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# 2875. Minimum Size Subarray in Infinite Array

## Problem Statement

<br>

You are given a **0-indexed** array `nums` and an integer `target`.

A **0-indexed** array `infinite_nums` is generated by infinitely appending the elements of `nums` to itself.

Return *the length of the **shortest** subarray of the array* `infinite_nums` *with a sum equal to* `target`*.* If there is no such subarray return `-1`.

&#x20;

**Example 1:**

<pre><code><strong>Input: nums = [1,2,3], target = 5
</strong><strong>Output: 2
</strong><strong>Explanation: In this example infinite_nums = [1,2,3,1,2,3,1,2,...].
</strong>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.
</code></pre>

**Example 2:**

<pre><code><strong>Input: nums = [1,1,1,2,3], target = 4
</strong><strong>Output: 2
</strong><strong>Explanation: In this example infinite_nums = [1,1,1,2,3,1,1,1,2,3,1,1,...].
</strong>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.
</code></pre>

**Example 3:**

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

&#x20;

**Constraints:**

* `1 <= nums.length <= 105`
* `1 <= nums[i] <= 105`
* `1 <= target <= 109`

<br>

## Intuition

```
Approach:

We use normal Sliding window as it would for finite array.
Now the catch is,
    if the sum of array < target
No need for extending array

But if the array_sum <= target
    Observation- We cannot have another array of less lenght of sum than
                    actual array, Without appending 

    Hence we just have to find that extra thing in the array,
Target is also modded appropriately

Like,

1 1 2 3 1   Target = 9

1 1 2 3 1 | 1 1 2 3 1
We just have to find that extra 1
9 mod Sum_arr(8)

```

### Links

<https://leetcode.com/problems/minimum-size-subarray-in-infinite-array/description/>

### Video Links

### Approach 1:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    int minSizeSubarray(vector<int>& nums, int target) {
        vector<int>temp=nums;
        long long sumNums=0;
        
        for(int i=0;i<nums.size();i++)
            temp.push_back(nums[i]);

        for(int i=0;i<nums.size();i++)
            sumNums+=nums[i];
        
        int q=target/sumNums;
        int rem=target%sumNums;
        if(!rem){
            return q*nums.size();
        }
        target=rem;
        int ans=1e9, j=0;
        long long sum=0;
        
        for(int i=0;i<temp.size();i++){
            sum+=temp[i];
            while(sum>target){
                sum-=temp[j++];
            }
            if(sum==target){
                ans=min(ans,i-j+1);
            }
        
        }
        if(ans!=1e9){
            return q*nums.size()+ans;
        }
        return -1;
    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###