LeetCode #768 — HARD

Max Chunks To Make Sorted II

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer array arr.

We split arr into some number of chunks (i.e., partitions), and individually sort each chunk. After concatenating them, the result should equal the sorted array.

Return the largest number of chunks we can make to sort the array.

Example 1:

Input: arr = [5,4,3,2,1]
Output: 1
Explanation:
Splitting into two or more chunks will not return the required result.
For example, splitting into [5, 4], [3, 2, 1] will result in [4, 5, 1, 2, 3], which isn't sorted.

Example 2:

Input: arr = [2,1,3,4,4]
Output: 4
Explanation:
We can split into two chunks, such as [2, 1], [3, 4, 4].
However, splitting into [2, 1], [3], [4], [4] is the highest number of chunks possible.

Constraints:

1 <= arr.length <= 2000
0 <= arr[i] <= 10⁸

Step 01

Brute Force Baseline

Problem summary: You are given an integer array arr. We split arr into some number of chunks (i.e., partitions), and individually sort each chunk. After concatenating them, the result should equal the sorted array. Return the largest number of chunks we can make to sort the array.

Baseline thinking

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

Pattern signal: Array · Stack · Greedy

Example 1

[5,4,3,2,1]

Example 2

[2,1,3,4,4]

Core Insight

What unlocks the optimal approach

Each k for which some permutation of arr[:k] is equal to sorted(arr)[:k] is where we should cut each chunk.

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

Largest constraint values

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 #768: Max Chunks To Make Sorted II
class Solution {
    public int maxChunksToSorted(int[] arr) {
        Deque<Integer> stk = new ArrayDeque<>();
        for (int v : arr) {
            if (stk.isEmpty() || stk.peek() <= v) {
                stk.push(v);
            } else {
                int mx = stk.pop();
                while (!stk.isEmpty() && stk.peek() > v) {
                    stk.pop();
                }
                stk.push(mx);
            }
        }
        return stk.size();
    }
}

// Accepted solution for LeetCode #768: Max Chunks To Make Sorted II
func maxChunksToSorted(arr []int) int {
	var stk []int
	for _, v := range arr {
		if len(stk) == 0 || stk[len(stk)-1] <= v {
			stk = append(stk, v)
		} else {
			mx := stk[len(stk)-1]
			stk = stk[:len(stk)-1]
			for len(stk) > 0 && stk[len(stk)-1] > v {
				stk = stk[:len(stk)-1]
			}
			stk = append(stk, mx)
		}
	}
	return len(stk)
}

# Accepted solution for LeetCode #768: Max Chunks To Make Sorted II
class Solution:
    def maxChunksToSorted(self, arr: List[int]) -> int:
        stk = []
        for v in arr:
            if not stk or v >= stk[-1]:
                stk.append(v)
            else:
                mx = stk.pop()
                while stk and stk[-1] > v:
                    stk.pop()
                stk.append(mx)
        return len(stk)

// Accepted solution for LeetCode #768: Max Chunks To Make Sorted II
impl Solution {
    pub fn max_chunks_to_sorted(arr: Vec<i32>) -> i32 {
        let mut stk = Vec::new();
        for &v in arr.iter() {
            if stk.is_empty() || v >= *stk.last().unwrap() {
                stk.push(v);
            } else {
                let mut mx = stk.pop().unwrap();
                while let Some(&top) = stk.last() {
                    if top > v {
                        stk.pop();
                    } else {
                        break;
                    }
                }
                stk.push(mx);
            }
        }
        stk.len() as i32
    }
}

// Accepted solution for LeetCode #768: Max Chunks To Make Sorted II
function maxChunksToSorted(arr: number[]): number {
    const stk: number[] = [];
    for (let v of arr) {
        if (stk.length === 0 || v >= stk[stk.length - 1]) {
            stk.push(v);
        } else {
            let mx = stk.pop()!;
            while (stk.length > 0 && stk[stk.length - 1] > v) {
                stk.pop();
            }
            stk.push(mx);
        }
    }
    return stk.length;
}

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

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

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.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.