915. Partition Array into Disjoint Intervals

1. Question:

915. Partition Array into Disjoint Intervals

https://leetcode.com/problems/partition-array-into-disjoint-intervals/

Given an array A, partition it into two (contiguous) subarrays left and right so that:

• Every element in left is less than or equal to every element in right.
• left and right are non-empty.
• left has the smallest possible size.

Return the length of left after such a partitioning.  It is guaranteed that such a partitioning exists.

 

Example 1:

Input: [5,0,3,8,6]
Output: 3
Explanation: left = [5,0,3], right = [8,6]

Example 2:

Input: [1,1,1,0,6,12]
Output: 4
Explanation: left = [1,1,1,0], right = [6,12]

 

Note:

1. 2 <= A.length <= 30000
2. 0 <= A[i] <= 10^6
3. It is guaranteed there is at least one way to partition A as described.

2. Solution:

class Solution:
    def partitionDisjoint(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        size = len(A)
        right_min = [A[-1] for v in range(size)]
        min_value = A[-1]
        for i in range(size - 1, -1, -1):
            min_value = min(min_value, A[i])
            right_min[i] = min_value

        left_max = A[0]
        for i in range(size - 1):
            left_max = max(left_max, A[i])
            if left_max <= right_min[i + 1]:
                return i + 1

3. Complexity Analysis

Time Complexity : O(N)

Space Complexity: O(N)

转载于:https://www.cnblogs.com/ordili/p/10004345.html