grid47 Exploring patterns and algorithms
Aug 22, 2024
5 min read
Solution to LeetCode 769: Max Chunks To Make Sorted Problem
You are given an integer array arr of length n, where the array is a permutation of integers in the range [0, n - 1]. Your task is to split this array into as many chunks (or partitions) as possible, where each chunk can be individually sorted. After sorting each chunk, when concatenated, the result should be the same as the fully sorted array. Return the largest number of chunks that you can create to achieve the sorted array.
Problem
Approach
Steps
Complexity
Input: The input consists of a single array arr, which represents a permutation of integers in the range [0, n - 1].
Example: Input: arr = [5, 3, 2, 1, 4]
Constraints:
• 1 <= arr.length <= 10
• arr is a permutation of integers in the range [0, arr.length - 1].
Output: Return the largest number of chunks that can be created such that after sorting each chunk individually, the resulting concatenated array is sorted.
Example: Output: 1
Constraints:
• The array should be partitioned in such a way that the sorted chunks, when concatenated, form the fully sorted array.
Goal: The goal is to partition the array into the maximum number of chunks such that after sorting each chunk, the concatenated result is sorted.
Steps:
• Iterate through the array and track the maximum value encountered so far.
• At each index, if the maximum value is equal to the current index, increment the chunk count.
• Return the total number of chunks.
Goal: The constraints ensure the array is a permutation of integers from 0 to n-1 and the size of the array is manageable (n <= 10).
Steps:
• The array is a permutation of integers in the range [0, n - 1].
• The size of the array is at most 10.
Assumptions:
• The input array will always be a valid permutation of integers from 0 to n-1.
• Input: Example 1: Input: arr = [5, 3, 2, 1, 4]
• Explanation: In this case, the array is not easily separable into smaller chunks that would result in the sorted array, so the maximum number of chunks is 1.
• Input: Example 2: Input: arr = [1, 0, 2, 3, 4]
• Explanation: In this case, the array can be split into 4 chunks: [1, 0], [2], [3], [4], all of which are sorted when processed separately.
Approach: To solve this problem, we can iterate through the array while keeping track of the maximum value encountered so far. When the maximum value equals the current index, it indicates a valid chunk boundary. We can increment the chunk count and continue.
Observations:
• We need to partition the array into chunks where each chunk can be sorted independently and when combined, it forms the fully sorted array.
• Tracking the maximum value up to each index helps in determining the correct chunk boundaries.
Steps:
• Iterate over the array while maintaining the maximum value encountered so far.
• Each time the maximum value is equal to the current index, it marks a valid chunk boundary, and we increment the chunk count.
Empty Inputs:
• The array will always have at least one element, as n >= 1.
Large Inputs:
• Ensure the solution handles the maximum constraint, where the array size is 10.
Special Values:
• If the array is already sorted, each element can be its own chunk.
Constraints:
• The array is a permutation of integers from 0 to n-1.
intmaxChunksToSorted(vector<int>& arr) {
int n = arr.size();
int res =0, mx = arr[0];
for(int i =0; i < n; i++) {
mx = max(mx, arr[i]);
res += (mx == i);
}
return res;
}
1 : Function Definition
intmaxChunksToSorted(vector<int>& arr) {
Define the function that takes a vector of integers as input.
2 : Variable Declaration
int n = arr.size();
Initialize the variable 'n' to store the size of the input array.
3 : Variable Initialization
int res =0, mx = arr[0];
Initialize 'res' (the result variable) to 0 and 'mx' (the maximum value seen so far) to the first element of the array.
4 : Loop Start
for(int i =0; i < n; i++) {
Start iterating through the array from the first element.
5 : Max Update
mx = max(mx, arr[i]);
Update the maximum value 'mx' by comparing it with the current element 'arr[i]'.
6 : Result Update
res += (mx == i);
Increment 'res' by 1 if the maximum value matches the current index 'i'. This checks if the current subarray can be sorted independently.
7 : Return Value
return res;
Return the final result, which is the number of chunks.
Best Case: O(n), where n is the length of the array.
Average Case: O(n), since we iterate over the array once.
Worst Case: O(n), as we only perform one pass through the array.
Description: The time complexity is linear because we only iterate through the array once.
Best Case: O(1), since we only need constant space to track the maximum value and chunk count.
Worst Case: O(1), as no additional space is required except for the variables used in the logic.
Description: The space complexity is constant because we only use a few variables for tracking.
