3379. Transformed Array
Description
You are given an integer array nums
that represents a circular array. Your task is to create a new array result
of the same size, following these rules:
i
(where 0 <= i < nums.length
), perform the following independent actions:
- If
nums[i] > 0
: Start at indexi
and movenums[i]
steps to the right in the circular array. Setresult[i]
to the value of the index where you land. - If
nums[i] < 0
: Start at indexi
and moveabs(nums[i])
steps to the left in the circular array. Setresult[i]
to the value of the index where you land. - If
nums[i] == 0
: Setresult[i]
tonums[i]
.
Return the new array result
.
Note: Since nums
is circular, moving past the last element wraps around to the beginning, and moving before the first element wraps back to the end.
Example 1:
Input: nums = [3,-2,1,1]
Output: [1,1,1,3]
Explanation:
- For
nums[0]
that is equal to 3, If we move 3 steps to right, we reachnums[3]
. Soresult[0]
should be 1. - For
nums[1]
that is equal to -2, If we move 2 steps to left, we reachnums[3]
. Soresult[1]
should be 1. - For
nums[2]
that is equal to 1, If we move 1 step to right, we reachnums[3]
. Soresult[2]
should be 1. - For
nums[3]
that is equal to 1, If we move 1 step to right, we reachnums[0]
. Soresult[3]
should be 3.
Example 2:
Input: nums = [-1,4,-1]
Output: [-1,-1,4]
Explanation:
- For
nums[0]
that is equal to -1, If we move 1 step to left, we reachnums[2]
. Soresult[0]
should be -1. - For
nums[1]
that is equal to 4, If we move 4 steps to right, we reachnums[2]
. Soresult[1]
should be -1. - For
nums[2]
that is equal to -1, If we move 1 step to left, we reachnums[1]
. Soresult[2]
should be 4.
Constraints:
1 <= nums.length <= 100
-100 <= nums[i] <= 100
Solutions
Solution 1
Python3
class Solution:
def constructTransformedArray(self, nums: List[int]) -> List[int]:
ans = []
n = len(nums)
for i, x in enumerate(nums):
ans.append(nums[(i + x + n) % n] if x else 0)
return ans
Java
class Solution {
public int[] constructTransformedArray(int[] nums) {
int n = nums.length;
int[] ans = new int[n];
for (int i = 0; i < n; ++i) {
ans[i] = nums[i] != 0 ? nums[(i + nums[i] % n + n) % n] : 0;
}
return ans;
}
}
C++
class Solution {
public:
vector<int> constructTransformedArray(vector<int>& nums) {
int n = nums.size();
vector<int> ans(n);
for (int i = 0; i < n; ++i) {
ans[i] = nums[i] ? nums[(i + nums[i] % n + n) % n] : 0;
}
return ans;
}
};
Go
func constructTransformedArray(nums []int) []int {
n := len(nums)
ans := make([]int, n)
for i, x := range nums {
if x != 0 {
ans[i] = nums[(i+x%n+n)%n]
}
}
return ans
}
TypeScript
function constructTransformedArray(nums: number[]): number[] {
const n = nums.length;
const ans: number[] = [];
for (let i = 0; i < n; ++i) {
ans.push(nums[i] ? nums[(i + (nums[i] % n) + n) % n] : 0);
}
return ans;
}