I want to create a function that counts pairs of an array and at the end count up the pairs, the expected result here should be:
{
1: 1,
10: 1,
90: 2
}, 4
But I am getting
{
1: 1,
90: 2
}, 3
Can someone give piece of advice how to rework this? It is not getting 10 as a pair because it occurs three times, but it should count it as a pair because it was up twice. Are there better methods to create this?
const pairs = [90, 10, 1, 2, 3, 4, 5, 10, 1, 90, 90, 90, 10, 22];
function checkPairs(arr) {
const result = {}
for(let i =0;i<pairs.length;i++) {
const counter = pairs.filter(item => item == arr[i]);
if(counter.length > 1 && counter.length % 2 == 0) {
result[pairs[i]] = counter.length/2;
}
}
return result;
}
function countPairs(obj) {
let result = 0;
for([i, t] of Object.entries(obj)) {
result+=t;
}
return result;
}
const countedPairs = checkPairs(pairs);
const counter = countPairs(countedPairs);
console.log(countedPairs, counter)Your check is saying there has to be exact number for pairs when you are doing counter.length % 2 == 0
You just need to make sure you have 2 or more and just round down with floor
const pairs = [90, 10, 1, 2, 3, 4, 5, 10, 1, 90, 90, 90, 10, 22];
function checkPairs(arr) {
const result = {}
for(let i =0;i<pairs.length;i++) {
const counter = pairs.filter(item => item == arr[i]);
if(counter.length > 1) {
result[pairs[i]] = Math.floor(counter.length/2);
}
}
return result;
}
function countPairs(obj) {
let result = 0;
for([i, t] of Object.entries(obj)) {
result+=t;
}
return result;
}
const countedPairs = checkPairs(pairs);
const counter = countPairs(countedPairs);
console.log(countedPairs, counter)Your filter code causes you to loop way more than you need to. You just have to loop once over the array and keep track of the times you have seen it.
const pairs = [90, 10, 1, 2, 3, 4, 5, 10, 1, 90, 90, 90, 10, 22];
const counts = pairs.reduce(function (o, val) {
// update how many times we have seen it.
o.count[val] = (o.count[val] || 0) + 1;
// is it the a pair yet?
if (o.count[val] ===2) {
// if it is, reset the count
o.count[val] = 0;
// update the count for the pair
o.pairsCount[val] = (o.pairsCount[val] || 0) + 1;
o.totalPairs++;
}
return o;
}, { count: {}, pairsCount: {}, totalPairs: 0});
console.log(counts.pairsCount, counts.totalPairs);How about you try this approach? Checking the length of counter array is even. If yes then do the regular calculation else just subtract 1 from counter's length.
const pairs = [90, 10, 1, 2, 3, 4, 5, 10, 1, 90, 90, 90, 10, 22];
function checkPairs(arr) {
const result = {}
for(let i =0;i<pairs.length;i++) {
const counter = pairs.filter(item => item == arr[i]);
if(counter.length > 1) {
result[pairs[i]] = counter.length % 2 == 0 ? counter.length / 2 : (counter.length -1) / 2 ;
}
}
return result;
}
function countPairs(obj) {
let result = 0;
for([i, t] of Object.entries(obj)) {
result+=t;
}
return result;
}
const countedPairs = checkPairs(pairs);
const counter = countPairs(countedPairs);
console.log(countedPairs, counter)You can also do it using Array.prototype.reduce.
1.total by 1.const
nums = [90, 10, 1, 2, 3, 4, 5, 10, 1, 90, 90, 90, 10, 22],
totalPairs = nums.reduce(
(acc, num) => {
acc[num] ??= 0;
acc[num] += 1;
if (!(acc[num] % 2)) {
acc.total += 1;
}
return acc;
},
{ total: 0 }
).total;
console.log(totalPairs);