Easy
Minimum Positive Sum Subarray — C++
Full explanation · Time O(nlogn) · Space O(n)
// Time: O(nlogn)
// Space: O(n)
// prefix sum, two pointers, sliding window, bst, binary search
class Solution {
public:
int minimumSumSubarray(vector<int>& nums, int l, int r) {
static const int INF = numeric_limits<int>::max();
vector<int> prefix(size(nums) + 1);
for (int i = 0; i < size(nums); ++i) {
prefix[i + 1] = prefix[i] + nums[i];
}
int result = INF;
multiset<int> bst;
for (int i = 0; i < size(nums); ++i) {
if (i - l + 1 >= 0) {
bst.emplace(prefix[i - l + 1]);
}
if (i - r >= 0) {
bst.erase(bst.find(prefix[i - r]));
}
const auto it = bst.lower_bound(prefix[i + 1]);
if (it != begin(bst)) {
result = min(result, prefix[i + 1] - *prev(it));
}
}
return result != INF ? result : -1;
}
};