1929 - Concatenation of Array (Easy)
Problem Link
https://leetcode.com/problems/concatenation-of-array/
Problem Statement
Given an integer array nums
of length n
, you want to create an array ans
of length 2n
where ans[i] == nums[i]
and ans[i + n] == nums[i]
for 0 <= i < n
(0-indexed).
Specifically, ans
is the concatenation of two nums
arrays.
Return the array ans
.
Example 1:
Input: nums = [1,2,1]
Output: [1,2,1,1,2,1]
Explanation: The array ans is formed as follows:
- ans = [nums[0],nums[1],nums[2],nums[0],nums[1],nums[2]]
- ans = [1,2,1,1,2,1]
Example 2:
Input: nums = [1,3,2,1]
Output: [1,3,2,1,1,3,2,1]
Explanation: The array ans is formed as follows:
- ans = [nums[0],nums[1],nums[2],nums[3],nums[0],nums[1],nums[2],nums[3]]
- ans = [1,3,2,1,1,3,2,1]
Constraints:
n == nums.length
1 <= n <= 1000
1 <= nums[i] <= 1000
Approach 1: Iteration
We need to look at nums
two times to create the desired result. The simplest approach is to perform two for-loops in nums
and copy the numbers to result
.
def getConcatenation(self, nums: List[int]) -> List[int]:
#initialize result
result = []
#first iteration of nums
for i in range(len(nums)):
result.append(nums[i])
#second iteration of nums
for i in range(len(nums)):
result.append(nums[i])
#return result
return result