Tengo esta matriz

var arr = ["s00","s01","s02","s03","s04","s05","s07","s08","s09","s10","s11","s12","s13","s14","s17","s19","s20","s21","s22","s24","s25","s26","s27","s28","s30","s32","s33","s34","s36","s38","s39","s41","s43","s44","s45","s46","s47","s48","s49","s50","s51","s52","s53","s54","s55","s56","s58","s60","s61","s62","s63","s64","s65","s67","s69","s70"];

staba tratando de encontrar un algoritmo que me dijera quésaltan @s. Como puede ver, la lista consta de @ consecutivss s1, s2, etc.).

Al principio fui con esta solución:

    var arr = ["s00","s01","s02","s03","s04","s05","s07","s08","s09","s10","s11","s12","s13","s14","s17","s19","s20","s21","s22","s24","s25","s26","s27","s28","s30","s32","s33","s34","s36","s38","s39","s41","s43","s44","s45","s46","s47","s48","s49","s50","s51","s52","s53","s54","s55","s56","s58","s60","s61","s62","s63","s64","s65","s67","s69","s70"];
for (var i=1;i<arr.length;i++){
    var thisI = parseInt(arr[i].toLowerCase().split("s")[1]);
    var prevI = parseInt(arr[i-1].toLowerCase().split("s")[1]);
    if (thisI != prevI+1)
      console.log(`Seems like ${prevI+1} is missing. thisI is ${thisI} and prevI is ${prevI}`)
}

Pero este método falla por faltar más de un número consecutivo s15, s16). Entonces agregué unwhile loop que funciona.

var arr = ["s00","s01","s02","s03","s04","s05","s07","s08","s09","s10","s11","s12","s13","s14","s17","s19","s20","s21","s22","s24","s25","s26","s27","s28","s30","s32","s33","s34","s36","s38","s39","s41","s43","s44","s45","s46","s47","s48","s49","s50","s51","s52","s53","s54","s55","s56","s58","s60","s61","s62","s63","s64","s65","s67","s69","s70"];
for (var i=1;i<arr.length;i++){
  var thisI = parseInt(arr[i].toLowerCase().split("s")[1]);
  var prevI = parseInt(arr[i-1].toLowerCase().split("s")[1]);
  if (thisI != prevI+1) {
    while(thisI-1 !== prevI++){
       console.log(`Seems like ${prevI} is missing. thisI is ${thisI} and prevI is ${prevI}`)
    }
   }
}

Sin embargo, siento que estoy complicando demasiado las cosas. Pensé en crear una matriz ideal:

var idealArray = [];
for (var i =0; i<200;i++) {
  idealArray.push(i)
}

Y luego, mientras lo verificas, manipulas mi matriz arr) para que el bucle verifique dos matrices de la misma longitud. Es decir, use esta solución:

var idealArray = [];
for (var i =0; i<200;i++) {
  idealArray.push(i)
}
var arr = ["s00","s01","s02","s03","s04","s05","s07","s08","s09","s10","s11","s12","s13","s14","s17","s19","s20","s21","s22","s24","s25","s26","s27","s28","s30","s32","s33","s34","s36","s38","s39","s41","s43","s44","s45","s46","s47","s48","s49","s50","s51","s52","s53","s54","s55","s56","s58","s60","s61","s62","s63","s64","s65","s67","s69","s70"];
for (let i = 0; i<idealArray.length;i++){
  if (parseInt(arr[i].toLowerCase().split("s")[1]) != idealArray[i]) {
    console.log(`Seems like ${idealArray[i]}is missing`);
    arr.splice(i,0,"dummyel")
  }
}

Pero, una vez más, tengo la sensación de que crear esta segunda matriz tampoco es muy eficiente (pensando en una lista grande, desperdiciaría espacio innecesario).

Entonces ... ¿cómo realizo eficientemente esta tarea en JavaScript? (Significa eficientemente lo más cercano posible a O (1) tanto para la complejidad del tiempo como para la complejidad del espacio).