leetcode-problemset/leetcode-cn/originData/number-of-changing-keys.json

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            "questionId": "3312",
            "questionFrontendId": "3019",
            "categoryTitle": "Algorithms",
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            "title": "Number of Changing Keys",
            "titleSlug": "number-of-changing-keys",
            "content": "<p>You are given a <strong>0-indexed </strong>string <code>s</code> typed by a user. Changing a key is defined as using a key different from the last used key. For example, <code>s = &quot;ab&quot;</code> has a change of a key while <code>s = &quot;bBBb&quot;</code> does not have any.</p>\n\n<p>Return <em>the number of times the user had to change the key. </em></p>\n\n<p><strong>Note: </strong>Modifiers like <code>shift</code> or <code>caps lock</code> won&#39;t be counted in changing the key that is if a user typed the letter <code>&#39;a&#39;</code> and then the letter <code>&#39;A&#39;</code> then it will not be considered as a changing of key.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;aAbBcC&quot;\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> \nFrom s[0] = &#39;a&#39; to s[1] = &#39;A&#39;, there is no change of key as caps lock or shift is not counted.\nFrom s[1] = &#39;A&#39; to s[2] = &#39;b&#39;, there is a change of key.\nFrom s[2] = &#39;b&#39; to s[3] = &#39;B&#39;, there is no change of key as caps lock or shift is not counted.\nFrom s[3] = &#39;B&#39; to s[4] = &#39;c&#39;, there is a change of key.\nFrom s[4] = &#39;c&#39; to s[5] = &#39;C&#39;, there is no change of key as caps lock or shift is not counted.\n\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;AaAaAaaA&quot;\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> There is no change of key since only the letters &#39;a&#39; and &#39;A&#39; are<!-- notionvc: 8849fe75-f31e-41dc-a2e0-b7d33d8427d2 --> pressed which does not require change of key.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 100</code></li>\n\t<li><code>s</code> consists of only upper case and lower case English letters.</li>\n</ul>\n",
            "translatedTitle": "按键变更的次数",
            "translatedContent": "<p>给你一个下标从<strong> 0</strong> 开始的字符串 <code>s</code> ，该字符串由用户输入。按键变更的定义是：使用与上次使用的按键不同的键。例如 <code>s = \"ab\"</code> 表示按键变更一次，而 <code>s = \"bBBb\"</code> 不存在按键变更。</p>\n\n<p>返回用户输入过程中按键变更的次数。</p>\n\n<p><strong>注意：</strong><code>shift</code> 或 <code>caps lock</code> 等修饰键不计入按键变更，也就是说，如果用户先输入字母 <code>'a'</code> 然后输入字母 <code>'A'</code> ，不算作按键变更。</p>\n\n<p>&nbsp;</p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"aAbBcC\"\n<strong>输出：</strong>2\n<strong>解释：</strong> \n从 s[0] = 'a' 到 s[1] = 'A'，不存在按键变更，因为不计入 caps lock 或 shift 。\n从 s[1] = 'A' 到 s[2] = 'b'，按键变更。\n从 s[2] = 'b' 到 s[3] = 'B'，不存在按键变更，因为不计入 caps lock 或 shift 。\n从 s[3] = 'B' 到 s[4] = 'c'，按键变更。\n从 s[4] = 'c' 到 s[5] = 'C'，不存在按键变更，因为不计入 caps lock 或 shift 。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>s = \"AaAaAaaA\"\n<strong>输出：</strong>0\n<strong>解释：</strong> 不存在按键变更，因为这个过程中只按下字母 'a' 和 'A' ，不需要进行按键变更。<!-- notionvc: 8849fe75-f31e-41dc-a2e0-b7d33d8427d2 -->\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= s.length &lt;= 100</code></li>\n\t<li><code>s</code> 仅由英文大写字母和小写字母组成。</li>\n</ul>\n",
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                "Change all the characters to lowercase and then return the number of indices where the character does not match with the last index character."
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