ripped books: find min/max pages that an array of ripped pages could represent.

gistfile1.txt

Sure, there's a way to use reduce and do this all in one line. This v2 is 477 bytes as is.
 
 test data:
 const sample_input = [[7,8,100,101,222,223],
                [ 2,3,88,89,90,103,177 ],
                [ 2,3,6,7,10,11 ],
                [ 1 ],
                [ 1,2 ]]

const sample_output = [ "4/5","5/6","3/6","1/1","1/2" ]

and test with:
console.log(app(sample_input).toString() === sample_output.toString() ? "success" : "fail")

ripped_books_v1.js

const app = function(input){
  const minMaxPages=function(ripped){
    let min=0
    let max=0
    let eos=0
    let oes=0
    for(let i=0;i<ripped.length;){
      min++
      max++
      if(ripped[i-2]!==undefined && ripped[i-2]===ripped[i-1]-1 && ripped[i]===ripped[i-1]+1){
        max-=1
      }
      if(ripped[i+1]!==undefined && ripped[i+1]===ripped[i]+1){
        ripped[i]%2===0?eos+=1:oes+=1
        max+=1
        i+=1
      }
      i+=1
    }
    eos>oes?(min+=oes,max-=oes):(min+-eos,max-=eos)
    return `${min}/${max}`
  }
  let out=[]
  for(let i = 0; i < input.length; i++) {
    let ripped=input[i]
    out.push(minMaxPages(ripped))
  }
  return out
}

ripped_books_v2.js

const mnMx=(ripd)=>{
  let min=0
  let max=0
  let eos=0
  let oes=0
  for(let i=0;i<ripd.length;){
    min++
    max++
    if(ripd[i-2]!==undefined && ripd[i-2]===ripd[i-1]-1 && ripd[i]===ripd[i-1]+1){
      max-=1
    }
    if(ripd[i+1]!==undefined && ripd[i+1]===ripd[i]+1){
      ripd[i]%2===0?eos+=1:oes+=1
      max+=1
      i+=1
    }
    i+=1
  }
  eos>oes?(min+=oes,max-=oes):(min+-eos,max-=eos)
  return `${min}/${max}`
}
const app=(pgs)=>pgs.map((ripd)=>mnMx(ripd))