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+---
+title: Leetcode 394. Decode String
+description: Explanation for Leetcode 394 - Decode String, and its solution in Python.
+date: 2025-01-14
+categories: [Leetcode, Stack, Medium]
+tags: [Leetcode, Python, Study, Stack, Medium]
+math: true
+---
+
+## Problem
+[Leetcode 394 - Decode String](https://leetcode.com/problems/decode-string/description/)
+
+Example:
+```
+Input: s = "3[a]2[bc]"
+Output: "aaabcbc"
+
+Input: s = "3[a2[c]]"
+Output: "accaccacc"
+
+Input: s = "2[abc]3[cd]ef"
+Output: "abcabccdcdcdef"
+```
+
+## Approach
+
+We can use stack to solve this problem while iterating through the string, there can be 4 options, $curr_string$ defines current string, and $k$ defines how many times we must multiply the string in brackets
+- if ch == '['
+    - Since we're now starting with '[', we must append whatever we had (curr_string, k) into stack
+    - initialize curr_string, k again to 0
+- if ch == ']'
+    - Since we've ended the bracket, we must update the curr_string by popping whatever we had in last_string, and adding multiplied value for curr_string
+- if ch is digit (a number)
+    - We must update the value k, since k could have multiple digits, we must update using $k = k * 10 +$ int(ch)
+- if ch is alpha (a character)
+    - We just update the curr_string 
+
+Visualization of the Approach:
+```
+s = 3[a]2[bc]
+
+stack = []
+curr_string = ''
+k = 0
+
+Since ch is digit, update k
+stack = []
+curr_string = ''
+k = k * 10 + int(ch) = 0 + 3 = 3
+
+Since ch is opening bracket, put curr_string, and k into stack
+stack = [('', 3)]
+curr_string = ''
+k = 0
+
+Since ch is alpha, update curr_string
+stack = [('', 3)]
+curr_string = 'a'
+k = 0
+
+Since ch is closing bracket, update curr_string by popping stack
+stack = []
+curr_string = last_string + last_k * curr_string = '' + 3 * 'a' = 'aaa'
+k = 0
+
+Since ch is digit, update k
+stack = []
+curr_string = 'aaa'
+k = k * 10 + int(ch) = 0 + 2 = 2
+
+Since ch is opening bracket, put curr_string, and k into stack
+stack = [('aaa', 2)]
+curr_string = ''
+k = 0
+
+Since ch is alpha, update curr_string
+stack = [('aaa', 2)]
+curr_string = 'b'
+k = 0
+
+Since ch is alpha, update curr_string
+stack = [('aaa', 2)]
+curr_string = 'bc'
+k = 0
+
+Since ch is closing bracket, update curr_string, and pop stack
+stack = []
+curr_string = last_string + last_k * curr_string = 'aaa' + 2 * 'bc' = 'aaabcbc'
+k = 0
+
+return curr_string = 'aaabcbc'
+```
+
+Here is the Python code for the solution:
+```python
+class Solution:
+    def decodeString(self, s: str) -> str:
+        stack = []
+        curr_string = ''
+        k = 0
+
+        for ch in s:
+            # if ch is opening bracket, then we add the curr_string and k to stack and initialize them
+            if ch == '[':
+                stack.append((curr_string, k))
+                curr_string = ''
+                k = 0
+            # if ch is closing bracket, then we pop from stack and add the multiplied string into stack
+            elif ch == ']':
+                last_string, last_k = stack.pop()
+                curr_string = last_string + last_k * curr_string
+            # if ch is digit, then update k
+            elif ch.isdigit():
+                k = k * 10 + int(ch)
+            # if ch is character, then update curr_string
+            else:
+                curr_string += char
+        return curr_string
+```
+## Time Complexity and Space Complexity
+
+Time Complexity: $O(n)$
+
+Space Complexity: $O(n)$
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