You are given an array nums of positive integers and an integer k.
In one operation, you can remove the last element of the array and add it to your collection.
\n\nReturn the minimum number of operations needed to collect elements 1, 2, ..., k.
\n
Example 1:
\n\n\nInput: nums = [3,1,5,4,2], k = 2\nOutput: 4\nExplanation: After 4 operations, we collect elements 2, 4, 5, and 1, in this order. Our collection contains elements 1 and 2. Hence, the answer is 4.\n\n\n
Example 2:
\n\n\nInput: nums = [3,1,5,4,2], k = 5\nOutput: 5\nExplanation: After 5 operations, we collect elements 2, 4, 5, 1, and 3, in this order. Our collection contains elements 1 through 5. Hence, the answer is 5.\n\n\n
Example 3:
\n\n\nInput: nums = [3,2,5,3,1], k = 3\nOutput: 4\nExplanation: After 4 operations, we collect elements 1, 3, 5, and 2, in this order. Our collection contains elements 1 through 3. Hence, the answer is 4.\n\n\n
\n
Constraints:
\n\n1 <= nums.length <= 501 <= nums[i] <= nums.length1 <= k <= nums.length1, 2, ..., k.给你一个正整数数组 nums 和一个整数 k 。
一次操作中，你可以将数组的最后一个元素删除，将该元素添加到一个集合中。
\n\n请你返回收集元素 1, 2, ..., k 需要的 最少操作次数 。
\n\n
示例 1：
\n\n\n输入：nums = [3,1,5,4,2], k = 2\n输出：4\n解释：4 次操作后，集合中的元素依次添加了 2 ，4 ，5 和 1 。此时集合中包含元素 1 和 2 ，所以答案为 4 。\n\n\n
示例 2：
\n\n\n输入：nums = [3,1,5,4,2], k = 5\n输出：5\n解释：5 次操作后，集合中的元素依次添加了 2 ，4 ，5 ，1 和 3 。此时集合中包含元素 1 到 5 ，所以答案为 5 。\n\n\n
示例 3：
\n\n\n输入：nums = [3,2,5,3,1], k = 3\n输出：4\n解释：4 次操作后，集合中的元素依次添加了 1 ，3 ，5 和 2 。此时集合中包含元素 1 到 3 ，所以答案为 4 。\n\n\n
\n\n
提示：
\n\n1 <= nums.length <= 501 <= nums[i] <= nums.length1 <= k <= nums.length1, 2, ..., k 。nums[i] is not marked in the occurrence array and nums[i] <= k, mark nums[i].",
"Keep track of how many integers you have marked.",
"Return the current index as soon as the number of marked integers becomes equal to k."
],
"solution": null,
"status": null,
"sampleTestCase": "[3,1,5,4,2]\n2",
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