<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 k) {
    int n = nums.size();
    long long total_sum = 0;
    for (int num : nums) {
        total_sum += num;
    }
    long long remainder = total_sum % k;
    if (remainder == 0) {
        return 0;
    }
    unordered_map<int, int> mod_index;
    mod_index[0] = -1;
    long long prefix_sum = 0;
    int min_length = n;
    for (int i = 0; i < n; i++) {
        prefix_sum += nums[i];
        int mod = prefix_sum % k;
        int target_mod = (mod - remainder + k) % k;
        if (mod_index.find(target_mod) != mod_index.end()) {
            min_length = min(min_length, i - mod_index[target_mod]);
        }
        mod_index[mod] = i;
    }
    return min_length == n ? -1 : min_length;
}
};
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