# JavaScript Challenge - The Knapsack Problem

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So my code ended up a little differently from the psuedocode in the Knapsack Problem article (it didn’t exactly make sense to me…). I’m curious about others’ thoughts on this:

``````function knapsack(weightCap, weights, values) {
const itemCt = weights.length;
const matrix = new Array(itemCt);

for (let row = 0; row <= itemCt; row++) {
matrix[row] = new Array(weightCap + 1).fill(0);

if (row > 0) {
for (let col = 1; col <= weightCap; col++) {
matrix[row][col] = Math.max(matrix[row - 1][col], matrix[row][col]);

if (col === weights[row - 1]) {
matrix[row][col] = Math.max(matrix[row][col], values[row - 1]);
}

if (col + weights[row - 1] <= weightCap && matrix[row - 1][col] > 0) {
matrix[row][col + weights[row - 1]] = Math.max(matrix[row][col + weights[row - 1]], matrix[row - 1][col] + values[row - 1]);
}
}
}
}

console.log(matrix);
return Math.max(...matrix[itemCt]);
};

// Use this to test your function:
const weightCap = 10;
const weights = [3, 6, 8, 1, 3];
const val = [50, 60, 100, 200, 70];
console.log(knapsack(weightCap, weights, val));

// Leave this so we can test your code:
module.exports = knapsack;
``````

I used an array of booleans to keep track of which of the items (the indices) I “stole”.
I created a function to see what next additional item I could “steal” given an array of booleans representing what I’d already “stolen”; (creating more arrays of booleans to keep track of those new possible sets of things I could “steal”).
And I used a different function (recursively) to either add one possibility of things I could “steal” to the list of possibilities for a set I can choose to “steal”, or call the function to check if I could “steal” more.

Afterward, I iterated through my list (array) of possibilities, and checked which of then had the highest value; and returned that value.

My solution is not efficient because it produces the same possibility more than one … since it generates the possibility in every possible order of what can be “stolen”. (Meaning it creates permutations, but combinations are all that is needed).

``````function getMore(weightCap, weights, used) {
const length = weights.length;
// if every item is used (used is all true) then return null;
if (used.every(value => value)) {
return null;
}
let weight = 0;
for (let i = 0; i < length; i++) {
if (used[i]) {
weight += weights[i];
}
}
for (let j = 0; j < length; j++) {
if (!used[j]) {
if (weight + weights[j] <= weightCap) {
let another = used.slice();
another[j] = true;
}
}
}
// if no possibilities to add, then return null
return null;
}
}

function knapsack(weightCap, weights, values) {
const possibilities = [];
const length = weights.length;
let used = new Array(length);
used.fill(false);

function getPossibilities(weightCap, weights, used) {
let more = getMore(weightCap, weights, used);
if (more == null) {
possibilities.push(used);
}
else {
for (let boolArrays of more) {
getPossibilities(weightCap, weights, boolArrays);
}
}
}

getPossibilities(weightCap, weights, used);
let max = 0;
let best = used;
let i = 0;
for (let possibility of possibilities) {
let value = 0;
let weight = 0;
for (let j = 0; j < length; j++) {
if (possibility[j]) {
value += values[j];
weight += weights[j];
}
}
if ((value > max) && (weight <= weightCap)) {
max = value;
best = possibility;
}
i++;
}
return max;
};
``````

My solution took a while but here it is

``````
function knapsack(weightCap, weights, values) {

let high = 0;

for(let i = 0; i < weights.length; i++){
let cap = weights[i];//3
let val = values[i]; //70
let newCap = weights[i]; //3
let newVal = values[i];
for(let j = i + 1; j < weights.length; j++){
if(cap < weightCap && cap + weights[j] <= weightCap){
cap += weights[j];
val += values[j];
}else if(newCap + weights[j] <= weightCap){
cap = newCap + weights[j];
val = newVal + values[j];
}
if(val > high){
high = val;
}
}
}

return high;
};

// Use this to test your function:
const weightCap = 10;
const weights = [3, 6, 3];
const val = [70, 60, 100];
console.log(knapsack(weightCap, weights, val));

module.exports = knapsack;

//knapsack(weightCap, weights, val)
``````

