LeetCode in Kotlin
3925. Concatenate Array With Reverse
Easy
You are given an integer array nums of length n.
Construct a new array ans of length 2 * n such that the first n elements are the same as nums, and the next n elements are the elements of nums in reverse order.
Formally, for 0 <= i <= n - 1:
ans[i] = nums[i]ans[i + n] = nums[n - i - 1]
Return an integer array ans.
Example 1:
Input: nums = [1,2,3]
Output: [1,2,3,3,2,1]
Explanation:
The first n elements of ans are the same as nums.
For the next n = 3 elements, each element is taken from nums in reverse order:
ans[3] = nums[2] = 3ans[4] = nums[1] = 2ans[5] = nums[0] = 1
Thus, ans = [1, 2, 3, 3, 2, 1].
Example 2:
Input: nums = [1]
Output: [1,1]
Explanation:
The array remains the same when reversed. Thus, ans = [1, 1].
Constraints:
1 <= nums.length <= 1001 <= nums[i] <= 100
Solution
class Solution {
fun concatWithReverse(nums: IntArray): IntArray {
val n = nums.size
val ans = IntArray(n * 2)
var count = 0
for (i in 0..<n) {
ans[i] = nums[i]
count++
}
for (i in n - 1 downTo 0) {
ans[count++] = nums[i]
}
return ans
}
}