Sort Colors Two Pointer Dutch Flag in DSA Rust - Time & Space Complexity

Time Complexity: Sort Colors Two Pointer Dutch Flag

O(n)

Understanding Time Complexity

We want to understand how the time needed to sort colors using two pointers changes as the list grows.

How does the number of steps grow when we have more colors to sort?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn sort_colors(nums: &mut Vec<i32>) {
    let (mut low, mut mid, mut high) = (0, 0, nums.len() - 1);
    while mid <= high {
        if nums[mid] == 0 {
            nums.swap(low, mid);
            low += 1;
            mid += 1;
        } else if nums[mid] == 1 {
            mid += 1;
        } else {
            nums.swap(mid, high);
            if high == 0 { break; }
            high -= 1;
        }
    }
}

This code sorts an array containing 0s, 1s, and 2s by moving pointers and swapping elements.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The while loop runs through the array with the mid pointer.
• How many times: Each element is visited at most once or twice, but the loop moves forward or backward carefully.

How Execution Grows With Input

As the list size grows, the number of steps grows roughly in direct proportion.

Input Size (n)	Approx. Operations
10	About 10 to 20 steps
100	About 100 to 200 steps
1000	About 1000 to 2000 steps

Pattern observation: The steps grow roughly linearly with input size.

Final Time Complexity

Time Complexity: O(n)

This means the time to sort grows directly with the number of elements in the list.

Common Mistake

[X] Wrong: "Because there are nested ifs and swaps, the time must be quadratic O(n²)."

[OK] Correct: The loop moves pointers carefully so each element is handled only a few times, not repeatedly nested.

Interview Connect

Understanding this linear time sorting method shows you can handle problems efficiently with clever pointer use, a skill valued in many coding challenges.

Self-Check

"What if we changed the input to have more than three colors? How would the time complexity change?"

Practice

1. What is the main goal of the Sort Colors Two Pointer Dutch Flag algorithm?

easy

A. To sort an array containing only 0s, 1s, and 2s in a single pass

B. To sort any array of integers in ascending order

C. To find the maximum value in an array

D. To reverse the elements of an array

Solution

1. Step 1: Understand the problem context

   The algorithm is designed specifically for arrays with only three distinct values: 0, 1, and 2.

2. Step 2: Identify the sorting goal

   The goal is to rearrange the array so that all 0s come first, then all 1s, and finally all 2s, in one pass without extra space.

3. Final Answer:

   To sort an array containing only 0s, 1s, and 2s in a single pass -> Option A

4. Quick Check:

   Sort Colors = Sort 0,1,2 in one pass [OK]

Hint: Remember: Dutch Flag sorts 0,1,2 in one pass [OK]

Common Mistakes:

• Thinking it sorts any integer array
• Confusing it with general sorting algorithms
• Assuming it reverses the array

2. Which of the following Rust code snippets correctly declares the pointers used in the Dutch Flag algorithm?

easy

A. let mut low = 0; let mut mid = 0; let mut high = nums.len() - 1;

B. let low = 0; let mid = 0; let high = nums.len();

C. let mut low = 1; let mut mid = 1; let mut high = nums.len() - 1;

D. let mut low = 0; let mut mid = 1; let mut high = nums.len();

Solution

1. Step 1: Identify pointer initialization

   In the Dutch Flag algorithm, low and mid start at 0, and high starts at the last index (len() - 1).

2. Step 2: Check mutability and bounds

   All pointers must be mutable to move during sorting, and high must be len() - 1 to avoid out-of-bounds errors.

