209. Minimum Size Subarray Sum

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.