I am trying this solution but it is not giving me the result I want. I don't want to count the number of pairs of 1 cause it is not really a pair as it appears 4 times. I need to count the "perfect" pairs, like in this case: 5 and 2. Right now this is giving me 4 as result and it should be 2.
How could I achieve that? I am stuck.
let ar1 = [12, 5, 5, 2, 1, 1, 1, 1, 2];
const countPairs = (ar) => {
let obj = {};
ar.forEach((item) => {
obj[item] = obj[item] ? obj[item] + 1 : 1;
});
return Object.values(obj).reduce((acc, curr) => {
acc += Math.floor(curr / 2);
return acc;
}, 0);
};
console.log( countPairs(ar1) )You can filter the object values by 2 and count the list
let ar1 = [12, 5, 5, 2, 1, 1, 1, 1, 2];
const countPairs = (ar) => {
let obj = {};
ar.forEach((item) => {
obj[item] = obj[item] ? obj[item] + 1 : 1;
});
return Object.values(obj).filter(e => e == 2).length;
};
console.log(countPairs(ar1))This can be one-liner using Map as:
const countPairs(arr) => [...arr.reduce((dict, n) => dict.set(n, (dict.get(n) ?? 0) + 1), new Map()).values(),].filter((n) => n === 2).length;
let ar1 = [12, 5, 5, 2, 1, 1, 1, 1, 2];
const countPairs = (arr) =>
[
...arr
.reduce((dict, n) => dict.set(n, (dict.get(n) ?? 0) + 1), new Map())
.values(),
].filter((n) => n === 2).length;
console.log(countPairs(ar1));or that
const
ar1 = [12, 5, 5, 2, 1, 1, 1, 1, 2]
, countPerfectPairs = arr => arr.reduce((r,val,i,{[i+1]:next})=>
{
if(!r.counts[val])
{
r.counts[val] = arr.filter(x=>x===val).length
if (r.counts[val]===2) r.pairs++
}
return next ? r : r.pairs
},{counts:{},pairs:0})
console.log( countPerfectPairs(ar1) )If you prefer Details:
const
ar1 = [12, 5, 5, 2, 1, 1, 1, 1, 2]
, countPerfectPairs = arr => arr.reduce((r,val)=>
{
if(!r.counts[val])
{
r.counts[val] = arr.filter(x=>x===val).length
if (r.counts[val]===2) r.pairs++
}
return r
},{counts:{},pairs:0})
console.log( countPerfectPairs(ar1) )