Given an array
A = [1,2,3,5,6,7,8,10,0,1,1,2,4,10,6,7,3], group the consecutive elements which are greater than 2 in to sub-arrays.The desired output is
result = [[2,3,5,6,7,8,10],[2,4,10,6,7,3]].Example:
A = [1,2,3,5,6,7,8,10,0,1,1,2,4,10,6,7,3]
Output:result = [[2,3,5,6,7,8,10],[2,4,10,6,7,3]]
Could anyone help with this question?
I have tried:
const a = [0, 1, 2, 5, 6, 9];
const result = a.reduce((acc, current) =>
const lastSubArray = acc[acc.length - 1];
console.log(`${acc}-${current}`)
console.log(lastSubArray[lastSubArray.length - 1]);
if (current >2 && lastSubArray[lastSubArray.length - 1]>2) {
acc.push(current);
}
acc[acc.length - 1].push(current);
console.log(acc);
return acc;
}, []);
console.log(result);
You can use a global variable to keep track of the indexes so that it's possible to detect when exists a gap between two numbers, which allows you to add another array. In combination, using ternary operators can make the code cleaner to test for conditions.
const arr = [1, 2, 3, 5, 6, 7, 8, 10, 0, 1, 1, 2, 4, 10, 6, 7, 3];
let lastIdx = -1
const results = arr.reduce((acc, current, idx) =>
(
current > 2 ? (
lastIdx + 1 === idx && acc.length ?
(
lastIdx = idx,
acc[acc.length - 1].push(current),
acc
)
:
(
lastIdx = idx,
[...acc, [current]]
)
) : acc
)
, [])
console.log(results);