Transformed Array

Difficulty: Easy | Acceptance: 70.50% | Paid: No Topics: Array, Simulation

You are given a 0-indexed integer array nums. You have to transform the array into a new array result of the same length.

For each index i in the original array:

• If nums[i] > 0, add nums[i] to nums[(i + 1) % n].
• If nums[i] < 0, add nums[i] to nums[(i - 1 + n) % n].
• If nums[i] == 0, do nothing.

The operations are performed based on the original values of nums. Return the transformed array.

• Examples
• Constraints
• Simulation (In-place)
• Simulation (New Array)

Examples

Example 1

Input:

nums = [1,2,3,4]

Output:

[5,3,5,7]

Explanation:

• i=0, val=1>0: add 1 to nums[1]. nums becomes [1,3,3,4].
• i=1, val=2>0: add 2 to nums[2]. nums becomes [1,3,5,4].
• i=2, val=3>0: add 3 to nums[3]. nums becomes [1,3,5,7].
• i=3, val=4>0: add 4 to nums[0]. nums becomes [5,3,5,7].

Example 2

Input:

nums = [-1,2,-3,4]

Output:

[3,-1,-1,3]

Explanation:

• i=0, val=-1<0: add -1 to nums[3]. nums becomes [-1,2,-3,3].
• i=1, val=2>0: add 2 to nums[2]. nums becomes [-1,2,-1,3].
• i=2, val=-3<0: add -3 to nums[1]. nums becomes [-1,-1,-1,3].
• i=3, val=4>0: add 4 to nums[0]. nums becomes [3,-1,-1,3].

Constraints

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

Simulation (In-place)

Intuition We iterate through the array once. Since the transformation depends on the original values, we first create a copy of the array. Then, for each element in the copy, we update the corresponding neighbor in the original array.

Steps

• Create a copy of the input array nums.
• Iterate through the copy using index i.
• If the value is positive, add it to nums[(i + 1) % n].
• If the value is negative, add it to nums[(i - 1 + n) % n].
• If the value is zero, do nothing.
• Return the modified nums.

class Solution:
    def constructTransformedArray(self, nums: List[int]) -> List[int]:
        n = len(nums)
        original = nums[:]
        for i in range(n):
            val = original[i]
            if val > 0:
                nums[(i + 1) % n] += val
            elif val < 0:
                nums[(i - 1 + n) % n] += val
        return nums

Complexity

• Time: O(n)
• Space: O(n)
• Notes: We use O(n) space to store the copy of the array.

Simulation (New Array)

Intuition Instead of modifying the input array in-place, we create a new result array initialized with zeros. We iterate through the original array and add the values to the correct indices in the result array. This avoids modifying the input directly.

Steps

• Initialize a result array res of size n with zeros.
• Iterate through nums using index i.
• If nums[i] is positive, add it to res[(i + 1) % n].
• If nums[i] is negative, add it to res[(i - 1 + n) % n].
• If nums[i] is zero, do nothing.
• Return res.

class Solution:
    def constructTransformedArray(self, nums: List[int]) -> List[int]:
        n = len(nums)
        res = [0] * n
        for i in range(n):
            val = nums[i]
            if val > 0:
                res[(i + 1) % n] += val
            elif val < 0:
                res[(i - 1 + n) % n] += val
        return res

Complexity

• Time: O(n)
• Space: O(n)
• Notes: We use O(n) space for the result array. This approach is cleaner as it does not mutate the input.