#### Zero One Knapsack

P
```/**
* A Dynamic Programming based solution for calculating Zero One Knapsack
* https://en.wikipedia.org/wiki/Knapsack_problem
*
* Time and Space Complexity: O(n*cap)
*/
const zeroOneKnapsack = (arr, n, cap, cache) => {
// Base Case: No capacity or no items
if (cap === 0 || n === 0) {
cache[n][cap] = 0
return cache[n][cap]
}

if (cache[n][cap] !== -1) {
return cache[n][cap]
}

// Profit when excluding the nth item
let notPick = zeroOneKnapsack(arr, n - 1, cap, cache)

// Profit when including the nth item
let pick = 0
if (arr[n - 1][0] <= cap) {
// If weight of the nth item is within the capacity
pick =
arr[n - 1][1] + zeroOneKnapsack(arr, n - 1, cap - arr[n - 1][0], cache)
}

cache[n][cap] = Math.max(pick, notPick) // maximize profit
return cache[n][cap]
}

const example = () => {
/*
Problem Statement:
You are a thief carrying a single bag with limited capacity S. The museum you stole had N artifact that you could steal. Unfortunately you might not be able to steal all the artifact because of your limited bag capacity.
You have to cherry pick the artifact in order to maximize the total value of the artifacts you stole.

*/
let input = `1
4 5
1 8
2 4
3 0
2 5
2 3`

input = input.trim().split('\n')
input.shift()
const length = input.length

const output = []

let i = 0
while (i < length) {
const cap = Number(input[i].trim().split(' ')[0])
const currlen = Number(input[i].trim().split(' ')[1])
let j = i + 1
const arr = []
while (j <= i + currlen) {
arr.push(input[j])
j++
}
const newArr = arr.map((e) => e.trim().split(' ').map(Number))
const cache = []
for (let i = 0; i <= currlen; i++) {
const temp = []
for (let j = 0; j <= cap; j++) {
temp.push(-1)
}
cache.push(temp)
}
const result = zeroOneKnapsack(newArr, currlen, cap, cache)
output.push(result)
i += currlen + 1
}

return output
}

export { zeroOneKnapsack, example }
```