find-occurrences-of-an-element-in-an-array

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Algorithms

You are given an integer array <code>nums</code>, an integer array <code>queries</code>, and an integer <code>x</code>.

For each <code>queries[i]</code>, you need to find the index of the <code>queries[i]th</code> occurrence of <code>x</code> in the <code>nums</code> array. If there are fewer than <code>queries[i]</code> occurrences of <code>x</code>, the answer should be -1 for that query.

Return an integer array <code>answer</code> containing the answers to all queries.

&nbsp;
Example 1:

<div class="example-block">
Input: nums = [1,3,1,7], queries = [1,3,2,4], x = 1

Output: [0,-1,2,-1]

Explanation:

<ul>
	<li>For the 1st query, the first occurrence of 1 is at index 0.</li>
	<li>For the 2nd query, there are only two occurrences of 1 in <code>nums</code>, so the answer is -1.</li>
	<li>For the 3rd query, the second occurrence of 1 is at index 2.</li>
	<li>For the 4th query, there are only two occurrences of 1 in <code>nums</code>, so the answer is -1.</li>
</ul>
</div>

Example 2:

<div class="example-block">
Input: nums = [1,2,3], queries = [10], x = 5

Output: [-1]

Explanation:

<ul>
	<li>For the 1st query, 5 doesn&#39;t exist in <code>nums</code>, so the answer is -1.</li>
</ul>
</div>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length, queries.length &lt;= 105</code></li>
	<li><code>1 &lt;= queries[i] &lt;= 105</code></li>
	<li><code>1 &lt;= nums[i], x &lt;= 104</code></li>
</ul>

occurrencesOfElement

Find Occurrences of an Element in an Array

Array

Hash Table

给你一个整数数组&nbsp;<code>nums</code>&nbsp;，一个整数数组&nbsp;<code>queries</code>&nbsp;和一个整数&nbsp;<code>x</code>&nbsp;。

对于每个查询&nbsp;<code>queries[i]</code>&nbsp;，你需要找到&nbsp;<code>nums</code>&nbsp;中第&nbsp;<code>queries[i]</code>&nbsp;个&nbsp;<code>x</code>&nbsp;的位置，并返回它的下标。如果数组中&nbsp;<code>x</code>&nbsp;的出现次数少于&nbsp;<code>queries[i]</code>&nbsp;，该查询的答案为 -1 。

请你返回一个整数数组&nbsp;<code>answer</code>&nbsp;，包含所有查询的答案。

&nbsp;

示例 1：

<div class="example-block">
输入：nums = [1,3,1,7], queries = [1,3,2,4], x = 1

输出：[0,-1,2,-1]

解释：

<ul>
	<li>第 1 个查询，第一个 1 出现在下标 0 处。</li>
	<li>第 2 个查询，<code>nums</code>&nbsp;中只有两个 1 ，所以答案为 -1 。</li>
	<li>第 3 个查询，第二个 1 出现在下标 2 处。</li>
	<li>第 4 个查询，<code>nums</code>&nbsp;中只有两个 1 ，所以答案为 -1 。</li>
</ul>
</div>

示例 2：

<div class="example-block">
输入：nums = [1,2,3], queries = [10], x = 5

输出：[-1]

解释：

<ul>
	<li>第 1 个查询，<code>nums</code>&nbsp;中没有 5 ，所以答案为 -1 。</li>
</ul>
</div>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length, queries.length &lt;= 105</code></li>
	<li><code>1 &lt;= queries[i] &lt;= 105</code></li>
	<li><code>1 &lt;= nums[i], x &lt;= 104</code></li>
</ul>