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Formatted question description: https://leetcode.ca/all/1929.html

1929. Concatenation of Array

Level

Easy

Description

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

Solution

Get the length of nums as n. Create an array ans of length n * 2. Then for each i from 0 to n - 1, let ans[i] = ans[i + n] = nums[i]. Finally, return ans.

  • class Solution {
    public int[] getConcatenation(int[] nums) {
        int n = nums.length;
        int[] concat = new int[n * 2];
        for (int i = 0; i < n; i++) {
            concat[i] = nums[i];
            concat[i + n] = nums[i];
        }
        return concat;
    }
}

  • // OJ: https://leetcode.com/problems/concatenation-of-array/
// Time: O(N)
// Space: O(1)
class Solution {
public:
    vector<int> getConcatenation(vector<int>& A) {
        vector<int> ans(A);
        int N = A.size();
        for (int i = 0; i < N; ++i) ans.push_back(A[i]);
        return ans;
    }
};

  • class Solution:
    def getConcatenation(self, nums: List[int]) -> List[int]:
        return nums + nums

############

# 1929. Concatenation of Array
# https://leetcode.com/problems/concatenation-of-array/

class Solution:
    def getConcatenation(self, nums: List[int]) -> List[int]:
        return nums + nums

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