Priority Queue Introduction and Concept in DSA Rust - Time & Space Complexity

Time Complexity: Priority Queue Introduction and Concept

O(n log n)

Understanding Time Complexity

We want to understand how the time needed to use a priority queue changes as we add or remove items.

How does the cost grow when the queue gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

struct PriorityQueue {
    data: Vec<i32>,
}

impl PriorityQueue {
    fn push(&mut self, value: i32) {
        self.data.push(value);
        self.data.sort_unstable_by(|a, b| b.cmp(a));
    }

    fn pop(&mut self) -> Option<i32> {
        self.data.pop()
    }
}
    

This code adds values to a list and keeps it sorted so the biggest value is always first for easy removal.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Sorting the list after each insertion.
• How many times: Sorting happens every time we add a new value.

How Execution Grows With Input

Sorting takes longer as the list grows because it compares more items.

Input Size (n)	Approx. Operations
10	About 50 comparisons
100	About 700 comparisons
1000	About 10,000 comparisons

Pattern observation: The work grows faster than the number of items, roughly like n times log n.

Final Time Complexity

Time Complexity: O(n log n)

This means adding one item takes longer as the list grows, but not as fast as checking every pair.

Common Mistake

[X] Wrong: "Adding an item is always fast because we just put it at the end."

[OK] Correct: We must sort after adding, which takes more time as the list grows.

Interview Connect

Knowing how priority queues work and their time costs helps you explain choices clearly in interviews.

Self-Check

"What if we used a binary heap instead of sorting the list every time? How would the time complexity change?"

Practice

1. What is the main feature of a Priority Queue compared to a regular queue?

easy

A. It only stores elements of the same value

B. Elements are always dequeued in the order they were inserted

C. Elements are dequeued based on priority, not just order of insertion

D. It does not allow duplicate elements

Solution

1. Step 1: Understand queue behavior

   A regular queue removes elements in the order they were added (FIFO).

2. Step 2: Understand priority queue behavior

   A priority queue removes elements based on their priority, not just insertion order.

3. Final Answer:

   Elements are dequeued based on priority, not just order of insertion -> Option C

4. Quick Check:

   Priority Queue = Priority-based removal [OK]

Hint: Priority queue removes highest priority first, not oldest [OK]

Common Mistakes:

• Confusing priority queue with FIFO queue
• Thinking priority queue sorts elements permanently
• Assuming priority queue disallows duplicates

2. Which of the following is the correct way to create a max-heap priority queue in Rust using std::collections::BinaryHeap?

easy

A. let mut pq = std::collections::BinaryHeap::min_heap();

B. let mut pq = std::collections::BinaryHeap::with_capacity(10i32);

C. let mut pq = std::collections::BinaryHeap::default(10);

D. let mut pq = std::collections::BinaryHeap::new();

Solution

1. Step 1: Recall Rust BinaryHeap creation

   The standard way to create a max-heap priority queue is using BinaryHeap::new().

2. Step 2: Check other options

   with_capacity(10i32) type mismatch (usize expected); default(10) invalid arguments; min_heap() does not exist.

3. Final Answer:

   let mut pq = std::collections::BinaryHeap::new(); -> Option D

4. Quick Check:

   BinaryHeap::new() creates max-heap [OK]

Hint: Use BinaryHeap::new() for max-heap priority queue [OK]

Common Mistakes:

• Using with_capacity with i32 instead of usize
• Passing arguments to default()
• Using non-existent min_heap() method

3. What will be the output of the following Rust code using a priority queue?

use std::collections::BinaryHeap;

fn main() {
    let mut pq = BinaryHeap::new();
    pq.push(10);
    pq.push(5);
    pq.push(20);
    while let Some(val) = pq.pop() {
        print!("{} ", val);
    }
}

medium

A. 20 10 5

B. 5 20 10

C. 10 5 20

D. 5 10 20

Solution

1. Step 1: Understand BinaryHeap behavior

   Rust's BinaryHeap is a max-heap by default, so pop() returns the largest element first.

2. Step 2: Trace the pop order

   Inserted values: 10, 5, 20. Popping returns 20, then 10, then 5.

3. Final Answer:

   20 10 5 -> Option A

4. Quick Check:

   Max-heap pop order = descending [OK]

Hint: BinaryHeap pops largest first, so output is descending [OK]

Common Mistakes:

• Assuming pop returns elements in insertion order
• Thinking BinaryHeap is min-heap by default
• Mixing up push and pop order

4. Identify the error in this Rust code that tries to create a min-heap priority queue:

use std::collections::BinaryHeap;

fn main() {
    let mut pq = BinaryHeap::new();
    pq.push(3);
    pq.push(1);
    pq.push(2);
    while let Some(val) = pq.pop() {
        println!("{}", val);
    }
}

medium

A. BinaryHeap does not support min-heap by default

B. pq.push() method does not exist

C. pop() returns elements in insertion order

D. println! macro is used incorrectly

Solution

1. Step 1: Check BinaryHeap default behavior

   Rust's BinaryHeap is a max-heap by default, so it does not create a min-heap without extra work.

2. Step 2: Verify code correctness

   push() and pop() methods exist and println! is correct; the issue is the heap type.

3. Final Answer:

   BinaryHeap does not support min-heap by default -> Option A

4. Quick Check:

   BinaryHeap default = max-heap, no min-heap [OK]

Hint: BinaryHeap is max-heap by default; min-heap needs wrapper [OK]

Common Mistakes:

• Assuming BinaryHeap is min-heap by default
• Thinking push() or pop() are missing
• Misreading println! syntax

5. You want to implement a task scheduler in Rust that always runs the highest priority task first. Which approach correctly uses a priority queue to achieve this?

use std::collections::BinaryHeap;

struct Task {
    priority: i32,
    name: String,
}

fn main() {
    let mut pq = BinaryHeap::new();
    pq.push(Task { priority: 2, name: String::from("Clean") });
    pq.push(Task { priority: 5, name: String::from("Cook") });
    pq.push(Task { priority: 1, name: String::from("Shop") });

    while let Some(task) = pq.pop() {
        println!("{}", task.name);
    }
}

What must you do to make this code work as intended?

hard

A. Use a Vec<Task> instead of BinaryHeap

B. Implement Ord and Eq traits for Task to order by priority

C. Change priority to u32 type

D. Replace pop() with peek() in the loop

Solution

1. Step 1: Understand BinaryHeap requirements

   BinaryHeap requires elements to implement Ord and Eq to compare and order them.

2. Step 2: Check code for ordering

   Task struct has no ordering traits implemented, so BinaryHeap cannot order tasks by priority.

3. Step 3: Correct approach

   Implement Ord and Eq for Task to order by priority, so highest priority pops first.

4. Final Answer:

   Implement Ord and Eq traits for Task to order by priority -> Option B

5. Quick Check:

   BinaryHeap needs Ord/Eq for custom types [OK]

Hint: Custom types need Ord/Eq traits for BinaryHeap ordering [OK]

Common Mistakes:

• Using Vec instead of priority queue
• Changing data type without ordering
• Using peek() instead of pop() to remove elements