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DSA Javascriptprogramming~10 mins

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

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

Complete the code to select the next best item greedily based on value.

DSA Javascript
function greedySelect(items) {
  items.sort((a, b) => b.value - a.value);
  let result = [];
  for (let i = 0; i < items.length; i++) {
    if (canAdd(items[i])) {
      result.push(items[i]);
    }
  }
  return result;
}

function canAdd(item) {
  return [1];
}
Drag options to blanks, or click blank then click option'
Afalse
Btrue
Citem.value > 0
Ditem.weight < 10
Attempts:
3 left
💡 Hint
Common Mistakes
Returning false prevents adding any item.
Using conditions that filter out items incorrectly.
2fill in blank
medium

Complete the code to check if adding an item exceeds capacity.

DSA Javascript
function canAdd(item, currentWeight, capacity) {
  return currentWeight [1] capacity - item.weight;
}
Drag options to blanks, or click blank then click option'
A<=
B>
C<
D>=
Attempts:
3 left
💡 Hint
Common Mistakes
Using '>=' causes wrong logic and rejects valid items.
Using '<=' allows equal weight which may or may not be allowed.
3fill in blank
hard

Fix the error in the greedy algorithm that fails to consider item weights properly.

DSA Javascript
function greedyKnapsack(items, capacity) {
  items.sort((a, b) => b.value / b.weight - a.value / a.weight);
  let totalWeight = 0;
  let result = [];
  for (let i = 0; i < items.length; i++) {
    if (totalWeight + items[i].weight [1] capacity) {
      result.push(items[i]);
      totalWeight += items[i].weight;
    }
  }
  return result;
}
Drag options to blanks, or click blank then click option'
A<
B<=
C>
D>=
Attempts:
3 left
💡 Hint
Common Mistakes
Using '>=' or '>' causes wrong acceptance of overweight items.
Using '<=' may allow total weight equal to capacity which might be acceptable but can cause off-by-one errors.
4fill in blank
hard

Fill both blanks to complete the function that checks if greedy fails for this input.

DSA Javascript
function greedyFails(items, capacity) {
  items.sort((a, b) => a.weight - b.weight);
  let totalValue = 0;
  let totalWeight = 0;
  for (let i = 0; i < items.length; i++) {
    if (totalWeight + items[i].weight [1] capacity) {
      totalValue += items[i].value;
      totalWeight += items[i].weight;
    }
  }
  return totalValue [2] 60;
}
Drag options to blanks, or click blank then click option'
A<=
B<
C>
D>=
Attempts:
3 left
💡 Hint
Common Mistakes
Using '>=' in weight check allows overweight items.
Using '<=' in value check reverses logic.
5fill in blank
hard

Fill all three blanks to complete the function that demonstrates greedy success condition.

DSA Javascript
function greedySuccess(items, capacity) {
  items.sort((a, b) => b.value / b.weight - a.value / a.weight);
  let totalWeight = 0;
  let totalValue = 0;
  for (let i = 0; i < items.length; i++) {
    if (totalWeight + items[i].weight [1] capacity) {
      totalWeight += items[i].weight;
      totalValue += items[i].value;
    }
  }
  return totalValue [2] 100 && totalWeight [3] capacity;
}
Drag options to blanks, or click blank then click option'
A<=
B<
C>=
D>
Attempts:
3 left
💡 Hint
Common Mistakes
Using '>=' in weight check causes overweight items.
Using '<' in value check reverses success condition.
Using '>' in weight check allows overweight.

Practice

(1/5)
1. What is the main reason a greedy algorithm works correctly for some problems?
easy
A. Because it makes the best choice at each step hoping for a global optimum
B. Because it tries all possible solutions before choosing one
C. Because it uses recursion to explore all paths
D. Because it sorts the input data first

Solution

  1. Step 1: Understand greedy choice property

    A greedy algorithm picks the best option at each step without reconsidering previous choices.
  2. Step 2: Connect greedy choice to global optimum

    This local best choice leads to a global optimum only if the problem has the greedy choice property.
  3. Final Answer:

    Because it makes the best choice at each step hoping for a global optimum -> Option A
  4. Quick Check:

    Greedy choice property = best local choice [OK]
Hint: Greedy picks best local choice, not all solutions [OK]
Common Mistakes:
  • Thinking greedy tries all solutions
  • Confusing greedy with recursion
  • Assuming sorting always solves the problem
2. Which of the following is the correct syntax to start a greedy algorithm loop in JavaScript to process an array arr?
easy
A. for (const item in arr) {}
B. while (arr.length) {}
C. for (let i = 0; i < arr.length; i++) {}
D. arr.forEach(i => {})

