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DSA Rustprogramming~10 mins

Dutch National Flag Three Way Partition in DSA Rust - Interactive Practice

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Practice - 5 Tasks
Answer the questions below
1fill in blank
easy

Complete the code to initialize the low pointer to zero.

DSA Rust
let mut low = [1];
Drag options to blanks, or click blank then click option'
ANone
B1
C-1
D0
Attempts:
3 left
💡 Hint
Common Mistakes
Initializing low to 1 or -1 causes index errors.
Using None is invalid for integer index.
2fill in blank
medium

Complete the code to swap elements at indices i and j.

DSA Rust
arr.swap([1], j);
Drag options to blanks, or click blank then click option'
Ai
Bj
Clow
Dhigh
Attempts:
3 left
💡 Hint
Common Mistakes
Swapping with j instead of i swaps the wrong elements.
Using low or high without context causes logic errors.
3fill in blank
hard

Fix the error in the loop condition to continue while i is less than or equal to high.

DSA Rust
while i [1] high {
Drag options to blanks, or click blank then click option'
A<=
B>
C<
D>=
Attempts:
3 left
💡 Hint
Common Mistakes
Using < excludes the element at high index.
Using > or >= causes the loop to never run or run incorrectly.
4fill in blank
hard

Fill both blanks to correctly update pointers after swapping with low.

DSA Rust
arr.swap(i, [1]);
low [2] 1;
Drag options to blanks, or click blank then click option'
Alow
Bhigh
C+=
D-=
Attempts:
3 left
💡 Hint
Common Mistakes
Swapping with high instead of low causes wrong partition.
Decrementing low instead of incrementing breaks the algorithm.
5fill in blank
hard

Fill all three blanks to correctly update pointers after swapping with high.

DSA Rust
arr.swap(i, [1]);
high [2] 1;
[3] -= 1;
Drag options to blanks, or click blank then click option'
Ahigh
Bi
C-=
D+=
Attempts:
3 left
💡 Hint
Common Mistakes
Incrementing high instead of decrementing causes infinite loop.
Not decrementing i causes skipping elements.

Practice

(1/5)
1. What is the main goal of the Dutch National Flag Three Way Partition algorithm?
easy
A. To sort an array in ascending order
B. To reverse the elements of an array
C. To find the maximum element in an array
D. To rearrange an array into three parts based on a pivot value

Solution

  1. Step 1: Understand the problem goal

    The algorithm divides the array into three parts: less than pivot, equal to pivot, and greater than pivot.
  2. Step 2: Compare with other options

    Sorting or finding max are different tasks; reversing is unrelated.
  3. Final Answer:

    To rearrange an array into three parts based on a pivot value -> Option D
  4. Quick Check:

    Partitioning by pivot = rearrange into three parts [OK]
Hint: Remember: Dutch Flag splits array into three color groups [OK]
Common Mistakes:
  • Confusing partition with full sorting
  • Thinking it finds max or min values
  • Assuming it reverses the array
2. Which of the following Rust code snippets correctly declares mutable indices for the Dutch National Flag algorithm?
easy
A. let mut low = 0; let mut mid = 0; let mut high = arr.len() - 1;
B. let low = 0; let mid = 0; let high = arr.len() - 1;
C. let mut low = 0; let mid = 0; let mut high = arr.len() - 1;
D. let low = 0; let mut mid = 0; let high = arr.len() - 1;

Solution

  1. Step 1: Identify mutability requirement

    Indices must be mutable to update during partitioning.
  2. Step 2: Check each declaration

    Only let mut low = 0; let mut mid = 0; let mut high = arr.len() - 1; declares all three indices as mutable.
  3. Final Answer:

    let mut low = 0; let mut mid = 0; let mut high = arr.len() - 1; -> Option A
  4. Quick Check:

    All indices mutable = correct declaration [OK]
Hint: Indices change, so declare all as mutable with 'mut' [OK]
Common Mistakes:
  • Forgetting 'mut' keyword for variables that change
  • Mixing mutable and immutable declarations incorrectly
  • Using immutable variables causing compile errors
3. What will be the output of the following Rust code after applying the Dutch National Flag partition with pivot = 1?
let mut arr = vec![2, 0, 2, 1, 1, 0];
let pivot = 1;
// After partitioning
// arr = ?
medium
A. [1, 1, 0, 0, 2, 2]
B. [0, 0, 1, 1, 2, 2]
C. [2, 2, 1, 1, 0, 0]
D. [0, 1, 0, 1, 2, 2]

Solution

  1. Step 1: Understand partitioning by pivot 1

    Elements less than 1 go first (0s), then equal to 1, then greater than 1 (2s).
  2. Step 2: Rearrange array accordingly

    Original: [2, 0, 2, 1, 1, 0] After partition: [0, 0, 1, 1, 2, 2]
  3. Final Answer:

    [0, 0, 1, 1, 2, 2] -> Option B
  4. Quick Check:

    Partitioned array = [0, 0, 1, 1, 2, 2] [OK]
Hint: Group elements less, equal, then greater than pivot [OK]
Common Mistakes:
  • Not grouping equal elements in the middle
  • Mixing order of less and greater groups
  • Assuming full sorting instead of partitioning
4. Identify the bug in this Rust code snippet implementing Dutch National Flag partition:
let mut low = 0;
let mut mid = 0;
let mut high = arr.len() - 1;
while mid <= high {
    if arr[mid] < pivot {
        arr.swap(low, mid);
        low += 1;
        mid += 1;
    } else if arr[mid] == pivot {
        mid += 1;
    } else {
        arr.swap(mid, high);
        high -= 1;
    }
}
medium
A. mid should be decremented in else block
B. Incorrect comparison operators for pivot
C. Missing increment of mid after swapping with low
D. high should be incremented instead of decremented

Solution

  1. Step 1: Analyze pointer updates after swaps

    When swapping arr[low] and arr[mid], mid must also move forward to avoid infinite loop.
  2. Step 2: Identify missing mid increment

    Code increments mid only when arr[mid] == pivot, but not after swapping with low.
  3. Final Answer:

    Missing increment of mid after swapping with low -> Option C
  4. Quick Check:

    mid must increment after swap with low to progress [OK]
Hint: Increment mid after swapping with low to avoid infinite loop [OK]
Common Mistakes:
  • Forgetting to increment mid after swap with low
  • Changing mid incorrectly in else block
  • Misunderstanding pointer movements
5. Given an array with elements 0, 1, and 2 representing colors red, white, and blue respectively, how would you modify the Dutch National Flag algorithm to sort the array so that all reds come first, whites second, and blues last? Choose the correct approach.
hard
A. Use three pointers low, mid, high; move mid forward, swap with low if 0, swap with high if 2.
B. Sort the array using Rust's built-in sort() method.
C. Use two pointers only, swapping elements when needed.
D. Count the number of 0s, 1s, and 2s, then overwrite the array accordingly.

Solution

  1. Step 1: Understand the Dutch National Flag approach

    It uses three pointers to partition the array into three parts efficiently in one pass.
  2. Step 2: Compare with other methods

    Built-in sort or counting are valid but not the Dutch National Flag algorithm; two pointers are insufficient for three groups.
  3. Final Answer:

    Use three pointers low, mid, high; move mid forward, swap with low if 0, swap with high if 2. -> Option A
  4. Quick Check:

    Three pointers and swaps = Dutch National Flag method [OK]
Hint: Three pointers: low, mid, high handle 0,1,2 in one pass [OK]
Common Mistakes:
  • Using sorting instead of partitioning
  • Trying two pointers for three groups
  • Ignoring efficient one-pass solution
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10 questions~5 mineasy hard