768. Max Chunks To Make Sorted II LeetCode Solution

In this guide, you will get 768. Max Chunks To Make Sorted II LeetCode Solution with the best time and space complexity. The solution to Max Chunks To Make Sorted II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Max Chunks To Make Sorted II solution in C++
Max Chunks To Make Sorted II solution in Java
Max Chunks To Make Sorted II solution in Python
Additional Resources

768. Max Chunks To Make Sorted II LeetCode Solution image

Problem Statement of Max Chunks To Make Sorted II

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

Complexity Analysis

Time Complexity: O(n)
Space Complexity: O(n)

768. Max Chunks To Make Sorted II LeetCode Solution in C++

class Solution {
 public:
  int maxChunksToSorted(vector<int>& arr) {
    const int n = arr.size();
    int ans = 0;
    vector<int> maxL(n);  // l[i] := max(arr[0..i])
    vector<int> minR(n);  // r[i] := min(arr[i..n))

    for (int i = 0; i < n; ++i)
      maxL[i] = i == 0 ? arr[i] : max(arr[i], maxL[i - 1]);

    for (int i = n - 1; i >= 0; --i)
      minR[i] = i == n - 1 ? arr[i] : min(arr[i], minR[i + 1]);

    for (int i = 0; i + 1 < n; ++i)
      if (maxL[i] <= minR[i + 1])
        ++ans;

    return ans + 1;
  }
};
/* code provided by PROGIEZ */

768. Max Chunks To Make Sorted II LeetCode Solution in Java

class Solution {
  public int maxChunksToSorted(int[] arr) {
    final int n = arr.length;
    int ans = 0;
    int[] maxL = new int[n]; // l[i] := max(arr[0..i])
    int[] minR = new int[n]; // r[i] := min(arr[i..n))

    for (int i = 0; i < n; ++i)
      maxL[i] = i == 0 ? arr[i] : Math.max(arr[i], maxL[i - 1]);

    for (int i = n - 1; i >= 0; --i)
      minR[i] = i == n - 1 ? arr[i] : Math.min(arr[i], minR[i + 1]);

    for (int i = 0; i + 1 < n; ++i)
      if (maxL[i] <= minR[i + 1])
        ++ans;

    return ans + 1;
  }
}
// code provided by PROGIEZ

768. Max Chunks To Make Sorted II LeetCode Solution in Python

class Solution:
  def maxChunksToSorted(self, arr: list[int]) -> int:
    n = len(arr)
    ans = 0
    mx = -math.inf
    mn = [arr[-1]] * n

    for i in reversed(range(n - 1)):
      mn[i] = min(mn[i + 1], arr[i])

    for i in range(n - 1):
      mx = max(mx, arr[i])
      if mx <= mn[i + 1]:
        ans += 1

    return ans + 1
 # code by PROGIEZ