Max Chunks To Make Sorted II leetCode

 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.

 

class Solution {

    public int maxChunksToSorted(int[] arr) {

       

          int n = arr.length;

        int[] leftmax = new int[n]; // Array to store the maximum values from left to current index

        int[] rightmin = new int[n]; // Array to store the minimum values from current index to right

       

        // Initialize leftmax array

        leftmax[0] = arr[0];

        for (int i = 1; i < n; i++) {

            // leftmax[i] stores the maximum value from index 0 to i

            leftmax[i] = Math.max(leftmax[i - 1], arr[i]);

        }

        // Initialize rightmin array

        rightmin[n - 1] = arr[n - 1];

        for (int i = n - 2; i >= 0; i--) {

            // rightmin[i] stores the minimum value from index i to n-1

            rightmin[i] = Math.min(rightmin[i + 1], arr[i]);

        }

        int chunk = 0; // Count the number of valid chunks

        // Iterate through the array to find valid chunking points

        for (int i = 0; i < n - 1; i++) {

            // If the max in the left partition is smaller than the min in the right partition,

            // it means we can form a valid chunk

            if (leftmax[i] <= rightmin[i + 1]) {

                chunk++;

            }

        }

        // The final chunk count is incremented by 1 because the last segment is always a chunk

        return chunk + 1;

    }

}