Algorithm Inventory Update
Problem Explanation:
- Look through an array of new products, in the format:
[quantity, name] - Return an array containing updated quantities for each item that already existed, and any new products
Hint: 1
- You need to work through each item of the new inventory to see if it exists in the current inventory or not.
- Remember that the product name is stored as the second element of each sub-array:
arr2[0][1] = "Bowling Ball"
Hint: 2
- If the item exists, you need to add the quantity from the new inventory
- If the item doesn't exist, you need to add the entire item
Hint: 3
- Return the completed inventory in alphabetical order
Spoiler Alert!
Solution ahead!
Code Solution:
First solution
function inventory(arr1, arr2) {
// Variable for location of product
var index;
// A helper method to return the index of a specified product (undefined if not found)
var getProductIndex = function (name) {
for (var i = 0; i < this.length; i++) {
if (this[i][1] === name) {
return i;
}
}
return undefined;
}
// For each item of the new Inventory
for (var i = 0; i < arr2.length; i++) {
// Invoke our helper function using arr1 as this
index = getProductIndex.call(arr1, arr2[i][1]);
// If the item doesn't exist
if (index === undefined) {
// Push the entire item
arr1.push(arr2[i]);
} else {
// Add the new quantity of the current item
arr1[index][0] += arr2[i][0];
}
}
// Sort alphabetically, by the product name of each item
arr1.sort(function (a, b) {
if (a[1] > b[1]) {
return 1;
}
if (a[1] < b[1]) {
return -1;
}
return 0;
});
return arr1;
}Second solution
function inventory(arr1, arr2) {
// All inventory must be accounted for or you're fired!
var index;
var arrCurInvName = []; // Names of arr1's items
var arrNeInvName = []; // Names of arr2's items
// Same as using two for loops, this takes care of increasing the number of stock quantity.
arr1.map(function(item1) {
return arr2.map(function(item2) {
if (item1[1] === item2[1]) {
item1[0] = item1[0] + item2[0]; //Increase number of stock
}
});
});
// Get item's name for new Inventory
arr2.map(function(item) {
arrNeInvName.push(item[1]);
});
// Get item's name for Current Inventory
arr1.map(function(item) {
arrCurInvName.push(item[1]);
});
// Add new inventory items to current inventory.
arrNeInvName.map(function(item) {
if (arrCurInvName.indexOf(item) === -1) {
index = arrNeInvName.indexOf(item);
arr1.push(arr2[index]);
}
});
// Sort the array alphabetically using the second element of the array as base.
arr1.sort(function(currItem, nextItem) {
//Ternary function to avoid using if else
return currItem[1] > nextItem[1] ? 1 : -1;
});
return arr1;
}Third solution
function inventory(arr1, arr2) {
var flag=0;
arr2.forEach(function(item){
flag=0;
arr1.forEach(function(list){
//If the product is already present, increase the quantity
if(item[1]===list[1]){ list[0]+=item[0]; flag=1;}
});
//If not already present, add the product
if(flag===0) arr1.push(item);
});
//Return the sorted inventory in alphabetical order wrt product name
return arr1.sort(function(a, b) {
return a[1] > b[1] ? 1 : -1;
});
}Fourth Solution
//jshint esversion: 6
function inventory(curInv, newInv) {
var inv = new Map();
[...curInv, ...newInv].forEach(item => {
if(inv.has(item[1]))
inv.set(item[1], inv.get(item[1]) + item[0]);
else
inv.set(item[1], item[0]);
});
return [...inv]
.map(item => [item[1], item[0]])
.sort((a, b) => a[1] > b[1] ? 1 : -1);
}Code Explanation:
First solution
- Start by creating a variable to store the index in. Define variables outside of loops
- Create a helper function to find the index of a product name
- The helper function iterates through each element of the array that it is called on, until it can either find the name parameter, or if it cannot find it then returns undefined
- Then, work through each item in the delivery, and set index to the result of invoking our helper function on the current inventory (IE: Search the new inventory for that product name, and return it's index)
- If we can't find that product, then we can add the entire product (Name and quantity) to the current inventory
- Otherwise, then we can add the quantity from the new inventory
- Then we sort the array by the product name (
arr1[x][1]holds the name) - Return the sorted array
- For Solution two, the explanation is on the code. Feel free to use different components on each solution to create your own if you like.
Second and Third solutions
- Read comments in code.
Fourth solution
- Start by creating the
invmap to store the updated inventory - Combine both
curInvandnewInvand iterate through - Check if
invhas the key, update it's value ifinvhas them. Otherwise store the new value - Use ES6 spread operator to convert
invmap to a 2d array - Map through the array to change it's location to be [value, key] as needed by the challenge
- Sort the array alphabetically
Related links
Credits:
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