My Solution to the classic napsack problem

napsack.js

// Dollars to notebooks
var valueMap = {
  5: 6,
  10: 13,
  1: 1
};

// How much money you are allowed
var money = 17;

// Save best values
best = 0;
list = [];
left = 0;


function canBuyAny(moneyLeft) {
  return Object.keys(valueMap).filter(function(x){
    return moneyLeft >= x;
  }).length > 0;
}

function napsack(moneyLeft, bought){
  if(canBuyAny(moneyLeft)){
    Object.keys(valueMap).forEach(function(x) {
      if(x <= moneyLeft) {
        return napsack(moneyLeft - x, [x].concat(bought.slice()));
      }
    });
  } else {
    let sum = bought.reduce((accum, x) => +accum + valueMap[x] , 0);

    if(sum > best){
      list = bought;
      best = sum;
      left = moneyLeft;
    }

    return bought;
  }
}

napsack(money, []);
console.log(`Money Left: ${left}; Prices paid: ${list.join(', ')}; Total Items: ${best}`);