Option A

4. Quick Check:

   Comparison needs '==' not '=' [OK]

Hint: Use '==' to compare values, '=' assigns [OK]

Common Mistakes:

• Using '=' instead of '==' in if conditions
• Removing break causing unnecessary iterations
• Changing variable types unnecessarily

5. You want to find all indices of a target value in a vector arr using linear search in Rust. Which approach correctly collects all matching indices?

hard

A. Use a loop over indices, push index to a vector when element equals target

B. Break the loop at first match and return that index

C. Sort the vector first, then use binary search for all matches

D. Use a hash map to store element counts and find indices

Solution

1. Step 1: Understand requirement to find all indices

   We must check every element and collect indices where element equals target.

2. Step 2: Evaluate options for correctness

   Breaking loop stops early, sorting changes order, hash map counts but doesn't store indices. Only pushing indices in loop works.

3. Final Answer:

   Use a loop over indices, push index to a vector when element equals target -> Option A

4. Quick Check:

   Collect all matches by looping and storing indices [OK]

Hint: Collect indices by checking all elements without breaking [OK]

Common Mistakes:

• Breaking loop too early to find all matches
• Sorting vector which changes original order
• Using hash map counts instead of indices