<p>Given an array of positive integers <code>nums</code>, remove the <strong>smallest</strong> subarray (possibly <strong>empty</strong>) such that the <strong>sum</strong> of the remaining elements is divisible by <code>p</code>. It is <strong>not</strong> allowed to remove the whole array.</p>

<p>Return <em>the length of the smallest subarray that you need to remove, or </em><code>-1</code><em> if it&#39;s impossible</em>.</p>

<p>A <strong>subarray</strong> is defined as a contiguous block of elements in the array.</p>

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

<pre>
<strong>Input:</strong> nums = [3,1,4,2], p = 6
<strong>Output:</strong> 1
<strong>Explanation:</strong> The sum of the elements in nums is 10, which is not divisible by 6. We can remove the subarray [4], and the sum of the remaining elements is 6, which is divisible by 6.
</pre>

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

<pre>
<strong>Input:</strong> nums = [6,3,5,2], p = 9
<strong>Output:</strong> 2
<strong>Explanation:</strong> We cannot remove a single element to get a sum divisible by 9. The best way is to remove the subarray [5,2], leaving us with [6,3] with sum 9.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3], p = 3
<strong>Output:</strong> 0
<strong>Explanation:</strong> Here the sum is 6. which is already divisible by 3. Thus we do not need to remove anything.
</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>9</sup></code></li>
	<li><code>1 &lt;= p &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
public:
    int minSubarray(vector<int>& nums, int p) {
        int n = nums.size();
        vector<int>pref(n+1,0),suff(n+1,0);
        pref[0] = nums[0] % p;
        for(int i = 1 ; i < n ; ++i)
        {
            pref[i] = (pref[i-1] + nums[i]) % p;
        }
        suff[n-1] = (nums[n-1]) % p;
        for(int i = n-2 ; i > -1 ; --i)
        {
            suff[i] = (suff[i+1] + nums[i]) % p;
        }

        unordered_map<int,int>mp;
        int ans = n;
        
        for(int i = 0 ; i < n ; ++i)
        {
            int target = p - (suff[i] == 0 ? p : suff[i]);
            if(mp.find(target) != mp.end())
            {
                ans = min(ans,i - mp[target] - 1);
            }
            mp[pref[i]] = i;
            
            if(!pref[i]) ans = min(ans, n - i - 1);

            if(!suff[i]) ans = min(ans, i);
        }

        return ans == n ? -1 : ans;
    }
};


// 3  1  4  2           p = 6
// 3  4  8  10  pref
// 10  7  6  2   suf


//  6   3  5   2           p = 9
//  6   9  14  16  pref
//  17  10  7  2   suf
74900