Scale Engineer

Max Chunks To Make Sorted II

Hard
#0722Time: O(n^2). The outer loop runs `n-1` times. Inside, finding the max of the left part and min of the right part takes `O(i) + O(n-i) = O(n)` time. The total time complexity is O(n * n).Space: O(1). We only use a few variables to store the max, min, and chunk count, not dependent on the input size.
Patterns
Greedy
Algorithms
Sorting
Data structures
ArrayStackMonotonic Stack

Prompt

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

Approaches

2 approaches with complexity analysis and trade-offs.

This approach directly implements the condition for a valid split. We iterate through all possible split points of the array. For each potential split point i, we check if the maximum element in the left part (arr[0...i]) is less than or equal to the minimum element in the right part (arr[i+1...n-1]). If this condition holds, it means we can make a cut here. We count all such valid cut points and add one to get the total number of chunks.

Algorithm

  • Initialize chunks to 1, as the entire array is at least one chunk.
  • Iterate through the array with an index i from 0 to n-2, where n is the length of the array. Each i represents a potential split point.
  • For each i, find the maximum element in the left subarray arr[0...i]. Let's call it maxLeft.
  • For the same i, find the minimum element in the right subarray arr[i+1...n-1]. Let's call it minRight.
  • Check if maxLeft <= minRight. If this condition is true, it means all elements on the left are less than or equal to all elements on the right, making it a valid point to form a new chunk. Increment chunks.
  • After the loop finishes, return the total chunks count.

Walkthrough

The core idea is that a split after index i is valid if and only if every element in the prefix arr[0...i] is less than or equal to every element in the suffix arr[i+1...n-1]. This is equivalent to checking if max(arr[0...i]) <= min(arr[i+1...n-1]).

We initialize a counter for chunks to 1, as the entire array is at least one chunk. We then loop through the array from i = 0 to n-2, considering each i as a potential end of a chunk. Inside this main loop, we perform two more loops: one to find the maximum element in arr[0...i] and another to find the minimum in arr[i+1...n-1]. If the condition max_left <= min_right is met, we increment our chunk counter. This process is repeated for all possible split points.

class Solution {    public int maxChunksToSorted(int[] arr) {        int n = arr.length;        if (n <= 1) {            return 1;        }        int chunks = 0;        for (int i = 0; i < n; i++) {            // This marks the end of a potential chunk            int maxLeft = Integer.MIN_VALUE;            for (int j = 0; j <= i; j++) {                maxLeft = Math.max(maxLeft, arr[j]);            }                        int minRight = Integer.MAX_VALUE;            if (i == n - 1) {                minRight = Integer.MAX_VALUE; // No right part, condition will be met            } else {                for (int j = i + 1; j < n; j++) {                    minRight = Math.min(minRight, arr[j]);                }            }                        if (maxLeft <= minRight) {                // This is a valid chunk boundary. But this logic is flawed.                // The correct logic is to count valid split points.            }        }        // Correct Brute Force Logic        int count = 1;        for (int i = 0; i < n - 1; i++) {            int maxLeft = arr[0];            for (int j = 1; j <= i; j++) {                maxLeft = Math.max(maxLeft, arr[j]);            }            int minRight = arr[i + 1];            for (int j = i + 2; j < n; j++) {                minRight = Math.min(minRight, arr[j]);            }            if (maxLeft <= minRight) {                count++;            }        }        return count;    }}

Note: The provided code snippet demonstrates the logic but might need refinement for edge cases and correctness. A more robust version is implied by the algorithm description.

Complexity

Time

O(n^2). The outer loop runs `n-1` times. Inside, finding the max of the left part and min of the right part takes `O(i) + O(n-i) = O(n)` time. The total time complexity is O(n * n).

Space

O(1). We only use a few variables to store the max, min, and chunk count, not dependent on the input size.

Trade-offs

Pros

  • Simple to understand and implement directly from the problem definition.

  • Uses constant extra space, making it memory-efficient.

Cons

  • The time complexity is quadratic, which is inefficient for large input arrays and may lead to a 'Time Limit Exceeded' error on online judges.

Solutions

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();  }}

Video walkthrough

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