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MaxChunksToMakeSortedII768.java
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MaxChunksToMakeSortedII768.java
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/**
* This question is the same as "Max Chunks to Make Sorted" except the integers
* of the given array are not necessarily distinct, the input array could be
* up to length 2000, and the elements could be up to 10**8.
*
* Given an array arr of integers (not necessarily distinct), 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 = [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.
*
* Note:
* arr will have length in range [1, 2000].
* arr[i] will be an integer in range [0, 10**8].
*/
public class MaxChunksToMakeSortedII768 {
public int maxChunksToSorted(int[] arr) {
Stack<Integer> stack = new Stack<>();
for (int i=0; i<arr.length; i++) {
if (stack.isEmpty() || stack.peek() <= arr[i]) {
stack.push(arr[i]);
} else {
int top = stack.peek();
while (!stack.isEmpty() && stack.peek() > arr[i]) {
stack.pop();
}
stack.push(top);
}
}
return stack.size();
}
/**
* https://leetcode.com/problems/max-chunks-to-make-sorted-ii/discuss/113462/Java-solution-left-max-and-right-min.
*/
public int maxChunksToSorted2(int[] arr) {
int n = arr.length;
int[] maxOfLeft = new int[n];
int[] minOfRight = new int[n];
maxOfLeft[0] = arr[0];
for (int i = 1; i < n; i++) {
maxOfLeft[i] = Math.max(maxOfLeft[i-1], arr[i]);
}
minOfRight[n - 1] = arr[n - 1];
for (int i = n - 2; i >= 0; i--) {
minOfRight[i] = Math.min(minOfRight[i + 1], arr[i]);
}
int res = 0;
for (int i = 0; i < n - 1; i++) {
if (maxOfLeft[i] <= minOfRight[i + 1]) res++;
}
return res + 1;
}
}