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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:

For each index i (where 0 <= i < nums.length), perform the following independent actions:

  • If nums[i] > 0: Start at index i and move nums[i] steps to the right in the circular array. Set result[i] to the value of the index where you land.
  • If nums[i] < 0: Start at index i and move abs(nums[i]) steps to the left in the circular array. Set result[i] to the value of the index where you land.
  • If nums[i] == 0: Set result[i] to nums[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.

Β 

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 reach nums[3]. So result[0] should be 1.
  • For nums[1] that is equal to -2, If we move 2 steps to left, we reach nums[3]. So result[1] should be 1.
  • For nums[2] that is equal to 1, If we move 1 step to right, we reach nums[3]. So result[2] should be 1.
  • For nums[3] that is equal to 1, If we move 1 step to right, we reach nums[0]. So result[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 reach nums[2]. So result[0] should be -1.
  • For nums[1] that is equal to 4, If we move 4 steps to right, we reach nums[2]. So result[1] should be -1.
  • For nums[2] that is equal to -1, If we move 1 step to left, we reach nums[1]. So result[2] should be 4.

Β 

Constraints:

  • 1 <= nums.length <= 100
  • -100 <= nums[i] <= 100

Solutions

Solution 1: Simulation

We create a result array \(\textit{ans}\). For each index, we move right or left \(|nums[i]|\) steps based on whether \(nums[i]\) is positive or negative, calculate the landing index, and assign the value at that index to \(\textit{ans}[i]\).

The time complexity is \(O(n)\) and the space complexity is \(O(n)\), where \(n\) is the length of the array \(nums\).

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class Solution:
    def constructTransformedArray(self, nums: List[int]) -> List[int]:
        n = len(nums)
        return [nums[(i + x % n + n) % n] for i, x in enumerate(nums)]

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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 + nums[i] % n + n) % n];
        }
        return ans;
    }
}

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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] % n + n) % n];
        }
        return ans;
    }
};

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func constructTransformedArray(nums []int) []int {
    n := len(nums)
    ans := make([]int, n)
    for i, x := range nums {
        ans[i] = nums[(i+x%n+n)%n]
    }
    return ans
}

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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] % n) + n) % n]);
    }
    return ans;
}

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impl Solution {
    pub fn construct_transformed_array(nums: Vec<i32>) -> Vec<i32> {
        let n = nums.len() as i32;
        let mut ans = vec![0; nums.len()];
        for (i, &x) in nums.iter().enumerate() {
            ans[i] = nums[(((i as i32 + x % n + n) % n) as usize)];
        }
        ans
    }
}

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