leetcode-problemset/leetcode-cn/originData/maximum-number-of-jumps-to-reach-the-last-index.json

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            "questionId": "2855",
            "questionFrontendId": "2770",
            "categoryTitle": "Algorithms",
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            "title": "Maximum Number of Jumps to Reach the Last Index",
            "titleSlug": "maximum-number-of-jumps-to-reach-the-last-index",
            "content": "<p>You are given a <strong>0-indexed</strong> array <code>nums</code> of <code>n</code> integers and an integer <code>target</code>.</p>\n\n<p>You are initially positioned at index <code>0</code>. In one step, you can jump from index <code>i</code> to any index <code>j</code> such that:</p>\n\n<ul>\n\t<li><code>0 &lt;= i &lt; j &lt; n</code></li>\n\t<li><code>-target &lt;= nums[j] - nums[i] &lt;= target</code></li>\n</ul>\n\n<p>Return <em>the <strong>maximum number of jumps</strong> you can make to reach index</em> <code>n - 1</code>.</p>\n\n<p>If there is no way to reach index <code>n - 1</code>, return <code>-1</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,3,6,4,1,2], target = 2\n<strong>Output:</strong> 3\n<strong>Explanation:</strong> To go from index 0 to index n - 1 with the maximum number of jumps, you can perform the following jumping sequence:\n- Jump from index 0 to index 1. \n- Jump from index 1 to index 3.\n- Jump from index 3 to index 5.\nIt can be proven that there is no other jumping sequence that goes from 0 to n - 1 with more than 3 jumps. Hence, the answer is 3. </pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,3,6,4,1,2], target = 3\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> To go from index 0 to index n - 1 with the maximum number of jumps, you can perform the following jumping sequence:\n- Jump from index 0 to index 1.\n- Jump from index 1 to index 2.\n- Jump from index 2 to index 3.\n- Jump from index 3 to index 4.\n- Jump from index 4 to index 5.\nIt can be proven that there is no other jumping sequence that goes from 0 to n - 1 with more than 5 jumps. Hence, the answer is 5. </pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,3,6,4,1,2], target = 0\n<strong>Output:</strong> -1\n<strong>Explanation:</strong> It can be proven that there is no jumping sequence that goes from 0 to n - 1. Hence, the answer is -1. \n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= nums.length == n &lt;= 1000</code></li>\n\t<li><code>-10<sup>9</sup>&nbsp;&lt;= nums[i]&nbsp;&lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= target &lt;= 2 * 10<sup>9</sup></code></li>\n</ul>\n",
            "translatedTitle": "达到末尾下标所需的最大跳跃次数",
            "translatedContent": "<p>给你一个下标从 <strong>0</strong> 开始、由 <code>n</code> 个整数组成的数组 <code>nums</code> 和一个整数 <code>target</code> 。</p>\n\n<p>你的初始位置在下标 <code>0</code> 。在一步操作中，你可以从下标 <code>i</code> 跳跃到任意满足下述条件的下标 <code>j</code> ：</p>\n\n<ul>\n\t<li><code>0 &lt;= i &lt; j &lt; n</code></li>\n\t<li><code>-target &lt;= nums[j] - nums[i] &lt;= target</code></li>\n</ul>\n\n<p>返回到达下标 <code>n - 1</code> 处所需的 <strong>最大跳跃次数</strong> 。</p>\n\n<p>如果无法到达下标 <code>n - 1</code> ，返回 <code>-1</code> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre><strong>输入：</strong>nums = [1,3,6,4,1,2], target = 2\n<strong>输出：</strong>3\n<strong>解释：</strong>要想以最大跳跃次数从下标 0 到下标 n - 1 ，可以按下述跳跃序列执行操作：\n- 从下标 0 跳跃到下标 1 。 \n- 从下标 1 跳跃到下标 3 。 \n- 从下标 3 跳跃到下标 5 。 \n可以证明，从 0 到 n - 1 的所有方案中，不存在比 3 步更长的跳跃序列。因此，答案是 3 。 </pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre><strong>输入：</strong>nums = [1,3,6,4,1,2], target = 3\n<strong>输出：</strong>5\n<strong>解释：</strong>要想以最大跳跃次数从下标 0 到下标 n - 1 ，可以按下述跳跃序列执行操作：\n- 从下标 0 跳跃到下标 1 。 \n- 从下标 1 跳跃到下标 2 。 \n- 从下标 2 跳跃到下标 3 。 \n- 从下标 3 跳跃到下标 4 。 \n- 从下标 4 跳跃到下标 5 。 \n可以证明，从 0 到 n - 1 的所有方案中，不存在比 5 步更长的跳跃序列。因此，答案是 5 。 </pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre><strong>输入：</strong>nums = [1,3,6,4,1,2], target = 0\n<strong>输出：</strong>-1\n<strong>解释：</strong>可以证明不存在从 0 到 n - 1 的跳跃序列。因此，答案是 -1 。 \n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= nums.length == n &lt;= 1000</code></li>\n\t<li><code>-10<sup>9</sup>&nbsp;&lt;= nums[i]&nbsp;&lt;= 10<sup>9</sup></code></li>\n\t<li><code>0 &lt;= target &lt;= 2 * 10<sup>9</sup></code></li>\n</ul>\n",
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                "Define a dynamic programming array dp of size n, where dp[i] represents the maximum number of jumps from index 0 to index i.",
                "For each j iterate over all i < j. Set dp[j] = max(dp[j], dp[i] + 1) if -target <= nums[j] - nums[i] <= target."
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