Here is an example for you to consider:
inventory = {'Apple':2, 'Orange' :1 , 'Mango' : 2}
In this inventory, both Apple and Mango have the maximum quantity. How can I create a function that returns both Apple and Mango as the answer?
I attempted something like this :
Object.keys(inventory).filter(function(item) { return inventory[item] === Math.max(...Object.values(inventory)); });
However, this only gives one result which is Apple.
To enhance efficiency, one could consider optimizing the process by calculating the maximum value in a more streamlined manner and then executing the filtering operation:
const data = {Apple: 2, Orange: 1, Mango: 2};
const maxValue = Object.keys(data).reduce((accumulator, currentValue) => Math.max(accumulator, data[currentValue]), -Infinity);
const filteredResults = Object.keys(data).filter(currentValue => data[currentValue] === maxValue);
console.log(filteredResults);This method maintains simplicity and readability while striving for greater operational efficiency.
To ensure your reduce function returns an array of values instead of a single value, initialize the result as an empty array.
Within the function, start by comparing each key with the current longest key to determine if it should replace the existing value in the array.
If the array is empty or all keys have the same value, add the key to the array.
Consider using more descriptive names for your arguments to improve readability and understanding.
Object.keys(hash).reduce(function(longestKeysArray, currentKey){
if(hash[currentKey] > hash[longestKeysArray[0]]){
longestKeysArray.length = 0;
}
if(longestKeysArray.length === 0 || hash[currentKey] === hash[longestKeysArray[0]]){
longestKeysArray.push(currentKey);
}
return longestKeysArray;
}, []);
To change the object into one with properties that match the count, where the values are the original property names in an array, and then retrieve the value of the property with the highest count:
const hash = {'Apple':2, 'Orange' :1 , 'Mango' : 2};
const indexedByCount = Object.entries(hash).reduce((a, [key, val]) => {
if (!a[val]) a[val] = [];
a[val].push(key);
return a;
}, {});
console.log(
indexedByCount[Math.max(...Object.keys(indexedByCount))]
);A method that is less functional but more efficient involves keeping track of a max variable representing the maximum value encountered so far:
const hash = {'Apple':2, 'Orange' :1 , 'Mango' : 2};
let max = -Infinity;
console.log(
Object.entries(hash).reduce((a, [key, val]) => {
if (val > max) {
max = val;
return [key];
}
if (val === max) a.push(key);
return a;
}, [])
);This method is effective. In case the value of max changes, it will clear the previous result and update with only the newly assigned maximum value.
var hash = {'Apple':2, 'Orange':1 , 'Mango':2, "Jackfruit":10, "Pineapple":5, "Tomato":-4};
var max="";
var result = Object.keys(hash).reduce(function(acc, val){
if(max < hash[val]) (max=hash[val], acc={});
if(hash[val]==max) acc[val] = hash[val];
return acc;
},{});
console.log(result)It is required to traverse the array only once and generate an array of keys as the result
const hash = {'Apple':2, 'Orange' :1 , 'Mango' : 2};
let max = 0;
let result = [];
Object.getOwnPropertyNames(hash).forEach(k => {
if (hash[k] > max) {
result = [k];
max = hash[k];
} else if (hash[k] === max) {
result.push(k);
}
});
console.log(result);
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