💡 Problem Formulation: The task is to partition a given array into disjoint (non-overlapping) intervals where no element in the first interval is greater than any element in the second. For example, given the input array [5, 0, 3, 8, 6], one possible partition could be [5, 0, 3] and [8, 6], with the respective outputs being the lengths of these intervals, i.e., (3, 2).
Method 1: Iterative Comparison
This method iterates through the array, dynamically adjusting the partition point where the maximum on the left side is no larger than the minimum on the right side. This allows for the array to be split into two disjoint intervals.
Here’s an example:
def partition_disjoint(arr):
left_max = global_max = arr[0]
partition_idx = 0
for i in range(1, len(arr)):
global_max = max(global_max, arr[i])
if arr[i] < left_max:
left_max = global_max
partition_idx = i
return partition_idx + 1 # Length of the left interval
# Example usage
print(partition_disjoint([5, 0, 3, 8, 6]))Output:
3
This snippet defines a function partition_disjoint which tracks the maximum values in the left and right partitions. It iterates through the array to determine the point at which the intervals become disjoint. Finally, it returns the length of the left interval. In the example, we get 3 indicating the array can be split after the third element.
Method 2: Prefix and Suffix Arrays
This method uses two additional arrays to store the prefix maximum and the suffix minimum. The partition index is then found by scanning both arrays for the point where the prefix maximum is less than or equal to the suffix minimum.
Here’s an example:
def partition_disjoint(arr):
size = len(arr)
prefix_max = [0] * size
suffix_min = [0] * size
prefix_max[0] = arr[0]
for i in range(1, size):
prefix_max[i] = max(prefix_max[i - 1], arr[i])
suffix_min[size - 1] = arr[size - 1]
for i in range(size - 2, -1, -1):
suffix_min[i] = min(suffix_min[i + 1], arr[i])
for i in range(1, size):
if prefix_max[i - 1] <= suffix_min[i]:
return i # Length of the left interval
# Example usage
print(partition_disjoint([5, 0, 3, 8, 6]))Output:
3
The function partition_disjoint defined here creates prefix and suffix arrays to track the maximum and minimum values respectively as we traverse the main array. Then, it finds the smallest index at which the maximum of the left side is less than or equal to the minimum on the right, indicating a disjoint partition. The example usage demonstrates the process and returns the length of the left partition as 3.
Method 3: In-Place Marking
This approach modifies the array in-place by marking the maximum to the left including the current index. Then, it scans backwards to identify where to partition the array.
Here’s an example:
def partition_disjoint(arr):
max_left = arr[0]
marker = 0
for i in range(len(arr)):
if arr[i] < max_left:
max_left = max(arr[:i+1])
marker = i
return marker + 1 # Length of the left interval
# Example usage
print(partition_disjoint([5, 0, 3, 8, 6]))Output:
3
In this code snippet, we define the function partition_disjoint which uses the in-place marking strategy. It updates the max_left value dynamically and uses a marker to keep track of the rightmost position where this update was necessary. This ensures that all elements to the left of the marker are smaller or equal to all elements to the right. The resulting length of the left interval is 3.
Method 4: Optimized Space Usage
This method optimizes space by using only a single extra variable to keep track of the required partition. It relies on the intuition gained from in-place marking but avoids storing all maximums.
Here’s an example:
def partition_disjoint(arr):
max_left, partition_idx = arr[0], 0
current_max = arr[0]
for i in range(len(arr)):
current_max = max(current_max, arr[i])
if arr[i] < max_left:
partition_idx = i
max_left = current_max
return partition_idx + 1 # Length of the left interval
# Example usage
print(partition_disjoint([5, 0, 3, 8, 6]))Output:
3
This snippet demonstrates an optimized approach where we keep a running tally of the current maximum and update the partition index and max_left when necessary. This avoids additional space complexity while maintaining effective partitioning logic. The code returns 3, the length of the left partition, for the given example.
Bonus One-Liner Method 5: Pythonic Way with Enumerate
Python’s built-in enumerate function can be used for a sleek one-liner solution by utilizing tuple unpacking and list comprehensions.
Here’s an example:
partition_disjoint = lambda arr: next(i for i, (x, y) in enumerate(zip(arr, arr[1:]), 1) if max(arr[:i]) <= min(arr[i:])) # Example usage print(partition_disjoint([5, 0, 3, 8, 6]))
Output:
3
This one-liner uses a generator expression along with enumerate to iterate through pairs of elements, checking the condition to find the disjoint interval boundary. It is very concise but may not be as readable or efficient, since it recalculates maxima and minima at each step. The output confirms the partition length as 3.
Summary/Discussion
- Method 1: Iterative Comparison. Provides an efficient, readable solution. May not be as efficient on very large arrays due to linear complexity in each iteration.
- Method 2: Prefix and Suffix Arrays. Precomputes necessary values for partitioning, offering quick lookup but increased space complexity due to auxiliary arrays.
- Method 3: In-Place Marking. Offers a good trade-off between space and time complexity by avoiding extra data structures but still iterating multiple times.
- Method 4: Optimized Space Usage. Reduces the need for extra data structures while maintaining an efficient algorithm, ideal for constraints on memory usage.
- Method 5: Pythonic One-Liner. Offers a compact solution, but at the cost of computational efficiency. Best used when quick prototyping is preferred over performance.