LeetCode #915 — MEDIUM

Partition Array into Disjoint Intervals

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

[5,0,3,8,6]

Example 2

[1,1,1,0,6,12]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #915: Partition Array into Disjoint Intervals
class Solution {
    public int partitionDisjoint(int[] nums) {
        int n = nums.length;
        int[] mi = new int[n + 1];
        mi[n] = nums[n - 1];
        for (int i = n - 1; i >= 0; --i) {
            mi[i] = Math.min(nums[i], mi[i + 1]);
        }
        int mx = 0;
        for (int i = 1;; ++i) {
            int v = nums[i - 1];
            mx = Math.max(mx, v);
            if (mx <= mi[i]) {
                return i;
            }
        }
    }
}

// Accepted solution for LeetCode #915: Partition Array into Disjoint Intervals
func partitionDisjoint(nums []int) int {
	n := len(nums)
	mi := make([]int, n+1)
	mi[n] = nums[n-1]
	for i := n - 1; i >= 0; i-- {
		mi[i] = min(nums[i], mi[i+1])
	}
	mx := 0
	for i := 1; ; i++ {
		v := nums[i-1]
		mx = max(mx, v)
		if mx <= mi[i] {
			return i
		}
	}
}

# Accepted solution for LeetCode #915: Partition Array into Disjoint Intervals
class Solution:
    def partitionDisjoint(self, nums: List[int]) -> int:
        n = len(nums)
        mi = [inf] * (n + 1)
        for i in range(n - 1, -1, -1):
            mi[i] = min(nums[i], mi[i + 1])
        mx = 0
        for i, v in enumerate(nums, 1):
            mx = max(mx, v)
            if mx <= mi[i]:
                return i

// Accepted solution for LeetCode #915: Partition Array into Disjoint Intervals
struct Solution;

impl Solution {
    fn partition_disjoint(a: Vec<i32>) -> i32 {
        let n = a.len();
        let mut left = vec![0; n];
        let mut right = vec![0; n];
        let mut min = std::i32::MAX;
        let mut max = std::i32::MIN;
        for i in (0..n).rev() {
            min = min.min(a[i]);
            right[i] = min;
        }

        for i in 0..n {
            max = max.max(a[i]);
            left[i] = max;
        }
        for i in 1..n {
            if right[i] >= left[i - 1] {
                return i as i32;
            }
        }
        0
    }
}

#[test]
fn test() {
    let a = vec![5, 0, 3, 8, 6];
    let res = 3;
    assert_eq!(Solution::partition_disjoint(a), res);
    let a = vec![1, 1, 1, 0, 6, 12];
    let res = 4;
    assert_eq!(Solution::partition_disjoint(a), res);
}

// Accepted solution for LeetCode #915: Partition Array into Disjoint Intervals
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #915: Partition Array into Disjoint Intervals
// class Solution {
//     public int partitionDisjoint(int[] nums) {
//         int n = nums.length;
//         int[] mi = new int[n + 1];
//         mi[n] = nums[n - 1];
//         for (int i = n - 1; i >= 0; --i) {
//             mi[i] = Math.min(nums[i], mi[i + 1]);
//         }
//         int mx = 0;
//         for (int i = 1;; ++i) {
//             int v = nums[i - 1];
//             mx = Math.max(mx, v);
//             if (mx <= mi[i]) {
//                 return i;
//             }
//         }
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.