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Partition Array into Disjoint Intervals

Medium

Asked by:

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Topics:

Arrays

Given an integer array nums, partition it into two (contiguous) subarrays left and right so that:

Every element in left is less than or equal to every element in right.
left and right are non-empty.
left has the smallest possible size.

Return the length of left after such a partitioning.

Test cases are generated such that partitioning exists.

Example 1:

Input: nums = [5,0,3,8,6]
Output: 3
Explanation: left = [5,0,3], right = [8,6]

Example 2:

Input: nums = [1,1,1,0,6,12]
Output: 4
Explanation: left = [1,1,1,0], right = [6,12]

Constraints:

2 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁶
There is at least one valid answer for the given input.

Try coding it in LeetCode

Partition Array into Disjoint Intervals

Medium

Asked by:

13 views

Topics:

Arrays

Given an integer array nums, partition it into two (contiguous) subarrays left and right so that:

Every element in left is less than or equal to every element in right.
left and right are non-empty.
left has the smallest possible size.

Return the length of left after such a partitioning.

Test cases are generated such that partitioning exists.

Example 1:

Input: nums = [5,0,3,8,6]
Output: 3
Explanation: left = [5,0,3], right = [8,6]

Example 2:

Input: nums = [1,1,1,0,6,12]
Output: 4
Explanation: left = [1,1,1,0], right = [6,12]

Constraints:

2 <= nums.length <= 10⁵
0 <= nums[i] <= 10⁶
There is at least one valid answer for the given input.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values in the array, and can it contain negative numbers or zeros?
What should I return if no such partition exists that satisfies the condition?
Are there any constraints on the size of the input array?
Are duplicate values allowed in the input array, and if so, how should they be handled in determining the 'left' and 'right' intervals?
Is it guaranteed that a valid partition always exists, or should I handle cases where no such partition is possible?

Brute Force Solution

Approach

The brute force strategy for splitting a list of numbers involves examining every possible split point. For each split, we verify if the numbers to the left are all smaller than or equal to the numbers to the right. If we find a valid split, we stop; otherwise, we check the next possible split location.

Here's how the algorithm would work step-by-step:

Start by considering a split where the first group has only the first number.
Check if the largest number in the first group is less than or equal to the smallest number in the second group.
If this condition is met, you've found a valid split. You are done.
If not, try a split where the first group has the first two numbers.
Again, check if the largest number in the first group is less than or equal to the smallest number in the second group.
Keep increasing the size of the first group and repeating the check until you find a valid split.

Code Implementation

def partition_disjoint(numbers):
    for partition_index in range(1, len(numbers)): 
        # Iterate through all possible partition indices.

        left_partition = numbers[:partition_index]
        right_partition = numbers[partition_index:]

        max_left = max(left_partition)
        min_right = min(right_partition)

        # Check if max of left <= min of right
        if max_left <= min_right:

            # We found a valid split so return
            return partition_index

    return len(numbers)

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through all possible split points of the input array of size n. For each split point, it finds the maximum element in the left subarray and the minimum element in the right subarray. Finding the maximum and minimum requires iterating through the elements of the left and right subarrays respectively. In the worst-case scenario, for each of the n possible splits, we might iterate through almost all n elements to find the min and max, resulting in nested loops. This leads to approximately n * n operations, which simplifies to O(n²).

Space Complexity

O(1) – The brute force strategy, as described, doesn't use any auxiliary data structures like arrays, lists, or hash maps. It iterates through possible split points and performs comparisons using the existing input array. The space used is limited to a few variables to track the current split and potentially the maximum and minimum values within each partition, but these are independent of the input size N (where N is the length of the input array). Therefore, the space complexity is constant.

Optimal Solution

Approach

The goal is to split a sequence of numbers into two parts, where every number in the first part is smaller than or equal to every number in the second part. We find the best split by keeping track of the largest number we've seen so far and the overall best spot to split the sequence.

Here's how the algorithm would work step-by-step:

Start at the beginning of the sequence and keep track of the largest number encountered up to the current position.
Also, keep track of the maximum value of the entire left partition as we scan, as well as our current split position.
Move through the sequence, checking if the current number is smaller than the largest number we've seen so far on the left.
If we find a number that breaks this rule (is smaller than the largest number on the left), we need to extend the left part to include this number.
When we extend the left side, we move our potential split point forward. The new potential split point becomes the current position.
Repeat steps 3-5 until we've scanned the entire sequence.
At the end, the split position we've saved is the optimal place to divide the sequence into two parts that follow the rules.

Code Implementation

def partition_disjoint(numbers):
    left_partition_maximum = numbers[0]
    overall_maximum = numbers[0]
    partition_index = 0

    for i in range(1, len(numbers)):
        # If current number is smaller, expand left partition
        if numbers[i] < left_partition_maximum:

            partition_index = i

            # Update left partition's maximum
            left_partition_maximum = overall_maximum

        else:

            overall_maximum = max(overall_maximum, numbers[i])

    # Return the length of the first partition.
    return partition_index + 1

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array of size n exactly once to determine the partition point. Inside the loop, comparisons and updates of maximum values are performed, but these operations take constant time, O(1). Since the loop iterates n times and each iteration takes constant time, the overall time complexity is O(n).

Space Complexity

O(1) – The provided algorithm only uses a few scalar variables to store the maximum value seen so far, the overall maximum of the left partition, and the current split position. The number of variables remains constant regardless of the input array's size (N). Therefore, the auxiliary space complexity is O(1), indicating constant space.

