Hard
Count Stable Subarrays — C++
Full explanation · Time O(n + q) · Space O(n)
// Time: O(n + q)
// Space: O(n)
// prefix sum
class Solution {
public:
vector<long long> countStableSubarrays(vector<int>& nums, vector<vector<int>>& queries) {
const auto& count = [](int64_t n) {
return (n + 1) * n / 2;
};
vector<int> right(size(nums));
iota(begin(right), end(right), 0);
for (int i = size(nums) - 2; i >= 0; --i) {
if (nums[i] <= nums[i + 1]) {
right[i] = right[i + 1];
}
}
vector<int64_t> prefix(size(nums) + 1);
for (int i = 0, curr = 0; i < size(nums); ++i) {
if (i - 1 >= 0 && nums[i - 1] > nums[i]) {
curr = 0;
}
++curr;
prefix[i + 1] = prefix[i] + curr;
}
vector<long long> result(size(queries));
for (int i = 0; i < size(queries); ++i) {
const int l = queries[i][0];
const int r = queries[i][1];
const int m = min(right[l], r);
result[i] = count(m - l + 1) + (prefix[r + 1] - prefix[m + 1]);
}
return result;
}
};