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            "title": "Divide a String Into Groups of Size k",
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            "content": "<p>A string <code>s</code> can be partitioned into groups of size <code>k</code> using the following procedure:</p>\n\n<ul>\n\t<li>The first group consists of the first <code>k</code> characters of the string, the second group consists of the next <code>k</code> characters of the string, and so on. Each character can be a part of <strong>exactly one</strong> group.</li>\n\t<li>For the last group, if the string <strong>does not</strong> have <code>k</code> characters remaining, a character <code>fill</code> is used to complete the group.</li>\n</ul>\n\n<p>Note that the partition is done so that after removing the <code>fill</code> character from the last group (if it exists) and concatenating all the groups in order, the resultant string should be <code>s</code>.</p>\n\n<p>Given the string <code>s</code>, the size of each group <code>k</code> and the character <code>fill</code>, return <em>a string array denoting the <strong>composition of every group</strong> </em><code>s</code><em> has been divided into, using the above procedure</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;abcdefghi&quot;, k = 3, fill = &quot;x&quot;\n<strong>Output:</strong> [&quot;abc&quot;,&quot;def&quot;,&quot;ghi&quot;]\n<strong>Explanation:</strong>\nThe first 3 characters &quot;abc&quot; form the first group.\nThe next 3 characters &quot;def&quot; form the second group.\nThe last 3 characters &quot;ghi&quot; form the third group.\nSince all groups can be completely filled by characters from the string, we do not need to use fill.\nThus, the groups formed are &quot;abc&quot;, &quot;def&quot;, and &quot;ghi&quot;.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = &quot;abcdefghij&quot;, k = 3, fill = &quot;x&quot;\n<strong>Output:</strong> [&quot;abc&quot;,&quot;def&quot;,&quot;ghi&quot;,&quot;jxx&quot;]\n<strong>Explanation:</strong>\nSimilar to the previous example, we are forming the first three groups &quot;abc&quot;, &quot;def&quot;, and &quot;ghi&quot;.\nFor the last group, we can only use the character &#39;j&#39; from the string. To complete this group, we add &#39;x&#39; twice.\nThus, the 4 groups formed are &quot;abc&quot;, &quot;def&quot;, &quot;ghi&quot;, and &quot;jxx&quot;.\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 lowercase English letters only.</li>\n\t<li><code>1 &lt;= k &lt;= 100</code></li>\n\t<li><code>fill</code> is a lowercase English letter.</li>\n</ul>\n",
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                    "code": "class Solution(object):\n    def divideString(self, s, k, fill):\n        \"\"\"\n        :type s: str\n        :type k: int\n        :type fill: str\n        :rtype: List[str]\n        \"\"\"\n        ",
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                    "code": "public class Solution {\n    public string[] DivideString(string s, int k, char fill) {\n        \n    }\n}",
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                    "code": "/**\n * @param {string} s\n * @param {number} k\n * @param {character} fill\n * @return {string[]}\n */\nvar divideString = function(s, k, fill) {\n    \n};",
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                    "code": "# @param {String} s\n# @param {Integer} k\n# @param {Character} fill\n# @return {String[]}\ndef divide_string(s, k, fill)\n    \nend",
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                    "code": "class Solution {\n    func divideString(_ s: String, _ k: Int, _ fill: Character) -> [String] {\n        \n    }\n}",
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                "Using the length of the string and k, can you count the number of groups the string can be divided into?",
                "Try completing each group using characters from the string. If there aren’t enough characters for the last group, use the fill character to complete the group."
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