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<h2><a href="https://leetcode.com/problems/make-sum-divisible-by-p">1694. Make Sum Divisible by P</a></h2><h3>Medium</h3><hr><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>
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<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>
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<p>A <strong>subarray</strong> is defined as a contiguous block of elements in the array.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [3,1,4,2], p = 6
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<strong>Output:</strong> 1
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<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.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [6,3,5,2], p = 9
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<strong>Output:</strong> 2
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<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.
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</pre>
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<p><strong class="example">Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,2,3], p = 3
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<strong>Output:</strong> 0
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<strong>Explanation:</strong> Here the sum is 6. which is already divisible by 3. Thus we do not need to remove anything.
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
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<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
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<li><code>1 &lt;= p &lt;= 10<sup>9</sup></code></li>
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</ul>

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