Leetcode 915. Partition Array into Disjoint Intervals

Problem:

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.

Solution:

　　不得不说，这是道很有意思的题目，虽然只是medium，但要想出O（n）时间复杂度和O（1）空间复杂度的解法也不容易。这道题要求我们划分数组，使得left数组里的最大值不大于right数组中的最小值。因此我们维护两个变量leftMax和maximal，leftMax是已经确定在左侧数组中的最大值，maximal是当前位置左侧的最大值。这里有几个关键点，什么时候maximal等于leftMax，如果A[i]小于leftMax时，如果A[i]是right数组中的，那么right中的A[i]会小于left数组中的leftMax，因此当A[i]小于leftMax时A[i]必然是left数组的右界限，因此i左侧的maximal此时赋值给leftMax。如果A[i]大于leftMax，我们是无法断定A[i]是属于left还是right的，因此我们只需要更新maximal即可。

Code:

 

class Solution {
public:
    int partitionDisjoint(vector<int>& A) {
        int result = 0;
        int leftMax = A[0];
        int maximal = A[0];
        for(int i = 1;i != A.size();++i){
            if(A[i] < leftMax){
                result = i;
                leftMax = maximal;
            }
            else{
                maximal = max(maximal,A[i]);
            }
        }
        return result+1;
    }
};