So, I've the following arrays of odd numbers:
const oddNums = [1, 3, 5, 7, 9];
And I want to fill it with the missing even numbers to obtain the result below:
[ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
I've done it like this and it works fine, but can it be done in a more concise manner, maybe using array methods?
const oddNums = [1, 3, 5, 7, 9];
const nums = [];
for (let i=0; i<oddNums.length; i++) {
nums.push(oddNums[i]);
nums.push(oddNums[i] + 1);
}
console.log(nums);
Note: The odd numbers would always be in sequence but might not begin with 1, for ex: [11, 13, 15] is a valid array. And the output for [11, 13, 15] should be [ 11, 12, 13, 14, 15, 16 ].
The only information you need is the first (odd) number and the size of the input:
const oddNums = [1, 3, 5, 7, 9];
const result = Array.from({length: oddNums.length*2}, (_, i) => i + oddNums[0]);
console.log(result);Using Array.prototype.flatMap:
const
oddNums = [1, 3, 5, 7, 9],
nums = oddNums.flatMap(n => [n, n + 1]);
console.log(nums);Using Array.prototype.reduce and Array.prototype.concat:
const
oddNums = [1, 3, 5, 7, 9],
nums = oddNums.reduce((r, n) => r.concat(n, n + 1), []);
console.log(nums);Using Array.prototype.reduce and Array.prototype.push:
const
oddNums = [1, 3, 5, 7, 9],
nums = oddNums.reduce((r, n) => (r.push(n, n + 1), r), []);
console.log(nums);