Class Solution
java.lang.Object
g1501_1600.s1590_make_sum_divisible_by_p.Solution
1590 - Make Sum Divisible by P.<p>Medium</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’s impossible</em>.</p>
<p>A <strong>subarray</strong> is defined as a contiguous block of elements in the array.</p>
<p><strong>Example 1:</strong></p>
<p><strong>Input:</strong> nums = [3,1,4,2], p = 6</p>
<p><strong>Output:</strong> 1</p>
<p><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.</p>
<p><strong>Example 2:</strong></p>
<p><strong>Input:</strong> nums = [6,3,5,2], p = 9</p>
<p><strong>Output:</strong> 2</p>
<p><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.</p>
<p><strong>Example 3:</strong></p>
<p><strong>Input:</strong> nums = [1,2,3], p = 3</p>
<p><strong>Output:</strong> 0</p>
<p><strong>Explanation:</strong> Here the sum is 6. which is already divisible by 3. Thus we do not need to remove anything.</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>
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Solution
public Solution()
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Method Details
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minSubarray
public int minSubarray(int[] nums, int p)
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