This is horribly long and convoluted. . . but it does work . . . pretty new to JS and this was my first try at this problem!

``````function knapsack(weightCap, weights, values) {
let maxVal = 0;
let weightHolder = 0;
let valHolder = 0;

const helperSum = (a, b) => {
return a + b;
}
for (let i=0; i<values.length; i++) {
let a = weights[i]
for (let j=0; j<values.length; j++) {
if (i === j){
continue;
}
let b = weights[j]
if (helperSum(a, b) <= weightCap) {
weightHolder = helperSum(a, b);
valHolder = helperSum(values[i], values[j])
for (let x=0; x<values.length; x++) {
if (x === j || x == i) {
continue;
}
if (helperSum(weightHolder, weights[x]) <= weightCap) {
weightHolder = helperSum(weightHolder, weights[x]);
valHolder += values[x];
}
} if (valHolder > maxVal) {
maxVal = valHolder
}

} else if (values[i] > maxVal) {
maxVal = values[i]
}
}
} return maxVal;
};

// Use this to test your function:
const weightCap = 10;
const weights = [3, 6, 8];
const val = [50, 60, 100];
console.log(knapsack(weightCap, weights, val));

// Leave this so we can test your code:
module.exports = knapsack;`````````
``````function knapsack(weightCap, weights, values) {
const list = weights
.map((w, i) => ({w, i, v: values[i], p: (w / values[i])}))
.sort((a, b) => a.p - b.p)
let remainW = 10
let result = 0
list.forEach(({w, i, v, p}) => {
if (w <= remainW) {
result += v
remainW -= w
}
})
console.log(list)
return result
};
``````

### Output

110

``````[ { w: 3, i: 0, v: 50, p: 0.06 }, { w: 8, i: 2, v: 100, p: 0.08 }, { w: 6, i: 1, v: 60, p: 0.1 } ]
``````

2/5 tests passed.

``````function knapsack(weightCap, weights, values) {
let result = 0
const sortValuable = (a, b) => b.v - a.v
const list = weights.map((w, i) => ({w, i, v: values[i]})).sort(sortValuable)
const getMaxValuable = (items, item) => {
let remainW = weightCap - item.w
let sum = item.v
for (let otherItem of items.sort(sortValuable)) {
if (otherItem.i === item.i) continue;
if (remainW <= 0) break;
if (otherItem.w <= remainW) {
sum += otherItem.v
remainW -= otherItem.w
}
}
return sum
}
list.forEach(item => {
item.sum = getMaxValuable(list, item)
})
return list.sort((a, b) => b.sum - a.sum).sum
};

// Use this to test your function:
const weightCap = 10;
const weights = [3, 6, 8];
const val = [50, 60, 100];
console.log(knapsack(weightCap, weights, val));
``````

test pass complete

Honestly shocked this worked lol

const knapsack = (weightCap, weights, values) => { const matrix = values // 1) Map values and weights to 2D Array [[value, weight], ...] // 2) Filter out any items with weight above cap // 3) Sort highest value to lowest .map((value, idx) => [value, weights[idx]]) .filter(item => item <= weightCap) .sort((a, b) => b - a) let bagValue = 0; let bagWeight = 0; let itemIdx = 0; while(bagWeight < weightCap && itemIdx <= matrix.length - 1){ let itemValue = matrix[itemIdx]; let itemWeight = matrix[itemIdx]; if(bagWeight + itemWeight <= weightCap){ bagValue += itemValue; bagWeight += itemWeight; } itemIdx ++; } return bagValue; }; // Use this to test your function: const weightCap = 10; const weights = [3, 7, 6, 8]; const val = [50, 60, 60, 100]; console.log(knapsack(weightCap, weights, val)); // Leave this so we can test your code: module.exports = knapsack;
``````function knapsack(weightCap, weights, values) {
let maxValue = 0;
function next(num, weight, value) {
for (let i = num; i < weights.length; i++) {
if (weight + weights[i] <= weightCap) {
next(i+1, weight + weights[i], value + values[i]);
}
}
if (value > maxValue) {
maxValue = value;
}
}
next(0, 0, 0);
return maxValue;
};
``````
``````function knapsack(weightCap, weights, values) {