CODEWITHSUNDEEP
javascriptgrouping
Kamil Kiełczewski
Kamil Kiełczewski

Reputation: 92447

Create string ranges from sorted array of integers

I have following sorted array of numbers (they can repeat e.g 21)

let a = [1,2,3,4,7,8,12,15,21,21,22,23]

And want to get following (sorted) array of strings with ranges

let r =["1-4","7-8","12","15","21-23"]

for consecutive numbers a,a+1,a+2,...,a+n=b w must create string "a-b" e.g for 6,7,8 we want to get "6-8", for "alone" numbers we want to get only that number e.g. for 12 in above example we get "12".

Here is my attempt but I stuck on it (and get headache)

let a = [1,2,3,6,7,8,12,15,21,21,22,23];
let right=a[0];
let left=a[0];
let result=[];

for(let i=1; i<a.length; i++) {
    for(let j=1; j<a.length; j++) {
      if(a[i]<a[j])result.push(`${a[i]}-${a[j]}`);
      
  }
}

console.log(JSON.stringify(result));

Update:

Here is "inversion" of this question

Upvotes: 12

Views: 493

Answers (5)

Kamil Kiełczewski
Kamil Kiełczewski

Reputation: 92447

Here is also my answer (inspired by others answers)

let r = (a,c='-',f=a[0],g=[]) => (a.map((x,i,a,b=a[i+1]) =>
        b<=x+1 ? 0 : (g.push( f-x ? f+c+x : f+'' ),f=b) ),g);


let a = [1,2,3,4,7,8,12,15,21,21,22,23];
let b = [-7,-5,-4,-3,-1,0,1,3,4,5,8]; 

let r = (a,c='-',f=a[0],g=[]) => (a.map((x,i,a,b=a[i+1]) =>
        b<=x+1 ? 0 : (g.push( f-x ? f+c+x : f+'' ),f=b) ),g);

let J= JSON.stringify;
console.log(J(a) + ' -> ' + J(r(a)));
console.log(J(b) + ' -> ' + J(r(b,'..')));

Explanation: g=[] contains result, first range element is f=a[0], at the begining we check does next element b=a[i+1] is equal or less than current element x plus one b<=x+1. If NO (which means that range ends or i is last index - b=undefined) then we push range to result g.push( f-x ? f+'-'+x : f+'' ) (we check here does x is not equal f by f-x -> f-x!=0 -> f!=x ), after push set first range element f to next range (element b).

Upvotes: 0

Nina Scholz
Nina Scholz

Reputation: 386654

You could store the next expected value in a closure and reduce the array.

function getRanges(array) {
    return array.reduce((l => (r, v, i, a) => {
        if (l[1] > v) return r;
        r.push(l[1] === v
            ? (r.pop(), l.join('-'))
            : (l = [v, v]).slice(0, 1).toString()
        );
        l[1]++;
        return r;
    })([]), []);
}

console.log(getRanges([-3, -2, -1, 2]));
console.log(getRanges([1, 2, 3, 4, 7, 8, 12, 15, 21, 21, 22, 23]));

Upvotes: 7

Jonas Wilms
Jonas Wilms

Reputation: 138277

You don't need two loops, just keep track were you started the group:

let array = [1,2,3,4,7,8,12,15,21,21,22,23]

const groups = [];
let start = array[0];

array.push(NaN);

for(let index = 1; index < array.length; index++) {
  const value = array[index], previous = array[index - 1];
  if(value === previous + 1 || value === previous) 
     continue;

  if(start === previous) {
    groups.push("" + previous);
  } else {
    groups.push(start + "-" + previous);
  }
  start = value;
}

console.log(groups);

Upvotes: 2

Sami Hult
Sami Hult

Reputation: 3082

Another way could be

let a = [1,2,3,4,7,8,12,15,21,21,22,23]

let r = a.reduce((acc, val) => {
  const lastGroup = acc.pop() || [];
  const lastValue = lastGroup.slice(-1)[0];
  if (val - lastValue > 1) {
    return [...acc, lastGroup, [val]];    
  }
  return [...acc, [...lastGroup, val]];
}, []).map(group => {
  const first = group[0];
  const last = group[group.length-1];
  return first !== last ? `${first}-${last}` : `${first}`;
});

console.log(r)

Upvotes: 2

jmcgriz
jmcgriz

Reputation: 3358

This should do it for you.

const a = [1,2,3,4,7,8,12,15,21,21,22,23,27]

let min = a[0], last = a[0]

const result = a.reduce((arr, n, i) => {
  if(n - last > 1){
    if(min == last) arr.push(""+min)
    else arr.push([min, last].join("-"))
    
    min = n
  }
  
  if(i == a.length - 1) {
    if(min == n) arr.push(""+min)
    else arr.push([min, n].join("-"))
  }
  
  last = n
  
  return arr
}, [])

console.log(result)

Upvotes: 3

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