Sponsored
Sponsored
This approach uses a sliding window technique to efficiently calculate the sum of required elements. By maintaining a running sum for the window and updating it as you slide, you can achieve the necessary transformation in linear time. The key is to account for the circular nature of the array using modulo operations to wrap around indices.
Time Complexity: O(n), where n is the length of the `code` array. Each element is processed once with constant-time window updates.
Space Complexity: O(1) auxiliary space (excluding the output array).
Solve with full IDE support and test cases
The Java solution carries on with a two-pointer approach to maintain a constant-time cumulative sum, updating as the window traverses the code array by adjusting indices with modulo operations. This facilitates handling of circular formats smoothly.
This approach is straightforward but less efficient, involving a direct sum computation for each index by wrapping around using the modulo operator. Each element's circular context is individually recalculated, following conditions for the sign of k.
Time Complexity: O(n*k) with n as length of `code` and k an absolute value.
Space Complexity: O(1) extra space beyond output.
1using System;
2
3public class Solution {
4 public int[] Decrypt(int[] code, int k) {
5 int n = code.Length;
6 int[] result = new int[n];
7
8 if (k == 0) return result;
9
10 for (int i = 0; i < n; i++) {
11 int sum = 0;
12 for (int j = 1; j <= Math.Abs(k); j++) {
13 int index = (k > 0) ? (i + j) % n : (i - j + n) % n;
14 sum += code[index];
15 }
16 result[i] = sum;
17 }
18
19 return result;
20 }
21
22 public static void Main() {
23 Solution sol = new Solution();
24 int[] code = {5, 7, 1, 4};
25 int k = 3;
26 Console.WriteLine(string.Join(", ", sol.Decrypt(code, k)));
27 }
28}
29This C# strategy uses basic loops to calculate code transformation values, performing sum calculations directly based on whether k is positive or negative, adapting indices with modulo operations to accommodate circular constraints.