AI can give you what you want, but you should know what you want to achieve anyway.
js
given array A and array B
B is array with objects like {id:string}
write a function which gets - A, changed-B, and previous-B , detects items reorder in B and applies that on a
function applyReorder(a, prevB, changedB) {
// Create a map of id to index for the previous b array
const prevBIndexMap = new Map()
prevB.forEach((item, index) => {
prevBIndexMap.set(item.id, index)
})
// Create a reordered array based on the changedB array
const reorderedA = new Array(a.length)
changedB.forEach((item, index) => {
const prevIndex = prevBIndexMap.get(item.id)
reorderedA[index] = a[prevIndex]
})
return reorderedA
}
now write that in a more functional style
function applyReorder(a, prevB, changedB) {
const prevBIndexMap = prevB.reduce((map, item, index) => {
map.set(item.id, index)
return map
}, new Map())
return changedB.map((item) => a[prevBIndexMap.get(item.id)])
}
use Map constractor
function applyReorder(a, prevB, changedB) {
const prevBIndexMap = new Map(
prevB.map((item, index) => [item.id, index])
)
return changedB.map((item) => a[prevBIndexMap.get(item.id)])
}
implement this in one line
const applyReorder = (a, prevB, changedB) =>
changedB.map(
(item) =>
a[
new Map(prevB.map((el, idx) => [el.id, idx])).get(
item.id
)
]
)
use one line implementation using findIndex
const applyReorder = (a, prevB, changedB) =>
changedB.map(
(item) => a[prevB.findIndex((el) => el.id === item.id)]
)
So progressively we got one line instead of 10 and only one implementation was non working
And the final implementation:
function applyReorder(
applyTo: any[],
prevAccordingTo: { id: string }[],
changedAccordingTo: { id: string }[]
) {
return changedAccordingTo.map(
(item) =>
applyTo[
prevAccordingTo.findIndex((el) => el.id === item.id)
]
)
}
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