Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Build confidence with an intuition-first walkthrough focused on array fundamentals.
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:
i (where 0 <= i < nums.length), perform the following independent actions:
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 at the index where you land.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 at the index where you land.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:
nums[0] that is equal to 3, If we move 3 steps to right, we reach nums[3]. So result[0] should be 1.nums[1] that is equal to -2, If we move 2 steps to left, we reach nums[3]. So result[1] should be 1.nums[2] that is equal to 1, If we move 1 step to right, we reach nums[3]. So result[2] should be 1.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:
nums[0] that is equal to -1, If we move 1 step to left, we reach nums[2]. So result[0] should be -1.nums[1] that is equal to 4, If we move 4 steps to right, we reach nums[2]. So result[1] should be -1.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] <= 100Problem summary: 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 at 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 at 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array
[3,-2,1,1]
[-1,4,-1]
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3379: Transformed Array
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 + nums[i] % n + n) % n];
}
return ans;
}
}
// Accepted solution for LeetCode #3379: Transformed Array
func constructTransformedArray(nums []int) []int {
n := len(nums)
ans := make([]int, n)
for i, x := range nums {
ans[i] = nums[(i+x%n+n)%n]
}
return ans
}
# Accepted solution for LeetCode #3379: Transformed Array
class Solution:
def constructTransformedArray(self, nums: List[int]) -> List[int]:
n = len(nums)
return [nums[(i + x % n + n) % n] for i, x in enumerate(nums)]
// Accepted solution for LeetCode #3379: Transformed Array
impl Solution {
pub fn construct_transformed_array(nums: Vec<i32>) -> Vec<i32> {
let n = nums.len() as i32;
let mut ans = vec![0; nums.len()];
for (i, &x) in nums.iter().enumerate() {
ans[i] = nums[(((i as i32 + x % n + n) % n) as usize)];
}
ans
}
}
// Accepted solution for LeetCode #3379: Transformed Array
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] % n) + n) % n]);
}
return ans;
}
Use this to step through a reusable interview workflow for this problem.
Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.
Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.