Is Partition Array into Disjoint Intervals easy or hard?

LeetCode classifies this problem as Medium difficulty. The main challenge is recognizing the condition max(left) ≤ min(right) and translating it into prefix/suffix tracking or a constant-space greedy scan.

Partition Array into Disjoint Intervals Python/Java solution

Both Python and Java implementations follow the same logic. Either compute prefixMax and suffixMin arrays and find the first valid split, or implement the optimized single-pass scan with two variables tracking the left partition maximum and the global maximum.

How to solve Partition Array into Disjoint Intervals in O(n)?

Track two values while iterating: the maximum of the left partition and the maximum value seen globally. If the current element is smaller than the left maximum, extend the partition boundary to the current index and update the left maximum using the global maximum. This ensures the left side always satisfies max(left) ≤ min(right).

What is the best approach for Partition Array into Disjoint Intervals?

The optimal approach is a single-pass prefix tracking method with O(n) time and O(1) space. Maintain the maximum value in the current left partition and the global maximum seen so far. If a smaller element appears, extend the partition boundary and update the left maximum. This avoids building auxiliary arrays while still scanning the array only once.

Is Partition Array into Disjoint Intervals asked at Google/Amazon/Meta?

Partition-based array problems are common in interviews at large tech companies such as Amazon, Google, and Meta. This problem specifically tests reasoning about prefix maximums, suffix minimums, and greedy partition logic using linear scans.

What data structure is used in Partition Array into Disjoint Intervals?

The problem primarily uses arrays and simple scalar variables. One approach builds prefix maximum and suffix minimum arrays, while the optimized solution tracks values during a single pass without extra data structures.

What is the time complexity of Partition Array into Disjoint Intervals?

The optimal solution runs in O(n) time because the array is scanned once while maintaining a few running values. A simpler version using prefix maximum and suffix minimum arrays also runs in O(n) time but requires O(n) additional space for the helper arrays.

Description Solution Video 19:40

915. Partition Array into Disjoint Intervals

Medium49.4% acceptance2 companies

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.

Asked at 2 companies

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Array

19:40

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915. Partition Array into Disjoint Intervals

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