915

915. Partition Array into Disjoint Intervals (Medium)

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.

 

Related Topics:
Array

Solution 1. Mono Deque

// OJ: https://leetcode.com/problems/partition-array-into-disjoint-intervals/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class Solution {
public:
    int partitionDisjoint(vector<int>& A) {
        deque<int> q; // index
        int N = A.size(), mx = 0;
        for (int i = 0; i < N; ++i) {
            while (q.size() && A[q.back()] >= A[i]) q.pop_back();
            q.push_back(i);
        }
        for (int i = 0; i < N; ++i) {
            if (q.front() == i) q.pop_front();
            mx = max(mx, A[i]);
            if (mx <= A[q.front()]) return i + 1;
        }
        return -1;
    }
};