标签：maximum iterator fas ref miss ESS data ssi array
Return the length of the shortest, non-empty, contiguous subarray of A with sum at least K.
If there is no non-empty subarray with sum at least K, return -1.
Example 1:
Input: A = [1], K = 1
Output: 1
Example 2:
Input: A = [1,2], K = 4
Output: -1
Example 3:
Input: A = [2,-1,2], K = 3
Output: 3
Note:
1 <= A.length <= 50000-10 ^ 5 <= A[i] <= 10 ^ 51 <= K <= 10 ^ 9
Approach #1: prefix sum. [Time Limit Exceeded]
class Solution {
public:
int shortestSubarray(vector<int>& A, int K) {
int len = A.size();
if (len == 1 && A[0] >= K) return 1;
int step = INT_MAX;
vector<int> prefixSum(len, 0);
prefixSum[0] = A[0];
for (int i = 1; i < len; ++i)
prefixSum[i] = prefixSum[i-1] + A[i];
for (int i = 0; i < len; ++i) {
if (prefixSum[i] >= K)
step = min(step, i+1);
for (int j = i+1; j < len; ++j) {
if (prefixSum[j]-prefixSum[i] >= K) {
step = min(step, j-i);
}
}
}
if (step == INT_MAX) return -1;
else return step;
}
};
Approach #2: deque.
class Solution {
public:
int shortestSubarray(vector<int>& A, int K) {
int len = A.size();
int res = len + 1;
vector<int> sum(len+1, 0);
for (int i = 0; i < len; ++i)
sum[i+1] = sum[i] + A[i];
deque<int> d;
for (int i = 0; i < len+1; ++i) {
while (!d.empty() && sum[i]-sum[d.front()] >= K)
res = min(res, i-d.front()), d.pop_front();
while (!d.empty() && sum[i] <= sum[d.back()])
d.pop_back();
d.push_back(i);
}
return res <= len ? res : -1;
}
};
Runtime: 144 ms, faster than 33.12% of C++ online submissions for Shortest Subarray with Sum at Least K.
Analysis:
862. Shortest Subarray with Sum at Least K
标签：maximum iterator fas ref miss ESS data ssi array
原文地址：https://www.cnblogs.com/ruruozhenhao/p/9939890.html
