How do you solve Transformed Array on LeetCode?

Transformed Array (LeetCode #3379) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding circular arrays is crucial for index calculations.

What is the brute force solution for Transformed Array?

The brute force approach for Transformed Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach directly simulates the movement for each index in the array. For each element, we calculate the new index based on its value and set the corresponding result. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Transformed Array?

Transformed Array uses the following concepts: Array, Simulation. The recommended patterns to study are: Array, Modulo operations.

What are the interview tips for Transformed Array?

Clarify the circular nature of the array with the interviewer. Discuss edge cases like empty arrays or arrays with all zeros. Explain your thought process while coding to showcase your understanding.

What are common mistakes when solving Transformed Array?

Not handling negative indices correctly. Forgetting to use modulo for circular behavior.

#3379

Transformed Array

Easy

ArraySimulationArrayModulo operations

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the fact that we can directly compute the new index for each element without needing to simulate the entire process multiple times. This reduces unnecessary computations.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty result array of the same length as nums.
2Step 2: Iterate through each index i in nums.
3Step 3: Calculate the new index based on the value of nums[i] using modulo to ensure it wraps around correctly.
4Step 4: Set result[i] to nums[new index].

solution.py7 lines

1def transformedArray(nums):
2    n = len(nums)
3    result = [0] * n
4    for i in range(n):
5        new_index = (i + nums[i]) % n
6        result[i] = nums[new_index]
7    return result

ℹ

Complexity note: The time complexity is O(n) because we make a single pass through the array, and the space complexity is O(1) as we are using the result array of the same size as nums.

1Understanding circular arrays is crucial for index calculations.
2Using modulo operations helps in wrapping around the array.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.