'Algorithm to guess which array element has its index (position) changed inside the same array
I've been looking for a while at this thread but all i could find there as results of comparaison between two arrays are which elements were added or removed.
What i need is to just guess which element has moved from an index to another.
Example: let's take the following array:
let array1 = ['A', 'B', 'C', 'D', 'E', 'F'];
The 'E' element will move from the 4th index to the 2nd index and we wil get :
let array2 = ['A', 'B', 'E', 'C', 'D', 'F'];
I need a function that returns which element has changed , its new index in the new array2 and its old index in the array1;
I've made such as algo in the past which roughly (from what i remember now) consists of looking for the first different element between both arrays then to check the sequence behind it to guess if the diff element found is itself the moved one or the one which took its place.
So before rewritting it , i wished i could find one ready to use ;)
Solution 1:[1]
The approach is as follows ...
- iterate one of the arrays, preferably the current one.
- for each current item/value get its index from the recent array ...
- ... and compare this item's/value's current index to its recent index.
- in case both indices do not equal (including a recent index of
-1due to a removed value/item) create a state like object with data of the value and the indices and push it into the result array.
The implementation is based on Array.prototype.reduce where on would process the current item array and would pass the recent item array as part of a accumulator/collector as the reducer function's initial value.
function collectPositionChange({ recent, result }, value, currentIdx) {
const recentIdx = recent.indexOf(value);
if (recentIdx !== currentIdx) {
result.push({ value, currentIdx, recentIdx });
}
return { recent, result };
}
const recentItems = ['A', 'B', 'E', 'C', 'D', 'F'];
const currentItems = ['A', 'B', 'C', 'D', 'E', 'F'];
console.log(
currentItems
.reduce(collectPositionChange, { recent: recentItems, result: [] })
.result
);
console.log(
['A', 'C', 'D', 'B', 'E', 'F']
.reduce(collectPositionChange, { recent: ['A', 'B', 'C', 'D', 'E', 'F'], result: [] })
.result
);
console.log(
['A', 'B', 'C', 'D', 'E', 'F']
.reduce(collectPositionChange, { recent: ['A', 'C', 'D', 'B', 'E', 'F'], result: [] })
.result
);
.as-console-wrapper { min-height: 100%!important; top: 0; }
Solution 2:[2]
let array1 = ['A', 'B', 'C', 'D', 'E', 'F'];
// test 1 : moving 'B' from index 1 to index 3
let array2 = ['A', 'C', 'D', 'B', 'E', 'F'];
// test 2: moving 'E' from 4 to 1
// let array2 = ['A', 'E', 'B', 'C', 'D', 'F'];
// test 3 : moving 'A' from 0 to 5
// let array2 = ['B', 'C', 'D', 'E', 'F', 'A'];
function getMovedElementInfos(array1, array2) {
let firstDiffElIndexInNewArray = array2.findIndex((el, elindx) => el !== array1[elindx]);
let firstDiffElInNewArray = array2[firstDiffElIndexInNewArray];
let nextElInNewArray = array2[firstDiffElIndexInNewArray + 1];
let firstDiffElIndexInOldArray = array1.findIndex(el => el === firstDiffElInNewArray);
let nextElInOldArray = array1[firstDiffElIndexInOldArray + 1];
let movedEl, movedElFrom, movedElTo;
if (nextElInNewArray === nextElInOldArray) {
movedEl = array1[firstDiffElIndexInNewArray];
movedElFrom = firstDiffElIndexInNewArray;
movedElTo = array2.findIndex(el => el === movedEl);
} else {
movedEl = firstDiffElInNewArray;
movedElFrom = firstDiffElIndexInOldArray;
movedElTo = firstDiffElIndexInNewArray;
}
return {
movedEl,
movedElFrom,
movedElTo
}
}
const {
movedEl,
movedElFrom,
movedElTo
} = getMovedElementInfos(array1, array2)
console.log('movedEl is: ', movedEl);
console.log('movedEl index in old array is: ', movedElFrom);
console.log('movedEl index in new array is: ', movedElTo);
console.log('array1[movedElFrom]: ', array1[movedElFrom]);
console.log('array2[movedElTo]: ', array2[movedElTo]);
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Peter Seliger |
| Solution 2 | Bardelman |
