Given an array of integers arr, you are initially positioned at the first index of the array.
In one step you can jump from index i to index:
i + 1 where: i + 1 < arr.length.i - 1 where: i - 1 >= 0.j where: arr[i] == arr[j] and i != j.Return the minimum number of steps to reach the last index of the array.
\n\nNotice that you can not jump outside of the array at any time.
\n\n\n
Example 1:
\n\n\nInput: arr = [100,-23,-23,404,100,23,23,23,3,404]\nOutput: 3\nExplanation: You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.\n\n\n
Example 2:
\n\n\nInput: arr = [7]\nOutput: 0\nExplanation: Start index is the last index. You do not need to jump.\n\n\n
Example 3:
\n\n\nInput: arr = [7,6,9,6,9,6,9,7]\nOutput: 1\nExplanation: You can jump directly from index 0 to index 7 which is last index of the array.\n\n\n
\n
Constraints:
\n\n1 <= arr.length <= 5 * 104-108 <= arr[i] <= 108给你一个整数数组 arr ，你一开始在数组的第一个元素处（下标为 0）。
每一步，你可以从下标 i 跳到下标 i + 1 、i - 1 或者 j ：
i + 1 需满足：i + 1 < arr.lengthi - 1 需满足：i - 1 >= 0j 需满足：arr[i] == arr[j] 且 i != j请你返回到达数组最后一个元素的下标处所需的 最少操作次数 。
\n\n注意：任何时候你都不能跳到数组外面。
\n\n\n\n
示例 1：
\n\n\n输入：arr = [100,-23,-23,404,100,23,23,23,3,404]\n输出：3\n解释：那你需要跳跃 3 次，下标依次为 0 --> 4 --> 3 --> 9 。下标 9 为数组的最后一个元素的下标。\n\n\n
示例 2：
\n\n\n输入：arr = [7]\n输出：0\n解释：一开始就在最后一个元素处，所以你不需要跳跃。\n\n\n
示例 3：
\n\n\n输入：arr = [7,6,9,6,9,6,9,7]\n输出：1\n解释：你可以直接从下标 0 处跳到下标 7 处，也就是数组的最后一个元素处。\n\n\n
\n\n
提示：
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"book": null,
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"dailyRecordStatus": null,
"editorType": "CKEDITOR",
"ugcQuestionId": null,
"style": "LEETCODE",
"exampleTestcases": "[100,-23,-23,404,100,23,23,23,3,404]\n[7]\n[7,6,9,6,9,6,9,7]",
"__typename": "QuestionNode"
}
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