Reputation: 237
In the below function I am attempting to get an output which resembles this:
[[1,1,1,1],[2,2,2], 4,5,10,[20,20], 391, 392,591].
I can see that the problem I have embedded is that I am always adding the temp array with a push to the functions return, as a result, all of the individual numbers apart from the last number in the for each function are being pushed into the target array with the array object also.
I feel as though I need a further conditonal check but for the life of me I am unable to come up with solution which works.
Any suggestions would be much appreciated.
const sortme = (unsortedArr)=> {
let tempArr = [];
let outputArr = [];
const reorderedArr = unsortedArr.sort((a,b) => a-b);
reorderedArr.forEach((number, i) => {
if ((i === 0) || (reorderedArr[i] === reorderedArr[i-1])) {
tempArr.push(number);
}
else {
outputArr.push(tempArr);
tempArr = [];
tempArr.push(number);
}
})
outputArr.push(tempArr[0]);
return outputArr;
}
const unsortedArr = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20];
sortme(unsortedArr);
Upvotes: 1
Views: 96
Reputation: 19
I hope the below one is quite simple;
function findSame(pos, sortedArr){
for(let i =pos; i<sortedArr.length; i++){
if(sortedArr[i] !== sortedArr[pos]){
return i
}
}
}
function clubSameNumbers(unsortedArr){
let sortedArr = unsortedArr.sort((a,b)=>a-b)
//[ 1, 1, 1, 1, 2, 2, 2, 4, 5, 10, 20, 20, 391, 392, 591 ]
let result = []
for(let i = 0; i < sortedArr.length; i = end){
let start = i
var end = findSame(i, sortedArr)
let arr = sortedArr.slice(i, end)
arr.length > 1 ? result.push(arr) : result.push(...arr)
}
return result
}
console.log(clubSameNumbers([1,2,4,591,392,391,2,5,10,2,1,1,1,20,20]))
//[ [ 1, 1, 1, 1 ], [ 2, 2, 2 ], 4, 5, 10, [ 20, 20 ], 391, 392, 591 ]
Upvotes: 0
Reputation: 30685
And another approach, just for fun:
const unsortedArr = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20];
var arr = unsortedArr.sort((a,b) => a-b).reduce((acc, value) => {
if (acc.length > 0 && acc[acc.length-1].includes(value)) {
acc[acc.length-1].push(value);
} else {
acc.push([value])
}
return acc;
}, []).map((v) => v.length > 1 ? v: v[0]);
console.log(arr);
Upvotes: 0
Reputation: 30685
This should work (using reduce):
const unsortedArr = [1,2,4,591,392,391,2,5,10,2,1,1,1,20,20];
let lastValue = null;
var newArr = unsortedArr.sort((a,b) => a-b).reduce((acc, value) => {
if (acc.length == 0 || ((acc.length > 0 || !acc[acc.length-1].length) && lastValue !== value)) {
acc.push(value);
} else if (acc.length > 0 && lastValue === value) {
acc[acc.length-1] = (acc[acc.length-1].length ? acc[acc.length-1].concat([value]): [value, value]);
}
lastValue = value;
return acc;
}, []);
console.log(newArr);
Upvotes: 1
Reputation: 17654
i would make a deduped copy and .map()
it to transform the values into arrays containing values from the original ( sorted ) array that you get using a .forEach
:
const unsortedArr = [1, 2, 4, 591, 392, 391, 2, 5, 10, 2, 1, 1, 1, 20, 20];
const sortMe = (arr) => {
arr = arr.sort((a, b) => a - b);
// a short way to dedupe an array
// results in : 1, 2, 4, 5, 10, 20, 391, 392, 591
let dedupe = [...new Set(arr)];
let tmpArr;
return dedupe.map(e => {
tmpArr = []; // empty tmpArr on each iteration
// for each element of the deduped array, look for matching elements in the original one and push them in the tmpArr
arr.forEach(a => {
if (a === e)
tmpArr.push(e);
})
if(tmpArr.length === 1)
return tmpArr[0]; // in case you have [4] , just return the 4
else
return tmpArr; // in case you have [1,1,1,1]
// shorthand for the if/else above
// return tmpArr.length === 1 ? tmpArr[0] : tmpArr;
});
}
const result = sortMe(unsortedArr);
console.log(result);
Upvotes: 2