Separate numbers from array and order them, if there are consecutive 0's, put X [closed]-CodePudding

There is an arrangement with a list of numbers, which can be of any size.
These are divided into blocks with the number zero, which marks the beginning and end of this and another block.
A zero at the beginning and end of the block would imply that there are no elements.
These blocks also need to be sorted in ascending order.
When printing the sequences must be separated by a space.
If there is a sequence of zeros when printed it should be represented by an X.

Example 1: [1,3,2,0,7,8,1,3,0,6,7,1]

The result: 123 1,378 167

Example 2: [2,1,0,0,3,4]

The result: 12X34

I need help in guiding how to solve this algorithm

var indices = [1,3,2,0,7,8,1,3,0,6,7,1] 

for (var i=0; i < indices.length; i  ) {
  
  if(indices[i] == 0) {
    console.log(i)
  }
  
}
console.log(indices);

CodePudding user response：

You have to go through series of consecutive steps to achieve the desired result.

function separateNumber(arr) {
  return arr
    .join("")
    .replace(/(0)\1 /g, "0X0")
    .split("0")
    .map((s) => s.split("").sort().join(""))
    .join(" ")
    .replace(/ X /g, "X");
}

console.log(separateNumber([1, 3, 2, 0, 7, 8, 1, 3, 0, 6, 7, 1]));
console.log(separateNumber([2, 1, 0, 0, 3, 4]));
console.log(separateNumber([2, 0, 1, 0, 0, 3, 4]));
console.log(separateNumber([2, 1, 0, 0, 3, 4, 0, 0, 0]));

CodePudding user response：

this way ?

const separate = arr => arr.reduce((r,v,i,{[i 1]:n})=>
  {
  if (v>0)   r.a.push(v)
  if ((v===0 || isNaN(n)) && r.a.length > 0)
    {
    r.a.sort((x,y)=>x-y)
    r.s.push( r.a.join(''))
    r.a.length = 0
    } 
  if (v===0) r.s.push(v)

  return isNaN(n) ? r.s.join('').replace(/00/g,'X').replace(/0/g,' ') : r
  },{s:[],a:[]})


test([1,3,2,0,7,8,1,3,0,6,7,1])
test([2,1,0,0,3,4])

function test ( arr)
  {
  console.log('\n',JSON.stringify(arr),'\n --->',separate(arr),'\n')
  }