Why Greedy Works and When It Fails in DSA Rust - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to select the next interval with the earliest finish time.

DSA Rust

intervals.sort_by_key(|interval| interval.[1]);

Drag options to blanks, or click blank then click option'

Aend

Bstart

Clength

Dindex

Attempts:

3 left

💡 Hint

Common Mistakes

Sorting by start time instead of end time.

Sorting by length which doesn't guarantee optimal selection.

Sorting intervals by their end time helps the greedy algorithm pick the interval that finishes earliest, allowing more intervals to fit later.

2fill in blank

medium

Complete the code to check if the current interval can be added without overlapping.

DSA Rust

if intervals[i].start >= [1] { selected.push(intervals[i]); last_end = intervals[i].end; }

Drag options to blanks, or click blank then click option'

Aselected.len()

Bintervals[i].end

Cintervals[i].start

Dlast_end

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing start with start instead of end.

Using the length of selected intervals incorrectly.

We compare the start of the current interval with the end time of the last selected interval to avoid overlaps.

3fill in blank

hard

Fix the error in the greedy coin change algorithm that fails for some coin sets.

DSA Rust

let mut amount = [1]; let mut coins_used = 0;

Drag options to blanks, or click blank then click option'

Acoins.len()

Btotal_amount

C0

Dcoins[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing amount to zero or coin count.

Using the first coin value as amount.

The variable 'amount' should start with the total amount to change, not zero or coin count.

4fill in blank

hard

Fill both blanks to complete the greedy interval scheduling loop.

DSA Rust

for i in 1..intervals.len() {
    if intervals[i].[1] >= [2] {
        selected.push(intervals[i]);
        last_end = intervals[i].end;
    }
}

Drag options to blanks, or click blank then click option'

Astart

Blast_end

Cend

Dstart_time

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'end' instead of 'start' for current interval.

Comparing with wrong variable for last end time.

We check if the start of the current interval is after or equal to the end of the last selected interval to avoid overlap.

5fill in blank

hard

Fill all three blanks to complete the greedy coin change selection.

DSA Rust

while amount > 0 {
    let coin = coins[[1]];
    if coin <= amount {
        amount -= coin;
        coins_used += [2];
    } else {
        [3];
    }
}

Drag options to blanks, or click blank then click option'

Aindex

B1

Cindex += 1

Dcoins.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variable names.

Not incrementing coin count correctly.

Not moving to next coin properly.

We use 'index' to track the current coin, add one coin when used, and move to next coin when current coin is too big.

Practice

1. What is the main reason a greedy algorithm works correctly for some problems?

easy

A. Because it tries all possible solutions and picks the best one

B. Because choosing the best option at each step leads to an optimal solution

C. Because it uses recursion to explore all paths

D. Because it sorts the input data before processing

Solution

1. Step 1: Understand greedy choice property

   A greedy algorithm works when making the best local choice at each step leads to a globally optimal solution.

2. Step 2: Differentiate from other methods

   Unlike exhaustive search or recursion, greedy does not try all solutions but relies on local optimal choices.

3. Final Answer:

   Because choosing the best option at each step leads to an optimal solution -> Option B

4. Quick Check:

   Greedy choice property = A [OK]

Hint: Greedy picks best local choice aiming for global best [OK]

Common Mistakes:

• Thinking greedy tries all solutions
• Confusing greedy with recursion or backtracking
• Assuming sorting alone makes greedy work

2. Which of the following Rust code snippets correctly implements a greedy approach to select intervals with earliest finishing time?

easy

A. intervals.sort_by_key(|x| x.start); let mut count = 0; let mut last_end = 0; for i in &intervals { if i.end >= last_end { count += 1; last_end = i.start; } }

B. intervals.sort_by_key(|x| x.end); let mut count = 0; for i in &intervals { count += 1; }

C. intervals.sort_by_key(|x| x.end); let mut count = 0; let mut last_end = 0; for i in &intervals { if i.start >= last_end { count += 1; last_end = i.end; } }

D. intervals.sort_by_key(|x| x.start); let mut count = 0; for i in &intervals { if i.start > last_end { count += 1; last_end = i.end; } }

Solution

1. Step 1: Identify correct sorting key

   The greedy interval scheduling algorithm sorts intervals by their finishing time (end).

