LeetCode 768: Max Chunks To Make Sorted II Solution

Master LeetCode problem 768 (Max Chunks To Make Sorted II), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

768. Max Chunks To Make Sorted II

Hard

Array Stack Greedy Sorting Monotonic Stack

Problem Explanation

Explanation:

To solve this problem, we can iterate through the array and keep track of the maximum value encountered so far. At each index, we compare the current maximum value with the sorted value at that index. If the sorted value at that index is less than or equal to the maximum value encountered so far, we increment the chunk count.

This approach works because if the maximum value encountered so far is greater than or equal to the sorted value at the current index, it means that we can form a chunk up to that index without violating the sorted order.

Time complexity: O(n) where n is the length of the input array arr.
Space complexity: O(1)

Solution Code

class Solution {
    public int maxChunksToSorted(int[] arr) {
        int n = arr.length;
        int[] maxLeft = new int[n];
        int[] minRight = new int[n];

        maxLeft[0] = arr[0];
        for (int i = 1; i < n; i++) {
            maxLeft[i] = Math.max(maxLeft[i - 1], arr[i]);
        }

        minRight[n - 1] = arr[n - 1];
        for (int i = n - 2; i >= 0; i--) {
            minRight[i] = i == n - 1 ? arr[i] : Math.min(minRight[i + 1], arr[i]);
        }

        int chunks = 1;
        for (int i = 0; i < n - 1; i++) {
            if (maxLeft[i] <= minRight[i + 1]) {
                chunks++;
            }
        }

        return chunks;
    }
}

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