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43 lines
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43 lines
2.0 KiB
HTML
<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code> of equal length. Every second, for all indices <code>0 <= i < nums1.length</code>, value of <code>nums1[i]</code> is incremented by <code>nums2[i]</code>. <strong>After</strong> this is done, you can do the following operation:</p>
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<ul>
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<li>Choose an index <code>0 <= i < nums1.length</code> and make <code>nums1[i] = 0</code>.</li>
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</ul>
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<p>You are also given an integer <code>x</code>.</p>
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<p>Return <em>the <strong>minimum</strong> time in which you can make the sum of all elements of </em><code>nums1</code><em> to be<strong> less than or equal</strong> to </em><code>x</code>, <em>or </em><code>-1</code><em> if this is not possible.</em></p>
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<p> </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> nums1 = [1,2,3], nums2 = [1,2,3], x = 4
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<strong>Output:</strong> 3
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<strong>Explanation:</strong>
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For the 1st second, we apply the operation on i = 0. Therefore nums1 = [0,2+2,3+3] = [0,4,6].
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For the 2nd second, we apply the operation on i = 1. Therefore nums1 = [0+1,0,6+3] = [1,0,9].
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For the 3rd second, we apply the operation on i = 2. Therefore nums1 = [1+1,0+2,0] = [2,2,0].
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Now sum of nums1 = 4. It can be shown that these operations are optimal, so we return 3.
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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> nums1 = [1,2,3], nums2 = [3,3,3], x = 4
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<strong>Output:</strong> -1
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<strong>Explanation:</strong> It can be shown that the sum of nums1 will always be greater than x, no matter which operations are performed.
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</pre>
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<p> </p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code><font face="monospace">1 <= nums1.length <= 10<sup>3</sup></font></code></li>
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<li><code>1 <= nums1[i] <= 10<sup>3</sup></code></li>
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<li><code>0 <= nums2[i] <= 10<sup>3</sup></code></li>
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<li><code>nums1.length == nums2.length</code></li>
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<li><code>0 <= x <= 10<sup>6</sup></code></li>
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</ul>
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