Given an array nums
, we call (i, j)
an important reverse pair if i < j
and nums[i] > 2*nums[j]
.
You need to return the number of important reverse pairs in the given array.
Example1:
Input: [1,3,2,3,1] Output: 2
Example2:
Input: [2,4,3,5,1] Output: 3
class Solution {
public:
int reversePairs(vector<int>& nums) {
vector<int> sortedNums;
int res = 0, target = 0, pos = 0;
for (int i = nums.size() - 1; i >= 0; --i) {
target = nums[i] > 0 ? (nums[i] - 1) / 2 : nums[i] / 2 - 1;
// find the target that is approximately nums[i]/2 in the sorted array
auto ptr = upper_bound(sortedNums.begin(), sortedNums.end(), target);
res += ptr - sortedNums.begin();
// the number of reverse pair for nums[i] (j > i elements in nums are already sorted in "sortedNums")
pos = upper_bound(sortedNums.begin(), sortedNums.end(), nums[i]) - sortedNums.begin();
// insert nums[i] in "sortedNums"
sortedNums.insert(sortedNums.begin() + pos, nums[i]);
}
return res;
}
};