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"titleSlug": "range-sum-query-mutable",
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"content": "<p>Given an integer array <code>nums</code>, handle multiple queries of the following types:</p>\n\n<ol>\n\t<li><strong>Update</strong> the value of an element in <code>nums</code>.</li>\n\t<li>Calculate the <strong>sum</strong> of the elements of <code>nums</code> between indices <code>left</code> and <code>right</code> <strong>inclusive</strong> where <code>left <= right</code>.</li>\n</ol>\n\n<p>Implement the <code>NumArray</code> class:</p>\n\n<ul>\n\t<li><code>NumArray(int[] nums)</code> Initializes the object with the integer array <code>nums</code>.</li>\n\t<li><code>void update(int index, int val)</code> <strong>Updates</strong> the value of <code>nums[index]</code> to be <code>val</code>.</li>\n\t<li><code>int sumRange(int left, int right)</code> Returns the <strong>sum</strong> of the elements of <code>nums</code> between indices <code>left</code> and <code>right</code> <strong>inclusive</strong> (i.e. <code>nums[left] + nums[left + 1] + ... + nums[right]</code>).</li>\n</ul>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input</strong>\n["NumArray", "sumRange", "update", "sumRange"]\n[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]\n<strong>Output</strong>\n[null, 9, null, 8]\n\n<strong>Explanation</strong>\nNumArray numArray = new NumArray([1, 3, 5]);\nnumArray.sumRange(0, 2); // return 1 + 3 + 5 = 9\nnumArray.update(1, 2); // nums = [1, 2, 5]\nnumArray.sumRange(0, 2); // return 1 + 2 + 5 = 8\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 3 * 10<sup>4</sup></code></li>\n\t<li><code>-100 <= nums[i] <= 100</code></li>\n\t<li><code>0 <= index < nums.length</code></li>\n\t<li><code>-100 <= val <= 100</code></li>\n\t<li><code>0 <= left <= right < nums.length</code></li>\n\t<li>At most <code>3 * 10<sup>4</sup></code> calls will be made to <code>update</code> and <code>sumRange</code>.</li>\n</ul>\n",
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"translatedTitle": "区域和检索 - 数组可修改",
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"translatedContent": "<p>给你一个数组 <code>nums</code> ,请你完成两类查询。</p>\n\n<ol>\n\t<li>其中一类查询要求 <strong>更新</strong> 数组 <code>nums</code> 下标对应的值</li>\n\t<li>另一类查询要求返回数组 <code>nums</code> 中索引 <code>left</code> 和索引 <code>right</code> 之间( <strong>包含 </strong>)的nums元素的 <strong>和</strong> ,其中 <code>left <= right</code></li>\n</ol>\n\n<p>实现 <code>NumArray</code> 类:</p>\n\n<ul>\n\t<li><code>NumArray(int[] nums)</code> 用整数数组 <code>nums</code> 初始化对象</li>\n\t<li><code>void update(int index, int val)</code> 将 <code>nums[index]</code> 的值 <strong>更新</strong> 为 <code>val</code></li>\n\t<li><code>int sumRange(int left, int right)</code> 返回数组 <code>nums</code> 中索引 <code>left</code> 和索引 <code>right</code> 之间( <strong>包含 </strong>)的nums元素的 <strong>和</strong> (即,<code>nums[left] + nums[left + 1], ..., nums[right]</code>)</li>\n</ul>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入</strong>:\n[\"NumArray\", \"sumRange\", \"update\", \"sumRange\"]\n[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]\n<strong>输出</strong>:\n[null, 9, null, 8]\n\n<strong>解释</strong>:\nNumArray numArray = new NumArray([1, 3, 5]);\nnumArray.sumRange(0, 2); // 返回 1 + 3 + 5 = 9\nnumArray.update(1, 2); // nums = [1,2,5]\nnumArray.sumRange(0, 2); // 返回 1 + 2 + 5 = 8\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 3 * 10<sup>4</sup></code></li>\n\t<li><code>-100 <= nums[i] <= 100</code></li>\n\t<li><code>0 <= index < nums.length</code></li>\n\t<li><code>-100 <= val <= 100</code></li>\n\t<li><code>0 <= left <= right < nums.length</code></li>\n\t<li>调用 <code>pdate</code> 和 <code>sumRange</code> 方法次数不大于 <code>3 * 10<sup>4</sup></code> </li>\n</ul>\n",
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"translatedContent": "<p>给你一个数组 <code>nums</code> ,请你完成两类查询。</p>\n\n<ol>\n\t<li>其中一类查询要求 <strong>更新</strong> 数组 <code>nums</code> 下标对应的值</li>\n\t<li>另一类查询要求返回数组 <code>nums</code> 中索引 <code>left</code> 和索引 <code>right</code> 之间( <strong>包含 </strong>)的nums元素的 <strong>和</strong> ,其中 <code>left <= right</code></li>\n</ol>\n\n<p>实现 <code>NumArray</code> 类:</p>\n\n<ul>\n\t<li><code>NumArray(int[] nums)</code> 用整数数组 <code>nums</code> 初始化对象</li>\n\t<li><code>void update(int index, int val)</code> 将 <code>nums[index]</code> 的值 <strong>更新</strong> 为 <code>val</code></li>\n\t<li><code>int sumRange(int left, int right)</code> 返回数组 <code>nums</code> 中索引 <code>left</code> 和索引 <code>right</code> 之间( <strong>包含 </strong>)的nums元素的 <strong>和</strong> (即,<code>nums[left] + nums[left + 1], ..., nums[right]</code>)</li>\n</ul>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入</strong>:\n[\"NumArray\", \"sumRange\", \"update\", \"sumRange\"]\n[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]\n<strong>输出</strong>:\n[null, 9, null, 8]\n\n<strong>解释</strong>:\nNumArray numArray = new NumArray([1, 3, 5]);\nnumArray.sumRange(0, 2); // 返回 1 + 3 + 5 = 9\nnumArray.update(1, 2); // nums = [1,2,5]\nnumArray.sumRange(0, 2); // 返回 1 + 2 + 5 = 8\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 3 * 10<sup>4</sup></code></li>\n\t<li><code>-100 <= nums[i] <= 100</code></li>\n\t<li><code>0 <= index < nums.length</code></li>\n\t<li><code>-100 <= val <= 100</code></li>\n\t<li><code>0 <= left <= right < nums.length</code></li>\n\t<li>调用 <code>update</code> 和 <code>sumRange</code> 方法次数不大于 <code>3 * 10<sup>4</sup></code> </li>\n</ul>\n",
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"difficulty": "Medium",
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"similarQuestions": "[{\"title\": \"Range Sum Query - Immutable\", \"titleSlug\": \"range-sum-query-immutable\", \"difficulty\": \"Easy\", \"translatedTitle\": \"\\u533a\\u57df\\u548c\\u68c0\\u7d22 - \\u6570\\u7ec4\\u4e0d\\u53ef\\u53d8\"}, {\"title\": \"Range Sum Query 2D - Mutable\", \"titleSlug\": \"range-sum-query-2d-mutable\", \"difficulty\": \"Hard\", \"translatedTitle\": \"\\u4e8c\\u7ef4\\u533a\\u57df\\u548c\\u68c0\\u7d22 - \\u53ef\\u53d8\"}]",
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