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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.
1#include <stdio.h>
2
3void shuffle(int* nums, int n, int* result) {
4 int i, j = 0;
5 for (i = 0; i < n; i++) {
6 result[j++] = nums[i];
7 result[j++] = nums[i + n];
8 }
9}
10
11int main() {
12 int nums[] = {2, 5, 1, 3, 4, 7};
13 int n = 3;
14 int result[sizeof(nums)/sizeof(nums[0])];
15 shuffle(nums, n, result);
16 for (int i = 0; i < 2 * n; i++) {
17 printf("%d ", result[i]);
18 }
19 return 0;
20}This C solution uses a function shuffle that accepts pointers to the input array and output array, along with n (half the size of the array). It utilizes two pointers within a single loop to construct the resulting shuffled array.
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#
This C implementation performs in-place reordering by moving each y value to its correct position through a series of swaps.