Recursion on Arrays and Strings in DSA Rust - Time & Space Complexity

Time Complexity: Recursion on Arrays and Strings

O(n)

Understanding Time Complexity

When we use recursion on arrays or strings, we want to know how the time needed grows as the input gets bigger.

We ask: How many steps does the computer take as the array or string length increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn sum_array(arr: &[i32], index: usize) -> i32 {
    if index == arr.len() {
        return 0;
    }
    arr[index] + sum_array(arr, index + 1)
}

This function adds all elements of an array by calling itself with the next index until it reaches the end.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive call to process one element at a time.
• How many times: Once for each element in the array, so n times for array length n.

How Execution Grows With Input

Each recursive call handles one element, so the total steps grow directly with the array size.

Input Size (n)	Approx. Operations
10	10 recursive calls
100	100 recursive calls
1000	1000 recursive calls

Pattern observation: The number of steps grows in a straight line as input size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to complete grows directly with the size of the array or string.

Common Mistake

[X] Wrong: "Recursion always means exponential time because it calls itself multiple times."

[OK] Correct: Here, each call only makes one recursive call, so the total steps grow linearly, not exponentially.

Interview Connect

Understanding how recursion steps grow helps you explain your code clearly and choose the best approach during interviews.

Self-Check

"What if the function made two recursive calls each time instead of one? How would the time complexity change?"

Practice

1. What does recursion on arrays typically involve when processing elements?

easy

A. Using loops to iterate over the array elements

B. Calling the same function with a smaller part of the array until a base case is reached

C. Sorting the array before processing

D. Creating a new array for each element

Solution

1. Step 1: Understand recursion on arrays

   Recursion on arrays means the function calls itself with a smaller portion of the array each time.

2. Step 2: Identify the base case

   The recursion stops when the array is empty or has one element (base case).

3. Final Answer:

   Calling the same function with a smaller part of the array until a base case is reached -> Option B

4. Quick Check:

   Recursion = smaller input calls [OK]

Hint: Recursion breaks problem into smaller parts until simple case [OK]

Common Mistakes:

• Confusing recursion with loops
• Not defining a base case
• Trying to process the whole array at once

2. Which of the following is the correct Rust function signature for a recursive function that takes a slice of integers?

easy

A. fn recurse(arr: &mut [i32]) -> i32

B. fn recurse(arr: Vec<i32>) -> i32

C. fn recurse(arr: i32) -> &[i32]

D. fn recurse(arr: &[i32]) -> i32

Solution

1. Step 1: Understand Rust slice parameter

   A recursive function processing an array usually takes a slice reference like &[i32] to avoid ownership issues.

2. Step 2: Check return type

   Returning i32 is common if the function calculates a value like sum or count.

3. Final Answer:

   fn recurse(arr: &[i32]) -> i32 -> Option D

4. Quick Check:

   Slice input with reference = fn recurse(arr: &[i32]) -> i32 [OK]

Hint: Use &[i32] for read-only array slices in recursion [OK]

Common Mistakes:

• Using Vec<i32> causes ownership moves
• Returning a reference to local data
• Using mutable slice when not needed

3. What is the output of this Rust recursive function when called with arr = &[1, 2, 3]?

fn sum(arr: &[i32]) -> i32 {
    if arr.is_empty() {
        0
    } else {
        arr[0] + sum(&arr[1..])
    }
}

println!("{}", sum(arr));

medium

A. 0

B. 3

C. 6

D. Compilation error

Solution

1. Step 1: Trace recursive calls

   sum(&[1,2,3]) = 1 + sum(&[2,3]) sum(&[2,3]) = 2 + sum(&[3]) sum(&[3]) = 3 + sum(&[]) sum(&[]) = 0 (base case)

2. Step 2: Calculate total sum

   sum = 1 + 2 + 3 + 0 = 6

3. Final Answer:

   6 -> Option C

4. Quick Check:

   1+2+3 = 6 [OK]

Hint: Sum adds first element plus recursive sum of rest [OK]

Common Mistakes:

• Forgetting base case returns 0
• Adding wrong slice range
• Confusing output with length

4. Identify the error in this recursive Rust function that tries to count occurrences of a number in an array:

fn count_occurrences(arr: &[i32], target: i32) -> i32 {
    if arr.len() == 0 {
        0
    } else if arr[0] == target {
        1 + count_occurrences(&arr[1..], target)
    } else {
        count_occurrences(arr, target)
    }
}

medium

A. Infinite recursion because the else branch calls with the same array slice

B. Missing base case for empty array

C. Incorrect comparison operator in if condition

D. Function should return bool instead of i32

Solution

1. Step 1: Analyze recursive calls

   In the else branch, the function calls itself with the same array slice, causing infinite recursion.

2. Step 2: Identify fix

   The else branch should call with &arr[1..] to move forward and avoid infinite loop.

3. Final Answer:

   Infinite recursion because the else branch calls with the same array slice -> Option A

4. Quick Check:

   Same slice call causes infinite recursion [OK]

Hint: Always reduce array size in recursive calls to avoid infinite loops [OK]

Common Mistakes:

• Not reducing slice in else branch
• Confusing base case condition
• Returning wrong type

5. Given a string slice s: &str, which recursive Rust function correctly reverses the string?

hard

A. fn reverse(s: &str) -> String {
if s.len() == 0 {
String::new()
} else {
format!("{}{}", reverse(&s[1..]), &s[0..1])
}
}

B. fn reverse(s: &str) -> String {
if s.len() == 1 {
s.to_string()
} else {
format!("{}{}", &s[0..1], reverse(&s[1..]))
}
}

C. fn reverse(s: &str) -> String {
if s.is_empty() {
String::new()
} else {
format!("{}{}", &s[0..1], reverse(&s[1..]))
}
}

D. }
}
))]..1[s&(esrever ,]1..0[s& ,"}{}{"(!tamrof
{ esle }
)(wen::gnirtS
{ )(ytpme_si.s fi
{ gnirtS >- )rts& :s(esrever nf

Solution

1. Step 1: Understand string slicing and recursion

   To reverse, recursively reverse substring from index 1 onward, then append first character.

2. Step 2: Check base case and concatenation order

   Base case is empty string returning empty String. Concatenate reversed tail + first char.

3. Step 3: Compare options

   fn reverse(s: &str) -> String { if s.len() == 0 { String::new() } else { format!("{}{}", reverse(&s[1..]), &s[0..1]) } } matches correct base case and concatenation order.

4. Final Answer:

   fn reverse(s: &str) -> String { if s.len() == 0 { String::new() } else { format!("{}{}", reverse(&s[1..]), &s[0..1]) } } -> Option A

5. Quick Check:

   Reverse tail + first char = correct recursion [OK]

Hint: Reverse substring then add first char at end for string reversal [OK]

Common Mistakes:

• Appending first char before recursive call
• Wrong base case condition
• Using incorrect string slices