Easy
Longest Subsequence With Limited Sum — C++
Full explanation · Time O(nlogn + qlogn) · Space O(1)
// Time: O(nlogn + qlogn)
// Space: O(1)
// greedy, sort, binary search
class Solution {
public:
vector<int> answerQueries(vector<int>& nums, vector<int>& queries) {
sort(begin(nums), end(nums));
for (int i = 0; i + 1 < size(nums); ++i) {
nums[i + 1] += nums[i];
}
vector<int> result;
for (const auto& q : queries) {
result.emplace_back(distance(cbegin(nums), upper_bound(cbegin(nums), cend(nums), q)));
}
return result;
}
};