2. Step 2: Check selection condition and update

   It selects intervals starting after or at the last selected interval's end, updating last_end accordingly.

3. Final Answer:

   intervals.sort_by_key(|x| x.end); let mut count = 0; let mut last_end = 0; for i in &intervals { if i.start >= last_end { count += 1; last_end = i.end; } } -> Option C

4. Quick Check:

   Sort by end, pick if start >= last_end = B [OK]

Hint: Sort by finish time, pick intervals starting after last end [OK]

Common Mistakes:

• Sorting by start time instead of end time
• Updating last_end incorrectly
• Counting all intervals without condition

3. Consider the Rust code below that tries to find the minimum number of coins for a given amount using greedy approach with coin denominations [25, 10, 5, 1]. What is the output for amount = 30?

let coins = [25, 10, 5, 1];
let mut amount = 30;
let mut count = 0;
for &coin in &coins {
    count += amount / coin;
    amount %= coin;
}
println!("{}", count);

medium

A. 4

B. 3

C. 5

D. 2

Solution

1. Step 1: Calculate coins for 25-cent

   30 / 25 = 1 coin, remainder 30 % 25 = 5

2. Step 2: Calculate coins for 10-cent, 5-cent, 1-cent

   5 / 10 = 0 coins, remainder 5; 5 / 5 = 1 coin, remainder 0; 0 / 1 = 0 coins

3. Step 3: Sum total coins

   1 (25-cent) + 0 (10-cent) + 1 (5-cent) + 0 (1-cent) = 2 coins

4. Final Answer:

   2 -> Option D

5. Quick Check:

   30 = 25 + 5 = 2 coins [OK]

Hint: Divide amount by largest coin, reduce remainder, sum counts [OK]

Common Mistakes:

• Adding coins incorrectly
• Using wrong coin order
• Ignoring remainder after division

4. The following Rust code attempts to solve the coin change problem greedily with coins [10, 6, 1] and amount 12. It outputs the number of coins used. What is the bug?

let coins = [10, 6, 1];
let mut amount = 12;
let mut count = 0;
for &coin in &coins {
    count += amount / coin;
    amount %= coin;
}
println!("{}", count);

medium

A. The greedy approach fails to find minimum coins for these denominations

B. The code has a syntax error in the for loop

C. The amount variable is not mutable

D. The coins array is not sorted

Solution

1. Step 1: Understand greedy failure for coin set

   Greedy picks 10 once, remainder 2, then two 1 coins = 3 coins total.

2. Step 2: Check for better solution

   Using two 6 coins gives 12 with only 2 coins, better than greedy's 3.

3. Step 3: Identify bug

   The bug is greedy approach itself fails to find minimum coins for this coin set.

4. Final Answer:

   The greedy approach fails to find minimum coins for these denominations -> Option A

5. Quick Check:

   Greedy fails on non-canonical coins = D [OK]

Hint: Greedy fails if coin set is non-canonical [OK]

Common Mistakes:

• Thinking code syntax is wrong
• Assuming sorting fixes greedy failure
• Ignoring greedy algorithm limitations

5. You want to select activities from a list with start and end times to maximize the number of activities chosen. However, some activities overlap partially. Which approach correctly applies greedy algorithm to solve this, and why might it fail if you choose activities by earliest start time instead of earliest finish time?

hard

A. Sort activities by earliest finish time and pick compatible ones; choosing by start time can miss optimal sets

B. Sort activities by earliest start time and pick all that start earliest; this always works

C. Pick activities randomly; greedy always finds optimal solution

D. Sort activities by longest duration and pick those first; this maximizes usage

Solution

1. Step 1: Identify correct greedy strategy

   Sorting by earliest finish time ensures maximum remaining time for other activities.

2. Step 2: Explain failure of sorting by start time

   Choosing by earliest start time can select long activities blocking many others, missing optimal count.

3. Step 3: Conclude correct approach and failure reason

   Greedy works with earliest finish time sorting; earliest start time sorting may fail to maximize activities.

4. Final Answer:

   Sort activities by earliest finish time and pick compatible ones; choosing by start time can miss optimal sets -> Option A

5. Quick Check:

   Earliest finish time sorting = A [OK]

Hint: Sort by finish time, not start time, to maximize activities [OK]

Common Mistakes:

• Assuming earliest start time sorting always works
• Picking longest duration first
• Ignoring overlap constraints