3651. Transformed Array¶

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3651. Transformed Array

Easy

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:

Example 2:

Constraints:

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

Solution¶

class Solution {
    public int[] constructTransformedArray(int[] nums) {
        int n = nums.length;
        int ans[] = new int[n];
        for (int i = 0; i < n; i++) {
            int ele = nums[i];
            if (ele >= 0) {
                if (i + ele < n) ans[i] = nums[i + ele];
                else ans[i] = nums[(i + ele) % n];
            }
            else {
                if (i - Math.abs(ele) >= 0) ans[i] = nums[i - Math.abs(ele)];
                else ans[i] = nums[(i - Math.abs(ele) % n + n) % n];
            }
        }
        return ans;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here