Edge Cases

Case	How to Handle
Empty input array	Return 0 immediately as no partition is possible.
Array with a single element	Return 1, as the entire array is the left partition.
Array sorted in ascending order	The left partition will always be just the first element.
Array sorted in descending order	The left partition will be the entire array minus the last element.
Array with all elements equal	The left partition will be just the first element.
Array with very large or very small integers (potential overflow)	Use long data type to accommodate large integer values and prevent overflow during calculations of max_left and min_right.
Array where no valid partition exists	Iterate through all possible partition points, and if no valid partition is found, return the size of the array.
Large array size affecting memory usage or time complexity	The solution should have a time complexity of O(n) and space complexity of O(1) (or O(n) if using prefix max and suffix min arrays), ensuring scalability.

Empty input array

How to Handle:

Return 0 immediately as no partition is possible.

Array with a single element

How to Handle:

Return 1, as the entire array is the left partition.

Array sorted in ascending order

How to Handle:

The left partition will always be just the first element.

Array sorted in descending order

How to Handle:

The left partition will be the entire array minus the last element.

Array with all elements equal

How to Handle:

The left partition will be just the first element.

Array with very large or very small integers (potential overflow)

How to Handle:

Use long data type to accommodate large integer values and prevent overflow during calculations of max_left and min_right.

Array where no valid partition exists

How to Handle:

Iterate through all possible partition points, and if no valid partition is found, return the size of the array.

Large array size affecting memory usage or time complexity

How to Handle:

The solution should have a time complexity of O(n) and space complexity of O(1) (or O(n) if using prefix max and suffix min arrays), ensuring scalability.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values in the array, and can it contain negative numbers or zeros?
What should I return if no such partition exists that satisfies the condition?
Are there any constraints on the size of the input array?
Are duplicate values allowed in the input array, and if so, how should they be handled in determining the 'left' and 'right' intervals?
Is it guaranteed that a valid partition always exists, or should I handle cases where no such partition is possible?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Start by considering a split where the first group has only the first number.
Check if the largest number in the first group is less than or equal to the smallest number in the second group.
If this condition is met, you've found a valid split. You are done.
If not, try a split where the first group has the first two numbers.
Again, check if the largest number in the first group is less than or equal to the smallest number in the second group.
Keep increasing the size of the first group and repeating the check until you find a valid split.

Code Implementation

def partition_disjoint(numbers):
    for partition_index in range(1, len(numbers)): 
        # Iterate through all possible partition indices.

        left_partition = numbers[:partition_index]
        right_partition = numbers[partition_index:]

        max_left = max(left_partition)
        min_right = min(right_partition)

        # Check if max of left <= min of right
        if max_left <= min_right:

            # We found a valid split so return
            return partition_index

    return len(numbers)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Start at the beginning of the sequence and keep track of the largest number encountered up to the current position.
Also, keep track of the maximum value of the entire left partition as we scan, as well as our current split position.
Move through the sequence, checking if the current number is smaller than the largest number we've seen so far on the left.
If we find a number that breaks this rule (is smaller than the largest number on the left), we need to extend the left part to include this number.
When we extend the left side, we move our potential split point forward. The new potential split point becomes the current position.
Repeat steps 3-5 until we've scanned the entire sequence.
At the end, the split position we've saved is the optimal place to divide the sequence into two parts that follow the rules.

Code Implementation

def partition_disjoint(numbers):
    left_partition_maximum = numbers[0]
    overall_maximum = numbers[0]
    partition_index = 0

    for i in range(1, len(numbers)):
        # If current number is smaller, expand left partition
        if numbers[i] < left_partition_maximum:

            partition_index = i

            # Update left partition's maximum
            left_partition_maximum = overall_maximum

        else:

            overall_maximum = max(overall_maximum, numbers[i])

    # Return the length of the first partition.
    return partition_index + 1

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 immediately as no partition is possible.
Array with a single element	Return 1, as the entire array is the left partition.
Array sorted in ascending order	The left partition will always be just the first element.
Array sorted in descending order	The left partition will be the entire array minus the last element.
Array with all elements equal	The left partition will be just the first element.
Array with very large or very small integers (potential overflow)	Use long data type to accommodate large integer values and prevent overflow during calculations of max_left and min_right.
Array where no valid partition exists	Iterate through all possible partition points, and if no valid partition is found, return the size of the array.
Large array size affecting memory usage or time complexity	The solution should have a time complexity of O(n) and space complexity of O(1) (or O(n) if using prefix max and suffix min arrays), ensuring scalability.

Empty input array

How to Handle:

Return 0 immediately as no partition is possible.

Array with a single element

How to Handle:

Return 1, as the entire array is the left partition.

Array sorted in ascending order

How to Handle:

The left partition will always be just the first element.

Array sorted in descending order

How to Handle:

The left partition will be the entire array minus the last element.

Array with all elements equal

How to Handle:

The left partition will be just the first element.

Array with very large or very small integers (potential overflow)

How to Handle:

Use long data type to accommodate large integer values and prevent overflow during calculations of max_left and min_right.

Array where no valid partition exists

How to Handle:

Iterate through all possible partition points, and if no valid partition is found, return the size of the array.

Large array size affecting memory usage or time complexity

How to Handle:

The solution should have a time complexity of O(n) and space complexity of O(1) (or O(n) if using prefix max and suffix min arrays), ensuring scalability.