Day 31 of Studying LeetCode until I Can Solve One on My Own: 1481. Least Number of Unique Integers after K Removals(M/JS)

Input: arr = [5,5,4], k = 1Output: 1Explanation: Remove the single 4, only 5 is left.

Input: arr = [4,3,1,1,3,3,2], k = 3Output: 2Explanation: Remove 4, 2 and either one of the two 1s or three 3s.1 and 3 will be left.

var findLeastNumOfUniqueInts = function(arr, k) {    let frequency = {}, uiqueInt=0//Initial a map to count the frequencies of the numbers in the//array. Also initial "uniqueInt" to keep count of unique integers.     for(let i = 0;i < arr.length;i++){//Loop (note 1) through each interger in the array        if(!frequency[arr[i]]){//If the ith element in the array does NOT exsit in the map            frequency[arr[i]] = 0//initial the count for the ith element in the map            uiqueInt++//When new element is added in the map, increate the count of the//unique element.        }        frequency[arr[i]]++//If the ith element exists in the map, increase the frequency of//that element.    }        let sortedFreq = Object.entries(frequency).sort((a,b)=>{return a[1]-b[1]})//sort the frequency map created from steps above in ascending//order    for(let i = 0;i < sortedFreq.length;i++){//Loop through the sorted frequency map        if(k >= sortedFreq[i][1]){//If k is greater or equal to the frequency of the element            k=k-sortedFreq[i][1]//Subtract the frequency of that element from k. That means we//made k moves to remove all of that element entirely.            uiqueInt--//Once removed, reduce the count for unique interger        }else{            break;//if update k is less than frequency than any of the remainding//element, break out of the loop. That means the element will//still exist in the array even if we make k moves.        }    }    return uiqueInt//Return the number of unique interger(s).};