How do you solve Partition Array into Disjoint Intervals on LeetCode?

Partition Array into Disjoint Intervals (LeetCode #915) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The maximum value of the left subarray must always be less than or equal to the minimum value of the right subarray.

What is the brute force solution for Partition Array into Disjoint Intervals?

The brute force approach for Partition Array into Disjoint Intervals is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every possible partition of the array. For each partition, we verify if all elements in the left subarray are less than or equal to all elements in the right subarray. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Partition Array into Disjoint Intervals?

Partition Array into Disjoint Intervals uses the following concepts: Array. The recommended patterns to study are: Two Pointers, Array.

What are the interview tips for Partition Array into Disjoint Intervals?

Always explain your thought process clearly as you work through the problem. Consider edge cases and test your solution against them. Practice both brute force and optimal solutions to understand the trade-offs.

What are common mistakes when solving Partition Array into Disjoint Intervals?

Failing to keep track of the maximum value of the left subarray, leading to incorrect partitioning. Not considering edge cases, such as when the array is already sorted.

#915

Partition Array into Disjoint Intervals

Medium

ArrayTwo PointersArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach uses a single pass to find the partition point by keeping track of the maximum value in the left subarray and the minimum value in the right subarray. This allows us to find the partition efficiently without nested loops.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to keep track of the maximum value of the left subarray.
2Step 2: Iterate through the array, updating the maximum value and checking if the current element is less than or equal to the minimum value of the right subarray.
3Step 3: When the condition is satisfied, return the current index + 1 as the size of the left subarray.

solution.py14 lines

1# Full working Python code
2
3def partition_disjoint(nums):
4    left_max = nums[0]
5    partition_index = 0
6    for i in range(1, len(nums)):
7        if nums[i] < left_max:
8            partition_index = i
9        else:
10            left_max = max(left_max, nums[i])
11    return partition_index + 1
12
13# Example usage
14print(partition_disjoint([5,0,3,8,6]))  # Output: 3

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array, updating our maximum and checking conditions without any nested loops.

1The maximum value of the left subarray must always be less than or equal to the minimum value of the right subarray.
2The partition point must be found efficiently to avoid unnecessary computations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.