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53 lines
4.0 KiB
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<p>You are given two <strong>0-indexed</strong> integer arrays <code>nums1</code> and <code>nums2</code>, each of length <code>n</code>, and a <strong>1-indexed 2D array</strong> <code>queries</code> where <code>queries[i] = [x<sub>i</sub>, y<sub>i</sub>]</code>.</p>
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<p>For the <code>i<sup>th</sup></code> query, find the <strong>maximum value</strong> of <code>nums1[j] + nums2[j]</code> among all indices <code>j</code> <code>(0 <= j < n)</code>, where <code>nums1[j] >= x<sub>i</sub></code> and <code>nums2[j] >= y<sub>i</sub></code>, or <strong>-1</strong> if there is no <code>j</code> satisfying the constraints.</p>
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<p>Return <em>an array </em><code>answer</code><em> where </em><code>answer[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query.</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 = [4,3,1,2], nums2 = [2,4,9,5], queries = [[4,1],[1,3],[2,5]]
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<strong>Output:</strong> [6,10,7]
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<strong>Explanation:</strong>
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For the 1st query <code node="[object Object]">x<sub>i</sub> = 4</code> and <code node="[object Object]">y<sub>i</sub> = 1</code>, we can select index <code node="[object Object]">j = 0</code> since <code node="[object Object]">nums1[j] >= 4</code> and <code node="[object Object]">nums2[j] >= 1</code>. The sum <code node="[object Object]">nums1[j] + nums2[j]</code> is 6, and we can show that 6 is the maximum we can obtain.
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For the 2nd query <code node="[object Object]">x<sub>i</sub> = 1</code> and <code node="[object Object]">y<sub>i</sub> = 3</code>, we can select index <code node="[object Object]">j = 2</code> since <code node="[object Object]">nums1[j] >= 1</code> and <code node="[object Object]">nums2[j] >= 3</code>. The sum <code node="[object Object]">nums1[j] + nums2[j]</code> is 10, and we can show that 10 is the maximum we can obtain.
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For the 3rd query <code node="[object Object]">x<sub>i</sub> = 2</code> and <code node="[object Object]">y<sub>i</sub> = 5</code>, we can select index <code node="[object Object]">j = 3</code> since <code node="[object Object]">nums1[j] >= 2</code> and <code node="[object Object]">nums2[j] >= 5</code>. The sum <code node="[object Object]">nums1[j] + nums2[j]</code> is 7, and we can show that 7 is the maximum we can obtain.
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Therefore, we return <code node="[object Object]">[6,10,7]</code>.
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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 = [3,2,5], nums2 = [2,3,4], queries = [[4,4],[3,2],[1,1]]
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<strong>Output:</strong> [9,9,9]
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<strong>Explanation:</strong> For this example, we can use index <code node="[object Object]">j = 2</code> for all the queries since it satisfies the constraints for each query.
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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> nums1 = [2,1], nums2 = [2,3], queries = [[3,3]]
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<strong>Output:</strong> [-1]
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<strong>Explanation:</strong> There is one query in this example with <code node="[object Object]">x<sub>i</sub></code> = 3 and <code node="[object Object]">y<sub>i</sub></code> = 3. For every index, j, either nums1[j] < <code node="[object Object]">x<sub>i</sub></code> or nums2[j] < <code node="[object Object]">y<sub>i</sub></code>. Hence, there is no solution.
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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>nums1.length == nums2.length</code> </li>
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<li><code>n == nums1.length </code></li>
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<li><code>1 <= n <= 10<sup>5</sup></code></li>
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<li><code>1 <= nums1[i], nums2[i] <= 10<sup>9</sup> </code></li>
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<li><code>1 <= queries.length <= 10<sup>5</sup></code></li>
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<li><code>queries[i].length == 2</code></li>
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<li><code>x<sub>i</sub> == queries[i][1]</code></li>
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<li><code>y<sub>i</sub> == queries[i][2]</code></li>
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<li><code>1 <= x<sub>i</sub>, y<sub>i</sub> <= 10<sup>9</sup></code></li>
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</ul>
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