mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-11-04 19:53:12 +08:00
40 lines
1.6 KiB
HTML
40 lines
1.6 KiB
HTML
<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's impossible</em>.</p>
|
|
|
|
<p>A <strong>subarray</strong> is defined as a contiguous block of elements in the array.</p>
|
|
|
|
<p> </p>
|
|
<p><strong>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>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>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> </p>
|
|
<p><strong>Constraints:</strong></p>
|
|
|
|
<ul>
|
|
<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>
|
|
<li><code>1 <= nums[i] <= 10<sup>9</sup></code></li>
|
|
<li><code>1 <= p <= 10<sup>9</sup></code></li>
|
|
</ul>
|