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whenKisEnnough.ts
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whenKisEnnough.ts
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/*
LEVEL: undefined
Given an Array of items, some of which may be duplicated, create and return a new Array that is otherwise the same as items, but only up to k occurrences of each element are kept, and all occurrences of that element after those first k are discarded. Note also the counterintuitive but still completely legitimate edge case of k == 0 that has a well defined answer of an empty list!
Input: An Array and a number.
Output: An Array.
Examples:
removeAfterKth([42, 42, 42, 42, 42, 42, 42], 3) == [42, 42, 42]
removeAfterKth([42, 42, 42, 99, 99, 17], 0) == []
removeAfterKth([1, 1, 1, 2, 2, 2], 5) == [1, 1, 1, 2, 2, 2]
removeAfterKth(["tom", 42, "bob", "bob", 99, "bob", "tom", "tom", 99], 2)
== ["tom",42,"bob","bob",99,"tom",99]
removeAfterKth([1, 2, 3, 4, 5, 4, 3, 2, 1, 2, 3, 4, 5, 4, 3, 2, 1]) == [1, 2, 3, 4, 5]
This task is taken from the course CCPS 109 Computer Science I, as taught by Ilkka Kokkarinen.
*/
//Answer//
interface Iobj {
[key: string | number]: number
}
function removeAfterKth(
items: (string | number)[],
k: number
): (string | number)[] {
const finalarray: (string | number)[] = []
let numberid: string | number = 0
const obj: Iobj = {}
items.map((item: string | number, index: number) => {
if (index == 0) {
finalarray.push(item)
obj[item] = 1
numberid = item
}
if (index != 0 && numberid == item && obj[item] < k) {
finalarray.push(item)
obj[item] += 1
}
if (item != numberid) {
if (obj[item]) {
if (obj[item] < k) {
finalarray.push(item)
obj[item] += 1
numberid = item
}
} else {
obj[item] = 1
finalarray.push(item)
}
}
})
return k == 0 ? [] : finalarray
}