Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 17.6 MB (67.82%)

Author: jimit105

0394-decode-string/README.md

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+<p>Given an encoded string, return its decoded string.</p>
+
+<p>The encoding rule is: <code>k[encoded_string]</code>, where the <code>encoded_string</code> inside the square brackets is being repeated exactly <code>k</code> times. Note that <code>k</code> is guaranteed to be a positive integer.</p>
+
+<p>You may assume that the input string is always valid; there are no extra white spaces, square brackets are well-formed, etc. Furthermore, you may assume that the original data does not contain any digits and that digits are only for those repeat numbers, <code>k</code>. For example, there will not be input like <code>3a</code> or <code>2[4]</code>.</p>
+
+<p>The test cases are generated so that the length of the output will never exceed <code>10<sup>5</sup></code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;3[a]2[bc]&quot;
+<strong>Output:</strong> &quot;aaabcbc&quot;
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;3[a2[c]]&quot;
+<strong>Output:</strong> &quot;accaccacc&quot;
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;2[abc]3[cd]ef&quot;
+<strong>Output:</strong> &quot;abcabccdcdcdef&quot;
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 30</code></li>
+	<li><code>s</code> consists of lowercase English letters, digits, and square brackets <code>&#39;[]&#39;</code>.</li>
+	<li><code>s</code> is guaranteed to be <strong>a valid</strong> input.</li>
+	<li>All the integers in <code>s</code> are in the range <code>[1, 300]</code>.</li>
+</ul>

0394-decode-string/solution.py

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+# Approach: Stack
+
+# n = length of decoded string, m = number of nested brackets
+# Time: O(n)
+# Space: O(m)
+
+class Solution:
+    def decodeString(self, s: str) -> str:
+        stack = [] # to keep track of nested brackets
+        curr_str = ''
+        curr_num = 0
+
+        for char in s:
+            if char.isdigit():
+                # Build the number
+                curr_num = curr_num * 10 + int(char)
+            elif char == '[':
+                # Push the curr num and str to stacj
+                stack.append((curr_str, curr_num))
+                # Reset current values
+                curr_str = ''
+                curr_num = 0
+            elif char == ']':
+                prev_str, num = stack.pop()
+                curr_str = prev_str + curr_str * num
+            else:
+                # Add characters to the current str
+                curr_str += char
+
+        return curr_str