CODEWITHSUNDEEP
javascriptfor-loopsearch
Waqar Ul Khaf
Waqar Ul Khaf

Reputation: 619

Search and replace value of object property from one array with value of object property from second array?

I have two arrays of objects say 1- variants and 2- inventoryLevels. Objects in both arrays share a property which is the id. So I want to search for each variant if it's id is matched with any inventoryLevel I want to change its property named shopify_inventory_quantity with matched inventoryLevel's property available ? My words are little but confusing but take a look at code below basically it's doing properly whats needed I just want to know can it be optimized right now it's nested for loop. So any help to make it efficient would be appreciated ? 

for (let i = 0; i < variants.length; i++) {
  for (let j = 0; j < inventorylevels.length; j++) {
    if (variants[i].id === inventorylevels[j].variant_id) {
      variants[i].shopify_inventory_quantity = inventorylevels[j].available;
    }
  }
}

Upvotes: 0

Views: 64

Answers (1)

Dominik Schreiber
Dominik Schreiber

Reputation: 2771

I understand you have a solution in O(n²). Assuming your ids are unique, you can reduce the time complexity to O(n) (basically what @Alireza commented):

var variants = [
  {id: 0, shopify_inventory_quantity: 0},
  {id: 1, shopify_inventory_quantity: 0},
  {id: 2, shopify_inventory_quantity: 0}
];
var inventoryLevels = [
  {id: 0, available: 10},
  {id: 1, available: 2},
  {id: 2, available: 3}
];

// O(n) + O(n) = O(n)
function getAvailableVariants(v, i) {
  // O(n)
  var inventoryLevels = i.reduce(function(inventoryLevels, inventoryLevel) {
    inventoryLevels[inventoryLevel.id] = inventoryLevel;
    return inventoryLevels;
  }, {});
  
  // O(n)
  return v.map(variant => Object.assign(variant, {shopify_inventory_quantity: inventoryLevels[variant.id].available}));
}

var results = document.createElement('pre');
results.textContent = JSON.stringify(getAvailableVariants(variants, inventoryLevels), null, '\t');
document.body.appendChild(results);

Upvotes: 2

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