The “Shuffle the Array” problem is a classic example of a coding challenge that focuses on array manipulation. This problem is relatively straightforward but important for building a strong foundation in array operations and indexing.
In this in-depth article, we’ll explore:
- The Problem Statement
- Initial Thoughts and Simple Approaches
- Further Optimizations
- Python Implementations
- Test Cases and Edge Cases
- Complexity Analysis
- Conclusion
1. The Problem Statement
You are given an array nums
consisting of 2n
elements in the form [x1, x2, ..., xn, y1, y2, ..., yn]
. Shuffle the array to make it in the form [x1, y1, x2, y2, ..., xn, yn]
. Return the shuffled array.
Example 1:
Input: nums = [2, 5, 1, 3, 4, 7], n = 3
Output: [2, 3, 5, 4, 1, 7]
Explanation: We take [2, 5, 1]
as x and [3, 4, 7]
as y. After shuffling, we get [2, 3, 5, 4, 1, 7]
.
Example 2:
Input: nums = [1, 1, 2, 2], n = 2
Output: [1, 2, 1, 2]
2. Initial Thoughts and Simple Approaches
The problem is fairly straightforward. The simplest approach would be to create a new array, iterate through the existing arrays x and y, and insert elements in the required order. This approach would use O(n) extra space.
3. Further Optimizations
The algorithm can be optimized for space by doing the shuffle in-place, but for clarity, the simple approach often suffices, especially given that the problem doesn’t have tight constraints.
4. Python Implementations
Simple Implementation
def shuffle(nums, n):
x, y = nums[:n], nums[n:]
result = []
for i in range(n):
result.extend([x[i], y[i]])
return result
In-Place Implementation (Optimized)
This implementation doesn’t require additional space.
def shuffle(nums, n):
for i in range(n):
nums.insert(2 * i + 1, nums.pop(n + i))
return nums
5. Test Cases and Edge Cases
To validate your solution, you need to consider multiple test cases, including edge cases:
# Test 1
assert shuffle([2, 5, 1, 3, 4, 7], 3) == [2, 3, 5, 4, 1, 7]
# Test 2
assert shuffle([1, 1, 2, 2], 2) == [1, 2, 1, 2]
# Test 3 (Edge Case)
assert shuffle([1], 1) == [1]
# Test 4 (Edge Case)
assert shuffle([1, 2], 1) == [1, 2]
6. Complexity Analysis
Time Complexity
For the simple implementation, the time complexity is O(n) as we loop through the array once. For the in-place implementation, the insert operation takes O(n) time, making it O(n^2) overall.
Space Complexity
The simple implementation uses O(n) additional space. The in-place implementation has a space complexity of O(1).
7. Conclusion
The “Shuffle the Array” problem from LeetCode is a great starting point for understanding array manipulations and serves as a building block for more complicated problems. Although the problem may not present the most challenging scenario, it’s a valuable exercise in understanding the basic principles of array operations and algorithmic efficiency.