768. Max Chunks To Make Sorted II

Description

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

Solutions

Solution 1: Monotonic Stack

According to the problem, we can find that from left to right, each chunk has a maximum value, and these maximum values are monotonically increasing (non-strictly increasing). We can use a stack to store these maximum values of the chunks. The size of the final stack is the maximum number of chunks that can be sorted.

Time complexity is \(O(n)\), where \(n\) represents the length of \(\textit{arr}\).

Python3JavaC++GoTypeScriptRust

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)

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();
    }
}

class Solution {
public:
    int maxChunksToSorted(vector<int>& arr) {
        stack<int> stk;
        for (int& v : arr) {
            if (stk.empty() || stk.top() <= v)
                stk.push(v);
            else {
                int mx = stk.top();
                stk.pop();
                while (!stk.empty() && stk.top() > v) stk.pop();
                stk.push(mx);
            }
        }
        return stk.size();
    }
};

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)
}

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;
}

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
    }
}

Solution 2: Prefix Maximums + Suffix Minimums

We would like to partition the array of length \(n\) into several chunks such that after sorting each chunk individually, the entire array remains sorted.

Consider two adjacent chunks:

left chunk: left_chunk
right chunk: right_chunk

If the following condition holds:

max(left_chunk) <= min(right_chunk)

it means:

every element in the left chunk is less than or equal to every element in the right chunk
therefore, after sorting both chunks independently, they can still be concatenated into a globally sorted array

Thus, for every index \(i\) satisfying:

\[ 1 \le i < n \]

we check whether:

\[ \max(arr[:i]) \le \min(arr[i:]) \]

holds.

If true, then index \(i\) can serve as a valid partition point.

To efficiently evaluate the above condition, we preprocess:

prefix_maxs[j]

which represents:

\[ \max(arr[:j + 1]) \]

i.e. the prefix maximum.

suffix_min[j]

which represents:

\[ \min(arr[j:]) \]

i.e. the suffix minimum.

Traverse the array from left to right to compute all prefix maximums
Traverse the array from right to left to compute all suffix minimums
For each index \(i\), check whether:

prefix_maxs[i - 1] <= suffix_min[i]

holds.

If true, then:

every element on the left side is less than or equal to every element on the right side
therefore, the array can be partitioned at index \(i\)

Finally, count all valid partition points.

Note that:

Even if the array is strictly decreasing, the entire array itself can still be treated as one valid chunk.

Therefore, our final answer is equal to all valid partition points + \(1\).

The time complexity is \(O(n)\), and the space complexity is \(O(n)\), where \(n\) represents the length of \(\textit{arr}\).

Python3C++

class Solution:
    def maxChunksToSorted(self, arr: list[int]) -> int:
        prefix_maxs = []  # Max of each arr[:i + 1] where 0 <= i < len(arr).

        for num in arr:
            if not prefix_maxs:
                prefix_maxs.append(num)
                continue

            prefix_maxs.append(max(num, prefix_maxs[-1]))

        max_chunks = 1  # Base case.
        suffix_min = arr[-1]  # Min of arr[i:] where 0 <= i < len(arr).

        for idx in range(len(arr) - 1, 0, -1):
            if arr[idx] < suffix_min:
                suffix_min = arr[idx]

            if prefix_maxs[idx - 1] <= suffix_min:
                max_chunks += 1

        return max_chunks

class Solution {
public:
    int maxChunksToSorted(vector<int>& arr) {
        vector<int> prefixMaxs; // Max of each arr[:i + 1]. 0 <= i < arr.size().

        for (const auto& num : arr) {
            if (prefixMaxs.empty()) {
                prefixMaxs.push_back(num);
                continue;
            }

            prefixMaxs.push_back(max(prefixMaxs.back(), num));
        }

        int maxChunks = 1; // Base case.
        int suffixMin = arr.back(); // Min of arr[i:]. 0 <= i < arr.size().

        for (int idx = arr.size() - 1; idx >= 1; idx--) {
            if (arr[idx] < suffixMin)
                suffixMin = arr[idx];

            if (prefixMaxs[idx - 1] <= suffixMin)
                maxChunks++;
        }

        return maxChunks;
    }
};

768. Max Chunks To Make Sorted II

Description

Solutions

Solution 1: Monotonic Stack

Solution 2: Prefix Maximums + Suffix Minimums

Comments