Sort 2D array by matching the last values with the first one in the next index
Solution 1:
For the example you gave the easiest solution would be to sort the 2d array by the second element of each inner array, as follow:
let array = [[1,0],[2,1],[0,3],[3,2]];
array.sort((a, b) => b[1] - a[1]);
In that way you can sort the array with the sort method according to the elements in the inner arrays.
let array = [[1,0],[2,1],[0,3],[3,2]];
array.sort((a, b) => b[1] - a[1]);
console.log(array);Solution 2:
You could take an object as reference and rebuild the array by taking the chained items.
const
getItems = (reference, value) => {
const a = reference[value];
return a ? [a, ...(a[1] === 0 ? [] : getItems(reference, a[1]))] : [];
},
array = [[1, 0], [2, 1], [0, 3], [3, 2]],
reference = array.reduce((r, a) => (r[a[0]] = a, r), {}),
result = getItems(reference, 0);
console.log(result);Solution 3:
Look at this method:
const array = [[1,0],[2,1],[0,3],[3,2]];
const result = [...array];
// Put the subarray with the zero element first
// You can use the sort() function but I guess this method
// performs better in terms of time
for (let i = 0; i < result.length; ++i) {
if (result[i][0] === 0) {
result.unshift(...result.splice(i, 1));
break;
}
}
// Now reorder the array so that the last index of a subarray
// matches with the first index of the other subarray
for (let i = 0; i < result.length; ++i) {
for (let j = i + 1; j < result.length; ++j) {
if (result[i][1] === result[j][0]) {
// recollocate the subarray so that it matches the way we want
result.splice(i + 1, 0, ...result.splice(j, 1));
break;
}
}
}
console.log(result)