Transformed Array

Transformed Array

What is this problem about?

The "Transformed Array coding problem" is a simulation task involving circular array traversal. You are given an integer array nums, and you need to create a new array result of the same length. For each index i, result[i] is the value found in nums at a new position calculated by moving nums[i] steps from i. If nums[i] is positive, you move right; if negative, you move left; and if zero, you stay at the same index. Since the move can exceed the array bounds, the movement is treated as circular.

Why is this asked in interviews?

This "Transformed Array interview question" is a great way to test a candidate's comfort with the modulo operator and array indexing. Companies like Meta use it to ensure developers can handle "edge-of-array" logic correctly, which is a common source of bugs in production code. It evaluates your ability to translate a simple movement rule into a robust mathematical formula that handles both positive and negative offsets within a circular boundary.

Algorithmic pattern used

The "Array, Simulation interview pattern" is used here. For each element nums[i], the new index is calculated as (i + nums[i]) % n. However, special care must be taken with the modulo operator in different programming languages, as it can return negative results for negative numbers. A more robust formula for circular indexing is (i + (nums[i] % n) + n) % n, which ensures the result is always a valid non-negative index between 0 and n-1.

Example explanation

Input: [3, -2, 1, 1], n=4.

1. i=0, nums[0]=3: New index = (0 + 3) % 4 = 3. result[0] = nums[3] = 1.
2. i=1, nums[1]=-2: New index = (1 - 2 + 4) % 4 = 3. result[1] = nums[3] = 1.
3. i=2, nums[2]=1: New index = (2 + 1) % 4 = 3. result[2] = nums[3] = 1.
4. i=3, nums[3]=1: New index = (3 + 1) % 4 = 0. result[3] = nums[0] = 3. Final result: [1, 1, 1, 3].

Common mistakes candidates make

One frequent error in the "Transformed Array coding problem" is not correctly handling negative jumps. If nums[i] is -10 and n is 3, the jump should land at index 2, not -1. Another mistake is using a loop to perform the jumps one by one, which is inefficient if the jump values are very large. Using the modulo operator is the intended $O(1)$ way to find the destination index.

Interview preparation tip

For the "Array, Simulation interview pattern," always remember the "universal modulo formula": (a % b + b) % b. This ensures that your circular array logic is always positive, regardless of whether a is positive or negative. Practice problems involving "circular queues" or "rotating arrays" to build intuition for this pattern.