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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.

 

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.

 

Constraints:

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

Solutions

Solution 1

• Java
• C++
• Python
• Go
• TypeScript

• 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;
      }
  }
  
  

• 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;
      }
  };
  
  

• 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
  
  

• 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
  }
  
  

• 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;
  }
  
  

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