What if you could pack your bag perfectly in minutes instead of hours by just slicing items smartly?
Why Fractional Knapsack Problem in DSA Javascript?
Imagine you have a backpack with limited space and many items to carry. Each item has a weight and a value. You want to fill your backpack to get the most value possible. Doing this by trying every combination by hand is like trying to solve a huge puzzle without any clues.
Manually checking every possible way to fill the backpack takes forever and is very confusing. You might miss better combinations or spend hours calculating. It's easy to make mistakes and waste time.
The Fractional Knapsack Problem lets you take parts of items instead of whole items only. By sorting items by value per weight and taking the best parts first, you quickly find the best way to fill the backpack with maximum value.
let maxValue = 0; // Try all combinations manually for (let i = 0; i < items.length; i++) { for (let j = i + 1; j < items.length; j++) { // Calculate value and weight for combination } }
items.sort((a, b) => (b.value / b.weight) - (a.value / a.weight)); let totalValue = 0; let capacityLeft = maxCapacity; for (const item of items) { if (item.weight <= capacityLeft) { totalValue += item.value; capacityLeft -= item.weight; } else { totalValue += item.value * (capacityLeft / item.weight); break; } }
This method lets you quickly find the best way to fill limited space with the highest total value, even if you can take parts of items.
Imagine packing a suitcase for a trip where you can only carry a certain weight. You want to pack the most valuable things, even if you only take part of some items like liquids or snacks.
Manual checking of all combinations is slow and error-prone.
Fractional Knapsack sorts items by value per weight for quick decisions.
Allows taking parts of items to maximize total value efficiently.
Practice
Fractional Knapsack Problem?Solution
Step 1: Understand the problem goal
The fractional knapsack problem allows taking parts of items to maximize total value within a weight limit.Step 2: Compare options
Maximize total value by taking fractions of items within weight limit correctly states maximizing value by taking fractions. Others ignore fractions or value maximization.Final Answer:
Maximize total value by taking fractions of items within weight limit -> Option AQuick Check:
Fractional knapsack = maximize value with fractions [OK]
- Thinking only whole items can be taken
- Ignoring value and focusing on weight
- Confusing with 0/1 knapsack problem
Solution
Step 1: Identify sorting criteria
We must sort items by descending value-to-weight ratio (value/weight) to pick best items first.Step 2: Check each code snippet
items.sort((a, b) => (b.value / b.weight) - (a.value / a.weight)); sorts descending by (value/weight). Others sort incorrectly by inverse or by value or weight alone.Final Answer:
items.sort((a, b) => (b.value / b.weight) - (a.value / a.weight)); -> Option AQuick Check:
Sort by descending value/weight ratio [OK]
- Sorting by weight or value alone
- Sorting ascending instead of descending
- Using inverse ratio (weight/value)
const items = [
{value: 60, weight: 10},
{value: 100, weight: 20},
{value: 120, weight: 30}
];
const capacity = 50;
function fractionalKnapsack(items, capacity) {
items.sort((a, b) => (b.value / b.weight) - (a.value / a.weight));
let totalValue = 0;
let remaining = capacity;
for (const item of items) {
if (item.weight <= remaining) {
totalValue += item.value;
remaining -= item.weight;
} else {
totalValue += item.value * (remaining / item.weight);
break;
}
}
return totalValue;
}
console.log(fractionalKnapsack(items, capacity));Solution
Step 1: Sort items by value/weight ratio
Ratios: 60/10=6, 100/20=5, 120/30=4. Sorted order: item1, item2, item3.Step 2: Add items until capacity 50
Take item1 fully (weight 10, value 60), remaining 40.
Take item2 fully (weight 20, value 100), remaining 20.
Take fraction of item3: (20/30)*120 = 80.
Total value = 60 + 100 + 80 = 240.Final Answer:
240 -> Option BQuick Check:
Sum values with fractions = 240 [OK]
- Not sorting by ratio correctly
- Taking items without fractions
- Miscomputing fraction value
function fractionalKnapsack(items, capacity) {
items.sort((a, b) => (a.value / a.weight) - (b.value / b.weight));
let totalValue = 0;
let remaining = capacity;
for (const item of items) {
if (item.weight <= remaining) {
totalValue += item.value;
remaining -= item.weight;
} else {
totalValue += item.value * (remaining / item.weight);
break;
}
}
return totalValue;
}Solution
Step 1: Analyze sorting order
The code sorts items ascending by value/weight ratio, but we need descending to pick best items first.Step 2: Check other parts
Return statement exists, fraction formula is correct, 'let' is appropriate for totalValue.Final Answer:
Sorting is ascending; should be descending by value/weight ratio -> Option DQuick Check:
Sort descending by ratio to maximize value [OK]
- Sorting ascending instead of descending
- Ignoring fraction calculation correctness
- Confusing variable declarations
items = [
{value: 40, weight: 2},
{value: 100, weight: 20},
{value: 50, weight: 10},
{value: 60, weight: 12}
];
capacity = 30;What is the maximum total value achievable using the fractional knapsack approach?
Solution
Step 1: Calculate value/weight ratios and sort descending
Ratios:
Item1: 40/2=20
Item2: 100/20=5
Item3: 50/10=5
Item4: 60/12=5
Sorted order: Item1, then Items 2,3,4 (equal ratio 5).Step 2: Pick items until capacity 30
Take Item1 fully: weight 2, value 40, remaining 28.
Next, take Item2 fully: weight 20, value 100, remaining 8.
Then take fraction of Item3: fraction = 8/10 = 0.8, value = 50 * 0.8 = 40.
Total value = 40 + 100 + 40 = 180.Step 3: Check if taking Item4 fraction is better
Item4 has same ratio as Item3, but we already used remaining capacity on Item3 fraction.
So total value remains 180.Final Answer:
180 -> Option CQuick Check:
Sum values with fractions = 180 [OK]
- Ignoring highest ratio item first
- Not taking fractions correctly
- Adding values without considering capacity
Ready to prove it?
Pass the test and this pattern moves unknown
