Given an array of positive integers nums and a positive integer target, return the minimal length of a subarray whose sum is greater than or equal to target. If there is no such subarray, return 0 instead.
Example 1:
Input: target = 7, nums = [2,3,1,2,4,3] Output: 2 Explanation: The subarray [4,3] has the minimal length under the problem constraint.
Example 2:
Input: target = 4, nums = [1,4,4] Output: 1
Example 3:
Input: target = 11, nums = [1,1,1,1,1,1,1,1] Output: 0
Constraints:
1 <= target <= 1091 <= nums.length <= 1051 <= nums[i] <= 104Follow up: If you have figured out the
O(n) solution, try coding another solution of which the time complexity is O(n log(n)).
class Solution {
public:
int minSubArrayLen(int target, vector<int>& nums) {
int ans=INT_MAX, runningSum=0, start=0, end=0;
bool valid = false;
for(; end<nums.size();){
runningSum+=nums[end++];
while(runningSum>=target){
valid = true;
ans = min(ans, end-start);
runningSum-=nums[start++];
}
}
return valid?ans:0;
}
};