# 394. Decode String

## Description

Given an encoded string, return its decoded string.

The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer.

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, k. For example, there will not be input like 3a or 2[4].

The test cases are generated so that the length of the output will never exceed 105.

Example 1:

Input: s = "3[a]2[bc]"
Output: "aaabcbc"


Example 2:

Input: s = "3[a2[c]]"
Output: "accaccacc"


Example 3:

Input: s = "2[abc]3[cd]ef"
Output: "abcabccdcdcdef"


Constraints:

• 1 <= s.length <= 30
• s consists of lowercase English letters, digits, and square brackets '[]'.
• s is guaranteed to be a valid input.
• All the integers in s are in the range [1, 300].

## Solutions

A recursive approach or using a stack are common solutions to this problem. I will provide a solution using a stack here:

1. A stack is used to keep track of the strings and numbers.
2. When a digit is encountered, it is accumulated in current_num.
3. On encountering ‘[’, the current context (current string and number) is saved onto the stack, and the current string and number are reset for a new context.
4. On encountering ‘]’, a previous context is popped from the stack, and the current string is updated by repeating it according to the popped number, then concatenating with the popped string.
5. If a non-digit, non-bracket character is encountered, it is added to the current string.
6. Finally, the current_string holds the decoded string.

### Walk through the example s = "3[a2[bc]]"

1. Initialization: num_stk = [], str_stk = [], num = 0, res = ''.

2. Parsing s:
• '3': num becomes 3.
• '[': Push 3 into num_stk and '' into str_stk. Reset num and res.
• 'a': res becomes 'a'.
• '2': num becomes 2.
• '[': Push 2 into num_stk and 'a' into str_stk. Reset num and res.
• 'b' and 'c': res becomes 'bc'.
• ']': Pop 2 from num_stk and 'a' from str_stk. Set res to 'a' + 'bc' * 2, which is 'abcbc'.
• ']' again: Pop 3 from num_stk and '' (initial value) from str_stk. Set res to '' + 'abcbc' * 3, which is 'abcbcabcbcabcbc'.
3. Completion: No more characters in s. res = 'abcbcabcbcabcbc' is the final result.
• class Solution {
public String decodeString(String s) {
Deque<Integer> s1 = new ArrayDeque<>();
Deque<String> s2 = new ArrayDeque<>();
int num = 0;
String res = "";
for (char c : s.toCharArray()) {
if ('0' <= c && c <= '9') {
num = num * 10 + c - '0';
} else if (c == '[') {
s1.push(num);
s2.push(res);
num = 0;
res = "";
} else if (c == ']') {
StringBuilder t = new StringBuilder();
for (int i = 0, n = s1.pop(); i < n; ++i) {
t.append(res);
}
res = s2.pop() + t.toString();
} else {
res += String.valueOf(c);
}
}
return res;
}
}

• class Solution:
def decodeString(self, s: str) -> str:
num_stk, str_stk = [], []
num, res = 0, ''
for c in s:
if c.isdigit():
num = num * 10 + int(c)
elif c == '[':
num_stk.append(num)
str_stk.append(res) # initial, res being pushed is empty str ''
num, res = 0, ''
elif c == ']':
res = str_stk.pop() + res * num_stk.pop()
else:
res += c
return res


• function decodeString(s: string): string {
let ans = '';
let stack = [];
let count = 0; // repeatCount
for (let cur of s) {
if (/[0-9]/.test(cur)) {
count = count * 10 + Number(cur);
} else if (/[a-z]/.test(cur)) {
ans += cur;
} else if ('[' == cur) {
stack.push([ans, count]);
// reset
ans = '';
count = 0;
} else {
// match ']'
let [pre, count] = stack.pop();
ans = pre + ans.repeat(count);
}
}
return ans;
}