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"question": {
"questionId": "1398",
"questionFrontendId": "1269",
"categoryTitle": "Algorithms",
"boundTopicId": 45599,
"title": "Number of Ways to Stay in the Same Place After Some Steps",
"titleSlug": "number-of-ways-to-stay-in-the-same-place-after-some-steps",
"content": "<p>You have a pointer at index <code>0</code> in an array of size <code>arrLen</code>. At each step, you can move 1 position to the left, 1 position to the right in the array, or stay in the same place (The pointer should not be placed outside the array at any time).</p>\n\n<p>Given two integers <code>steps</code> and <code>arrLen</code>, return the number of ways such that your pointer is still at index <code>0</code> after <strong>exactly</strong> <code>steps</code> steps. Since the answer may be too large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> steps = 3, arrLen = 2\n<strong>Output:</strong> 4\n<strong>Explanation: </strong>There are 4 differents ways to stay at index 0 after 3 steps.\nRight, Left, Stay\nStay, Right, Left\nRight, Stay, Left\nStay, Stay, Stay\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> steps = 2, arrLen = 4\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> There are 2 differents ways to stay at index 0 after 2 steps\nRight, Left\nStay, Stay\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> steps = 4, arrLen = 2\n<strong>Output:</strong> 8\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= steps &lt;= 500</code></li>\n\t<li><code>1 &lt;= arrLen &lt;= 10<sup>6</sup></code></li>\n</ul>\n",
"translatedTitle": "停在原地的方案数",
"translatedContent": "<p>有一个长度为 <code>arrLen</code> 的数组,开始有一个指针在索引 <code>0</code> 处。</p>\n\n<p>每一步操作中,你可以将指针向左或向右移动 1 步,或者停在原地(指针不能被移动到数组范围外)。</p>\n\n<p>给你两个整数 <code>steps</code> 和 <code>arrLen</code> ,请你计算并返回:在恰好执行 <code>steps</code> 次操作以后,指针仍然指向索引 <code>0</code> 处的方案数。</p>\n\n<p>由于答案可能会很大,请返回方案数 <strong>模</strong> <code>10^9 + 7</code> 后的结果。</p>\n\n<p> </p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<strong>输入:</strong>steps = 3, arrLen = 2\n<strong>输出:</strong>4\n<strong>解释:</strong>3 步后,总共有 4 种不同的方法可以停在索引 0 处。\n向右向左不动\n不动向右向左\n向右不动向左\n不动不动不动\n</pre>\n\n<p><strong>示例  2</strong></p>\n\n<pre>\n<strong>输入:</strong>steps = 2, arrLen = 4\n<strong>输出:</strong>2\n<strong>解释:</strong>2 步后,总共有 2 种不同的方法可以停在索引 0 处。\n向右向左\n不动不动\n</pre>\n\n<p><strong>示例 3</strong></p>\n\n<pre>\n<strong>输入:</strong>steps = 4, arrLen = 2\n<strong>输出:</strong>8\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 <= steps <= 500</code></li>\n\t<li><code>1 <= arrLen <= 10<sup>6</sup></code></li>\n</ul>\n",
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"hints": [
"Try with Dynamic programming, dp(pos,steps): number of ways to back pos = 0 using exactly \"steps\" moves.",
"Notice that the computational complexity does not depend of \"arrlen\"."
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