// Source : https://leetcode.com/problems/minimum-size-subarray-sum/
// Author : Hao Chen
// Date   : 2015-06-09

/********************************************************************************** 
 * 
 * Given an array of n positive integers and a positive integer s, find the minimal length of a subarray 
 * of which the sum ≥ s. If there isn't one, return 0 instead.
 * 
 * For example, given the array [2,3,1,2,4,3] and s = 7,
 * the subarray [4,3] has the minimal length under the problem constraint.
 * 
 * click to show more practice.
 * 
 * More practice:
 * 
 * If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
 * 
 * Credits:Special thanks to @Freezen for adding this problem and creating all test cases.
 *               
 **********************************************************************************/


class Solution {
public:
    int minSubArrayLen(int s, vector<int>& nums) {
        int min = nums.size();
        int begin=0, end=0;
        int sum = 0;
        bool has = false;
        
        for (int i=0; i<nums.size(); i++, end++){
            
            sum += nums[i];
            
            while (sum >= s && begin <= end) {
                //the 1 is minial length, just return
                if (begin == end) return 1;
                
                if (end-begin+1 < min) {
                    min = end - begin + 1;
                    has = true;
                }
                //move the begin
                sum -= nums[begin++];
            }
 
        }
        
        return has ? min : 0; 
    }
};