deleteNth.js

/**
 * Enough is enough!
*   Alice and Bob were on a holiday. Both of them took many pictures of the places they've been, 
*   and now they want to show Charlie their entire collection. However, Charlie doesn't like this sessions, 
*   since the motive usually repeats. He isn't fond of seeing the Eiffel tower 40 times. He tells them that 
*   he will only sit during the session if they show the same motive at most N times. Luckily, Alice and Bob
*   are able to encode the motive as a number. Can you help them to remove numbers such that their list
*   contains each number only up to N times, without changing the order?

 *  Task
*     Given a list lst and a number N, create a new list that contains each number of lst at most N times 
*     without reordering. For example if N = 2, and the input is [1,2,3,1,2,1,2,3], you take [1,2,3,1,2], 
*     drop the next [1,2] since this would lead to 1 and 2 being in the result 3 times, and then take 3,
*     which leads to [1,2,3,1,2,3].

*   Example
*     deleteNth ([1,1,1,1],2) // return [1,1]
*     deleteNth ([20,37,20,21],1) // return [20,37,21]
*/

const _updateHashMap = (hashMap, val) => {
  if (hashMap.has(val)) {
     let value = hashMap.get(val)
     hashMap.set(val, value+1)
   } else {
     hashMap.set(val, 1)
   }
 }
   
 function deleteNth(arr,n){
   // ...
   const hashMap = new Map()
   const newArray = []
   return arr.filter(a=>{
     _updateHashMap(hashMap, a)
     if (hashMap.get(a) <= n) {
       return a
     }
   })
   return newArray
 }