<p>You are given a <strong>0-indexed</strong> array <code>nums</code> and an integer <code>target</code>.</p>

<p>A <strong>0-indexed</strong> array <code>infinite_nums</code> is generated by infinitely appending the elements of <code>nums</code> to itself.</p>

<p>Return <em>the length of the <strong>shortest</strong> subarray of the array </em><code>infinite_nums</code><em> with a sum equal to </em><code>target</code><em>.</em> If there is no such subarray return <code>-1</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3], target = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> In this example infinite_nums = [1,2,3,1,2,3,1,2,...].
The subarray in the range [1,2], has the sum equal to target = 5 and length = 2.
It can be proven that 2 is the shortest length of a subarray with sum equal to target = 5.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,1,1,2,3], target = 4
<strong>Output:</strong> 2
<strong>Explanation:</strong> In this example infinite_nums = [1,1,1,2,3,1,1,1,2,3,1,1,...].
The subarray in the range [4,5], has the sum equal to target = 4 and length = 2.
It can be proven that 2 is the shortest length of a subarray with sum equal to target = 4.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,4,6,8], target = 3
<strong>Output:</strong> -1
<strong>Explanation:</strong> In this example infinite_nums = [2,4,6,8,2,4,6,8,...].
It can be proven that there is no subarray with sum equal to target = 3.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= target &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution(object):
    
    def findMinSubArray(self, nums, target):
        l = 0
        r = 0
        
        arrLen = len(nums)
        ansFound = False
        curSum = 0
        
        while(r<len(nums)):
            curSum += nums[r]
            while(curSum > target):
                curSum -= nums[l]
                l+=1
            if(curSum == target):
                arrLen = min(arrLen, r - l + 1)
                ansFound = True
            r+=1
        if(curSum == target):
            arrLen = min(arrLen, (r-1) - l + 1)
            ansFound = True
        
        if ansFound:
            return arrLen 
        else:
            return -1
            
        
        
    def minSizeSubarray(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: int
        """
        resultLen = 0
        totalSum = sum(nums)
        
        # print("arrSum# ",totalSum)
        
        mul = 0
        if(target >= totalSum):
            mul = target//totalSum
            target = target % totalSum
        
       
        resultLen = mul * len(nums)
        ret = 0
        if target>0:
            tmpNums = nums + nums
            ret = self.findMinSubArray(tmpNums, target)
            if(ret < 0):
                return -1
        
        # print("partial#1 ",resultLen)
        # print("partial#2 ",ret)
        return ret + resultLen
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