Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead.
Example:
Input:s = 7, nums = [2,3,1,2,4,3]Output: 2 Explanation: the subarray[4,3]has the minimal length under the problem constraint.
Follow up:
If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
Approach #1:
class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
int len = nums.size();
if (len == 0) return 0;
int ans = INT_MAX;
vector<int> sum(len+1, 0);
for (int i = 1; i <= len; ++i)
sum[i] = sum[i-1] + nums[i-1];
for (int i = 0; i < len; ++i) {
int to_find = s + sum[i-1];
auto bound = lower_bound(sum.begin(), sum.end(), to_find);
if (bound != sum.end()) {
ans = min(ans, static_cast<int>(bound - (sum.begin() + i - 1)));
}
}
return (ans != INT_MAX) ? ans : 0;
}
};
Runtime: 8 ms, faster than 98.81% of C++ online submissions for Minimum Size Subarray Sum.
Approach #2: Using two pointer:
class Solution {
public:
int minSubArrayLen(int s, vector<int>& nums) {
int len = nums.size();
if (len == 0) return 0;
int ans = INT_MAX;
int sum = 0;
int left = 0;
for (int i = 0; i < len; ++i) {
sum += nums[i];
while (sum >= s) {
ans = min(ans, i+1-left);
sum -= nums[left++];
}
}
return (ans != INT_MAX) ? ans : 0;
}
};
Runtime: 8 ms, faster than 98.81% of C++ online submissions for Minimum Size Subarray Sum.