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Formatted question description: https://leetcode.ca/all/769.html

769. Max Chunks To Make Sorted (Medium)

Given an array arr that is a permutation of [0, 1, ..., arr.length - 1], we split the array into some number of "chunks" (partitions), and individually sort each chunk.  After concatenating them, the result equals the sorted array.

What is the most number of chunks we could have made?

Example 1:

Input: arr = [4,3,2,1,0]
Output: 1
Explanation:
Splitting into two or more chunks will not return the required result.
For example, splitting into [4, 3], [2, 1, 0] will result in [3, 4, 0, 1, 2], which isn't sorted.

Example 2:

Input: arr = [1,0,2,3,4]
Output: 4
Explanation:
We can split into two chunks, such as [1, 0], [2, 3, 4].
However, splitting into [1, 0], [2], [3], [4] is the highest number of chunks possible.

Note:

• arr will have length in range [1, 10].
• arr[i] will be a permutation of [0, 1, ..., arr.length - 1].

 

Related Topics:
Array

Similar Questions:

• Max Chunks To Make Sorted II (Hard)

Solution 1.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public int maxChunksToSorted(int[] arr) {
          int curMax = 0;
          int chunksCount = 0;
          int length = arr.length;
          for (int i = 0; i < length; i++) {
              curMax = Math.max(curMax, arr[i]);
              if (curMax == i)
                  chunksCount++;
          }
          return chunksCount;
      }
  }
  
  ############
  
  class Solution {
      public int maxChunksToSorted(int[] arr) {
          int ans = 0, mx = 0;
          for (int i = 0; i < arr.length; ++i) {
              mx = Math.max(mx, arr[i]);
              if (i == mx) {
                  ++ans;
              }
          }
          return ans;
      }
  }
  

• // OJ: https://leetcode.com/problems/max-chunks-to-make-sorted/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      int maxChunksToSorted(vector<int>& A) {
          int N = A.size(), ans = 0, sum = 0;
          for (int i = 0; i < N; ++i) {
              if ((sum += A[i] - i) == 0) ++ans;
          }
          return ans;
      }
  };
  

• class Solution:
      def maxChunksToSorted(self, arr: List[int]) -> int:
          mx = ans = 0
          for i, v in enumerate(arr):
              mx = max(mx, v)
              if i == mx:
                  ans += 1
          return ans
  
  ############
  
  class Solution:
      def maxChunksToSorted(self, arr):
          """
          :type arr: List[int]
          :rtype: int
          """
          chunks = 0
          pre_max = 0
          for i, num in enumerate(arr):
              if num > pre_max:
                  pre_max = num
              if pre_max == i:
                  chunks += 1
          return chunks
  

• func maxChunksToSorted(arr []int) int {
  	ans, mx := 0, 0
  	for i, v := range arr {
  		mx = max(mx, v)
  		if i == mx {
  			ans++
  		}
  	}
  	return ans
  }
  
  func max(a, b int) int {
  	if a > b {
  		return a
  	}
  	return b
  }
  

• function maxChunksToSorted(arr: number[]): number {
      const n = arr.length;
      let ans = 0;
      let max = 0;
      for (let i = 0; i < n; i++) {
          max = Math.max(arr[i], max);
          if (max == i) {
              ans++;
          }
      }
      return ans;
  }
  
  

• impl Solution {
      pub fn max_chunks_to_sorted(arr: Vec<i32>) -> i32 {
          let mut res = 0;
          let mut max = 0;
          for i in 0..arr.len() {
              max = max.max(arr[i]);
              if max == i as i32 {
                  res += 1;
              }
          }
          res
      }
  }
  
  

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