Question

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(分割). 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.

Difficulty:Medium

Category:Array

Analyze

Solution

Solution: Search

class Solution {
 public:
  int partitionDisjoint(vector<int>& A) {
    // 1) Begin with the second
    multiset<int> record(A.begin() + 1, A.end());
    // 2) Use the first element to init the left_max
    int left_max = A[0];
    for (int i = 1; i < A.size(); ++i) {
      // 3) When the record.begin() > left_max,
      // it means all the other elements is biger than the left elements.
      if (*record.begin() >= left_max) return i;
      record.erase(record.equal_range(A[i]).first);
      left_max = max(left_max, A[i]);
    }
    return -1;
  }
};