i’m trying to create a function that gives me an array of numbers with a limitation of repetition for each number. for example
([1,1,3,3,7,2,2,2,2], 3)
should give me
[1, 1, 3, 3, 7, 2, 2, 2]
it deletes a [2] because the max repetition of numbers is 3.
here is my code but i don’t know why it doesn’t work:
function deleteNth(arr,n){
var results = [];
for(var i=0; i<arr.length; i++){
if (count(results, arr[i])<=n) {
results.push(arr[i]);
}
}
return results;
}
function count(array, what){
var count =0;
for (var i=0; i<array.length; i++){
if (array[i]===what){
count++;
}
}
return count;
}
console.log(deleteNth([1,1,3,3,7,2,2,2,2], 3));
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Answer
I would use a reduce to iterate over all elements of the array and a dictionary to keep track of the number of times I found an element.
Here’s an example:
const filterReps = (arr, maxReps) => {
return arr.length ? arr.reduce((acc, num, i) => {
// Add this number to our dictionary,
// if already present add +1 to it's count
acc.found[num] = acc.found[num] ? ++acc.found[num] : 1
// If the dictionary says the number has been found less or equal
// times to our max repetitions, push it into the accumulating array
if (acc.found[num] <= maxReps)
acc.arr.push(num)
// If this is the final iteration, just return only the result array
// and not the dictionary
return i === nums.length - 1 ? acc.arr : acc
}, { found: {}, arr: [] }) : arr
}
const nums = [1, 1, 1, 1, 2, 2, 2, 2]
console.log(filterReps(nums, 3))