Solution

  1. Step 1: Identify standard loop for arrays

    The standard way to loop over array elements by index is for (let i = 0; i < arr.length; i++).
  2. Step 2: Check other options

    while (arr.length) loops infinitely unless modified; for...in loops keys, not values; forEach is valid but less common for greedy step control.
  3. Final Answer:

    for (let i = 0; i < arr.length; i++) {} -> Option C
  4. Quick Check:

    Standard for loop = for (let i = 0; i < arr.length; i++) {} [OK]
Hint: Use for loop with index for greedy steps [OK]
Common Mistakes:
  • Using for...in for arrays (loops keys)
  • Using while without proper exit
  • Confusing forEach with for loop syntax
3. What will be the output of the following JavaScript code that tries to select intervals greedily by earliest finish time?
const intervals = [[1,4],[2,3],[3,5]];
intervals.sort((a,b) => a[1] - b[1]);
const result = [];
let lastEnd = 0;
for (const [start, end] of intervals) {
  if (start >= lastEnd) {
    result.push([start, end]);
    lastEnd = end;
  }
}
console.log(result);
medium
A. [[1,4]]
B. [[1,4],[3,5]]
C. [[2,3]]
D. [[2,3],[3,5]]

Solution

  1. Step 1: Sort intervals by finish time

    Sorted intervals: [[2,3],[1,4],[3,5]] because 3 < 4 < 5.
  2. Step 2: Select intervals greedily

    Start with lastEnd=0. Check [2,3]: 2 >= 0, select it, lastEnd=3. Next [1,4]: 1 < 3, skip. Next [3,5]: 3 >= 3, select it, lastEnd=5.
  3. Final Answer:

    [[2,3],[3,5]] -> Option D
  4. Quick Check:

    Greedy interval selection = [[2,3],[3,5]] [OK]
Hint: Sort by finish time, pick non-overlapping intervals [OK]
Common Mistakes:
  • Not sorting intervals by finish time
  • Selecting overlapping intervals
  • Updating lastEnd incorrectly
4. The following code tries to find the minimum number of coins for a given amount using greedy approach with coins [1, 3, 4]. Why does it fail for amount = 6?
const coins = [4, 3, 1];
let amount = 6;
let count = 0;
for (const coin of coins) {
  while (amount >= coin) {
    amount -= coin;
    count++;
  }
}
console.log(count);
medium
A. It outputs 3 but the minimum coins needed is 2
B. It outputs 2 which is correct
C. It causes an infinite loop
D. It outputs 6 which is incorrect

Solution

  1. Step 1: Trace greedy coin selection

    Greedy picks 4 once (amount=2), then picks 1 twice (amount=0), total 3 coins.
  2. Step 2: Check optimal solution

    Optimal is 2 coins: 3 + 3 = 6, fewer than greedy's 3 coins.
  3. Final Answer:

    It outputs 3 but the minimum coins needed is 2 -> Option A
  4. Quick Check:

    Greedy fails for coin sets without canonical property [OK]
Hint: Greedy may fail if coin set is non-standard [OK]
Common Mistakes:
  • Assuming greedy always gives minimum coins
  • Not checking smaller coin combinations
  • Ignoring coin order importance
5. You want to schedule the maximum number of non-overlapping meetings from the list [[0, 30], [5, 10], [15, 20]]. Which approach correctly applies greedy algorithm and what is the maximum number of meetings?
hard
A. Sort by start time, pick earliest start; maximum meetings = 2
B. Sort by finish time, pick earliest finish; maximum meetings = 2
C. Sort by duration, pick shortest; maximum meetings = 3
D. Sort by finish time, pick earliest finish; maximum meetings = 3

Solution

  1. Step 1: Sort meetings by finish time

    Sorted: [5,10], [15,20], [0,30] (finish times 10, 20, 30).
  2. Step 2: Select meetings greedily

    Pick [5,10], next [15,20] starts after 10, pick it; [0,30] overlaps, skip.
  3. Final Answer:

    Sort by finish time, pick earliest finish; maximum meetings = 2 -> Option B
  4. Quick Check:

    Greedy by finish time = 2 meetings [OK]
Hint: Sort by finish time for max non-overlapping meetings [OK]
Common Mistakes:
  • Sorting by start time instead of finish time
  • Assuming shortest duration always works
  • Counting overlapping meetings
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10 questions~5 mineasy hard