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# Minimum Size Subarray Sum LeetCode Solution

## Problem – Minimum Size Subarray Sum

Given an array of positive integers `nums` and a positive integer `target`, return the minimal length of a contiguous subarray `[numsl, numsl+1, ..., numsr-1, numsr]` of which the 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 <= 109`
• `1 <= nums.length <= 105`
• `1 <= nums[i] <= 104`

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))`.

### Minimum Size Subarray Sum LeetCode Solution in Java

``````    public int minSubArrayLen(int s, int[] A) {
int i = 0, n = A.length, res = n + 1;
for (int j = 0; j < n; ++j) {
s -= A[j];
while (s <= 0) {
res = Math.min(res, j - i + 1);
s += A[i++];
}
}
return res % (n + 1);
}
``````

### Minimum Size Subarray Sum LeetCode Solution in C++

``````    int minSubArrayLen(int s, vector<int>& A) {
int i = 0, n = A.size(), res = n + 1;
for (int j = 0; j < n; ++j) {
s -= A[j];
while (s <= 0) {
res = min(res, j - i + 1);
s += A[i++];
}
}
return res % (n + 1);
}
``````

### Minimum Size Subarray Sum LeetCode Solution in Python

``````    def minSubArrayLen(self, s, A):
i, res = 0, len(A) + 1
for j in xrange(len(A)):
s -= A[j]
while s <= 0:
res = min(res, j - i + 1)
s += A[i]
i += 1
return res % (len(A) + 1)
``````
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