Make Sum Divisible by P
<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>
Array; Hash Table; Prefix Sum
use std::collections::HashMap;

impl Solution {
    pub fn min_subarray(nums: Vec<i32>, p: i32) -> i32 {
        let mut k = 0;
        for &x in &nums {
            k = (k + x) % p;
        }
        if k == 0 {
            return 0;
        }

        let mut last = HashMap::new();
        last.insert(0, -1);
        let n = nums.len();
        let mut ans = n as i32;
        let mut cur = 0;

        for i in 0..n {
            cur = (cur + nums[i]) % p;
            let target = (cur - k + p) % p;
            if let Some(&prev_idx) = last.get(&target) {
                ans = ans.min(i as i32 - prev_idx);
            }
            last.insert(cur, i as i32);
        }

        if ans == n as i32 {
            -1
        } else {
            ans
        }
    }
}
