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This approach uses a two-pointer technique to shuffle the array. The first pointer starts at the beginning of the array and represents the x values. The second pointer starts at the middle of the array and represents the y values. By iterating over the array and adding elements from these two pointers alternately to a new list, we can achieve the desired shuffle.
Time Complexity: O(n), where n is half the length of the array. We iterate through the array once.
Space Complexity: O(n), for storing the shuffled result.
1using System;
2
3class Program {
4 static int[] Shuffle(int[] nums, int n) {
5 int[] result = new int[2 * n];
6 for (int i = 0; i < n; i++) {
7 result[2 * i] = nums[i];
8 result[2 * i + 1] = nums[i + n];
9 }
10 return result;
11 }
12
13 static void Main() {
14 int[] nums = {2, 5, 1, 3, 4, 7};
15 int n = 3;
16 int[] result = Shuffle(nums, n);
17 Console.WriteLine(string.Join(", ", result));
18 }
19}This C# solution employs an array to store the shuffled result. A loop iterates through the indices to alternate between x and y elements.
An alternate approach is to reorder the array in-place without using additional space. However, given the elements might need to be accessed multiple times, a good understanding of index manipulation and mathematics is required. The complexity to deduce the correct indices for swapping can increase the difficulty.
Time Complexity: O(n^2) in the worst case as the array is shifted multiple times.
Space Complexity: O(1), as the shuffling is done in-place.
1
class Program {
static void ShuffleInPlace(int[] nums, int n) {
for (int i = n; i < 2 * n; i++) {
int y = nums[i];
for (int j = i; j > 2 * (i - n); j--) {
nums[j] = nums[j - 1];
}
nums[2 * (i - n)] = y;
}
}
static void Main() {
int[] nums = {2, 5, 1, 3, 4, 7};
int n = 3;
ShuffleInPlace(nums, n);
Console.WriteLine(string.Join(", ", nums));
}
}In the C# solution, we shuffle the input array in place by assigning elements through systematic index shifts, avoiding extra space use.