##### Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/768.html

# 768. Max Chunks To Make Sorted II (Hard)

This question is the same as "Max Chunks to Make Sorted" except the integers of the given array are not necessarily distinct, the input array could be up to length 2000, and the elements could be up to 10**8.

Given an array arr of integers (not necessarily distinct), we split the array into some number of "chunks" (partitions), and individually sort each chunk.  After concatenating them, the result equals the sorted array.

What is the most number of chunks we could have made?

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.


Note:

• arr will have length in range [1, 2000].
• arr[i] will be an integer in range [0, 10**8].

Related Topics:
Array

Similar Questions:

## Solution 1. Monoqueue

Use a deque<int> q to store the min values between A[i] and A[N-1], i = N-1, ..., 0.

Example, A = [3, 1, 2], q = [1, 2].

When we visit 3, q.front() == 1 which means that there are smaller values after 3, so we shouldn’t split here.

When we visit 1, q.front() == 1 which means that we are visiting the smallest element among the elements we’ve seen thus far.

Should we split here? We need to take a look at the next element in q.

• If the next element in q is greater than or equal to the maximum value we’ve seen thus far, we can split at this element. (If there is no next element, we should split as well) Examples: A = [3, 1, 4], q = [1, 4]; A = [3, 1, 3], q = [1, 3].
• otherwise, we shouldn’t split here.
Example: A = [3, 1, 2], q = [1, 2]

Another example, A = [2, 1, 1], q = [1, 1].

When we visit the first 1, q.front() == 1 but the next element is also 1. And 1 is smaller than the max value 2 we’ve seen thus far, so we shouldn’t split after the first 1.

Another example, A = [1, 2, 2], q = [1, 2, 2].

When we visit the first 2, q.front() == 2 and the next element is also 2.

// OJ: https://leetcode.com/problems/max-chunks-to-make-sorted-ii/
// Time: O(N)
// Space: O(N)
class Solution {
public:
int maxChunksToSorted(vector<int>& A) {
deque<int> q;
int N = A.size(), ans = 0, mx = A[0];
for (int i = N - 1; i >= 0; --i) {
if (q.empty() || A[i] <= q.front()) q.push_front(A[i]);
}
for (int i = 0; i < N; ++i) {
mx = max(mx, A[i]);
if (q.front() != A[i]) continue;
q.pop_front();
if (q.empty() || mx <= q.front()) ++ans;
}
return ans;
}
};


## Solution 2.

If we can split between A[i - 1] and A[i], it means that the max value in A[0..(i-1)] is smaller than or equal to the max value in A[i..(N-1)].

// OJ: https://leetcode.com/problems/max-chunks-to-make-sorted-ii/
// Time: O(N)
// Space: O(N)
class Solution {
public:
int maxChunksToSorted(vector<int>& A) {
int N = A.size(), ans = 1;
vector<int> mx(N), mn(N);
mx[0] = A[0];
mn[N - 1] = A[N - 1];
for (int i = 1; i < N; ++i) mx[i] = max(mx[i - 1], A[i]);
for (int i = N - 2; i >= 0; --i) mn[i] = min(mn[i + 1], A[i]);
for (int i = 1; i < N; ++i) {
if (mn[i] >= mx[i - 1]) ++ans;
}
return ans;
}
};

• class Solution {
public int maxChunksToSorted(int[] arr) {
int length = arr.length;
int[] maxLeft = new int[length];
maxLeft[0] = arr[0];
for (int i = 1; i < length; i++)
maxLeft[i] = Math.max(maxLeft[i - 1], arr[i]);
int[] minRight = new int[length];
minRight[length - 1] = arr[length - 1];
for (int i = length - 2; i >= 0; i--)
minRight[i] = Math.min(minRight[i + 1], arr[i]);
int chunksCount = 0;
int max = -1;
for (int i = 0; i < length; i++) {
int curMax = maxLeft[i], curMin = minRight[i];
if (curMin >= max)
chunksCount++;
max = Math.max(max, curMax);
}
return chunksCount;
}
}

############

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();
}
}

• // OJ: https://leetcode.com/problems/max-chunks-to-make-sorted-ii/
// Time: O(N)
// Space: O(N)
class Solution {
public:
int maxChunksToSorted(vector<int>& A) {
deque<int> q;
int N = A.size(), ans = 0, mx = A[0];
for (int i = N - 1; i >= 0; --i) {
if (q.empty() || A[i] <= q.front()) q.push_front(A[i]);
}
for (int i = 0; i < N; ++i) {
mx = max(mx, A[i]);
if (q.front() != A[i]) continue;
q.pop_front();
if (q.empty() || mx <= q.front()) ++ans;
}
return ans;
}
};

• 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(object):
def maxChunksToSorted(self, arr):
"""
:type arr: List[int]
:rtype: int
"""
asort = sorted(arr)
res = 0
sum1 = 0
sum2 = 0
for i, a in enumerate(arr):
sum1 += a
sum2 += asort[i]
if sum1 == sum2:
res += 1
sum1 = 0
sum2 = 0
return res

• 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 stack = [];
for (const num of arr) {
if (stack.length !== 0 && num < stack[stack.length - 1]) {
const max = stack.pop();
while (stack.length !== 0 && num < stack[stack.length - 1]) {
stack.pop();
}
stack.push(max);
} else {
stack.push(num);
}
}
return stack.length;
}


• impl Solution {
pub fn max_chunks_to_sorted(arr: Vec<i32>) -> i32 {
let mut stack = vec![];
for num in arr.iter() {
if !stack.is_empty() && num < stack.last().unwrap() {
let max = stack.pop().unwrap();
while !stack.is_empty() && num < stack.last().unwrap() {
stack.pop();
}
stack.push(max)
} else {
stack.push(*num);
}
}
stack.len() as i32
}
}