Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 17.8 MB (63.40%)

Author: jimit105

2260-divide-a-string-into-groups-of-size-k/README.md

@@ -1,7 +1,7 @@
 <p>A string <code>s</code> can be partitioned into groups of size <code>k</code> using the following procedure:</p>
 
 <ul>
-	<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>
+	<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 element can be a part of <strong>exactly one</strong> group.</li>
 	<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>
 </ul>
 

2260-divide-a-string-into-groups-of-size-k/solution.py

@@ -1,9 +1,16 @@
+# Time: O(max(n, k))
+# Space: O(1)
+
 class Solution:
     def divideString(self, s: str, k: int, fill: str) -> List[str]:
-        additional_chars = len(s) % k
-        if additional_chars != 0:
-            additional_chars = k - additional_chars
-        s += fill * additional_chars
-          
-        return [s[i: i+k] for i in range(0, len(s), k)]
-        
+        res = []
+        n = len(s)
+        curr = 0
+
+        while curr < n:
+            res.append(s[curr : curr + k])
+            curr += k
+
+        res[-1] += fill * (k - len(res[-1]))
+        return res
+