3. Final Answer:

   let mut low = 0; let mut mid = 0; let mut high = nums.len() - 1; -> Option A

4. Quick Check:

   Correct pointer setup = let mut low = 0; let mut mid = 0; let mut high = nums.len() - 1; [OK]

Hint: Pointers start at 0 except high at last index [OK]

Common Mistakes:

• Setting high to nums.len() causing out-of-bounds
• Not using mutable variables
• Starting pointers at wrong indices

3. What will be the output of the following Rust code after applying the Dutch Flag algorithm?

let mut nums = vec![2, 0, 2, 1, 1, 0];
let mut low = 0;
let mut mid = 0;
let mut high = nums.len() - 1;

while mid <= high {
    match nums[mid] {
        0 => {
            nums.swap(low, mid);
            low += 1;
            mid += 1;
        },
        1 => {
            mid += 1;
        },
        2 => {
            nums.swap(mid, high);
            if high == 0 { break; }
            high -= 1;
        },
        _ => {}
    }
}
println!("{:?}", nums);

medium

A. [0, 1, 0, 1, 2, 2]

B. [2, 2, 1, 1, 0, 0]

C. [0, 0, 1, 1, 2, 2]

D. [1, 0, 2, 0, 1, 2]

Solution

1. Step 1: Trace initial swaps and pointer moves

   Starting with nums = [2,0,2,1,1,0], the algorithm swaps 2 at mid with 0 at high, then moves pointers accordingly.

2. Step 2: Continue until mid passes high

   After all swaps and increments, the array becomes sorted as [0, 0, 1, 1, 2, 2].

3. Final Answer:

   [0, 0, 1, 1, 2, 2] -> Option C

4. Quick Check:

   Sorted 0,1,2 array = [0,0,1,1,2,2] [OK]

Hint: Swap 0s left, 2s right, 1s stay middle [OK]

Common Mistakes:

• Not updating pointers correctly after swap
• Swapping with wrong indices
• Stopping loop too early

4. Identify the bug in this Rust code snippet implementing the Dutch Flag algorithm:

let mut nums = vec![2, 0, 1];
let mut low = 0;
let mut mid = 0;
let mut high = nums.len();

while mid <= high {
    match nums[mid] {
        0 => {
            nums.swap(low, mid);
            low += 1;
            mid += 1;
        },
        1 => {
            mid += 1;
        },
        2 => {
            nums.swap(mid, high);
            high -= 1;
        },
        _ => {}
    }
}

medium

A. The match statement is missing a case for 3

B. mid pointer is not incremented after swapping 2

C. low pointer is incremented incorrectly

D. high is initialized to nums.len(), causing out-of-bounds access

Solution

1. Step 1: Check pointer initialization

   high is set to nums.len(), which is one past the last valid index, causing out-of-bounds when accessing nums[high].

2. Step 2: Understand valid index range

   Valid indices are 0 to nums.len() - 1, so high should be initialized to nums.len() - 1.

3. Final Answer:

   high is initialized to nums.len(), causing out-of-bounds access -> Option D

4. Quick Check:

   high must be last index, not length [OK]

Hint: high = length - 1 to avoid out-of-bounds [OK]

Common Mistakes:

• Using length instead of last index for high
• Not incrementing mid after swapping 2
• Ignoring pointer bounds

5. Given an array nums = [1, 2, 0, 1, 2, 0, 1], how many swaps will the Dutch Flag algorithm perform to sort it?

hard

A. 3

B. 4

C. 5

D. 6

Solution

1. Step 1: Trace swaps during sorting

   Initial array: [1, 2, 0, 1, 2, 0, 1]. Swaps occur when 0 or 2 is found at mid.

2. Step 2: Count swaps step-by-step

   Swaps: - Swap 0 at index 2 with 1 at low (index 0) -> 1 swap - Swap 2 at index 1 with 1 at high (index 6) -> 1 swap - Swap 0 at index 5 with 1 at low (index 1) -> 1 swap - Swap 2 at index 1 with 1 at high (index 5) -> 1 swap Total swaps = 4.

3. Final Answer:

   4 -> Option B

4. Quick Check:

   Count swaps for 0 and 2 moves = 4 [OK]

Hint: Count swaps only when swapping 0 or 2 [OK]

Common Mistakes:

• Counting pointer moves as swaps
• Missing swaps when 2 is swapped
• Assuming swaps equal array length