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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.
1def shuffle(nums, n):
2 result = []
3 for i in range(n):
4 result.append(nums[i])
5 result.append(nums[i + n])
6 return result
7
8nums = [2, 5, 1, 3, 4, 7]
9n = 3
10print(shuffle(nums, n))This Python solution utilizes list to achieve the shuffled form by iterating from 0 to n and appending elements alternately from each half of the nums list.
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#include <vector>
void shuffleInPlace(std::vector<int>& nums, int n) {
for (int i = n; i < 2 * n; i++) {
int y = nums[i];
int j = i;
while (j > 2 * (i - n)) {
nums[j] = nums[j - 1];
j--;
}
nums[j] = y;
}
}
int main() {
std::vector<int> nums = {2, 5, 1, 3, 4, 7};
int n = 3;
shuffleInPlace(nums, n);
for (int num : nums) {
std::cout << num << " ";
}
return 0;
}This C++ solution shifts elements in-place to achieve the same result as reordering the array without additional space.