LC394. Decode String¶

Problem Description¶

LeetCode Problem 394: 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.

Clarification¶

Go through encoded rule
Always have a valid input?
Only contains english letters, digits, and square brackets []?
So no 3a. Instead, 3[a]

Assumption¶

-

Solution¶

Approach - Stack¶

Uses a stack to handle nested patterns in the encoded string. It loops through the characters of the string,

when encountering open bracket [, push the current string and number onto the stack.
when encountering a closing bracket ], it pops the stack to retrieve the previous string and repeats the current string according to the last number, then concatenates it.
digits are processed to form numbers.
regular characters are appended directly to the current string.

Python

class Solution:
def decodeString(self, s: str) -> str:
    stack = deque()
    curr_string = ""
    curr_num = 0

    for c in s:
        if c == "[":
            stack.append(curr_string)
            stack.append(curr_num)
            curr_string = ""
            curr_num = 0
        elif c == "]":
            num = stack.pop()
            prev_string = stack.pop()
            curr_string = prev_string + curr_string * num
        elif c.isdigit():
            curr_num = curr_num * 10 + int(c)
        else:
            curr_string += c
    return curr_string

Complexity Analysis of Approach 1¶

Time complexity: \(O(n)\)
The algorithm processes each character in the input string exactly once, when \(n\) is the length of the input string. For stack pushing and pop, the time complexity is \(O(1)\). For string concatenation, it concatenates at most \(n\) letters, so the time complexity is within \(O(n)\).
Space complexity: \(O(m)\), where \(m\) is the size of the expanded string, which can be larger than \(n\) if there are many repeated segments.