Decode String

 

1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

  • We are given a string that represents an encoded string.
  • The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets should be repeated k times.
  • We need to decode the given string and return the decoded string.

2. Analyze the problem: To solve this problem, we can use a stack data structure. We iterate through the characters of the input string and perform the following operations:

  • If we encounter a digit, we convert it to an integer and push it onto the stack.
  • If we encounter an opening bracket [, we push an empty string onto the stack to start building the encoded string.
  • If we encounter a closing bracket ], we start decoding the substring enclosed by the brackets by repeating it the appropriate number of times. We pop the top element from the stack, which is the number of repetitions, and append the decoded substring to the string below it on the stack.
  • If we encounter any other character, we append it to the current string on the stack.

3. Design an algorithm: Here is the algorithm to solve the problem:

  1. Initialize an empty stack.
  2. Iterate through each character ch in the input string:
    • If ch is a digit:
      • Convert ch to an integer num.
      • Push num onto the stack.
    • If ch is an opening bracket [:
      • Push an empty string "" onto the stack.
    • If ch is a closing bracket ]:
      • Pop the top element top from the stack, which represents the current encoded string.
      • Pop the element below it, which represents the number of repetitions num_repetitions.
      • Decode top by repeating it num_repetitions times.
      • Append the decoded string to the string below it on the stack.
    • If ch is any other character:
      • Append ch to the current string on the stack.
  3. The decoded string is the top element on the stack. Return it.

4. Explain your approach: The approach involves using a stack to decode the given encoded string. We iterate through the characters of the input string and perform different operations based on the type of character encountered. We push digits onto the stack, push an empty string when encountering an opening bracket, pop the top elements and decode the substring when encountering a closing bracket, and append other characters to the current string on the stack.

5. Write clean and readable code:

python
def decodeString(s):
    stack = []
    stack.append(["", 1])  # Initialize stack with an empty string and 1 repetition

    num = ""
    for ch in s:
        if ch.isdigit():
            num += ch
        elif ch == "[":
            stack.append(["", int(num)])
            num = ""
        elif ch == "]":
            decoded_str, repetitions = stack.pop()
            stack[-1][0] += decoded_str * repetitions
        else:
            stack[-1][0] += ch

    return stack[0][0]

6. Test your code: Let's test the code with some test cases:

  • Test case 1:

    • s = "3[a]2[bc]"
    • The expected output is "aaabcbc" because we repeat "a" 3 times and repeat "bc" 2 times.

  • Test case 2:

    • s = "3[a2[c]]"
    • The expected output is "accaccacc" because we repeat "c" 2 times, then repeat "ac" 3 times.

  • Test case 3:

    • s = "2[abc]3[cd]ef"
    • The expected output is "abcabccdcdcdef" because we repeat "abc" 2 times and repeat "cd" 3 times.

We chose these test cases because they cover different aspects of the problem, including nested encodings and multiple repetitions.

python
s1 = "3[a]2[bc]"
print(decodeString(s1))  # Expected output: "aaabcbc"

s2 = "3[a2[c]]"
print(decodeString(s2))  # Expected output: "accaccacc"

s3 = "2[abc]3[cd]ef"
print(decodeString(s3))  # Expected output: "abcabccdcdcdef"

7. Optimize if necessary: The current solution has a time complexity of O(n), where n is the length of the input string s. We iterate through each character of s once. The space complexity is also O(n) as we use a stack to store the intermediate results.

8. Handle error cases: The code assumes that the input s is a valid encoded string. If the input is None or an empty string, the code will return an empty string as the decoded result.

9. Discuss complexity analysis: The time complexity of the solution is O(n), where n is the length of the input string s. We iterate through each character of s once. The space complexity is also O(n) as we use a stack to store the intermediate results. The solution has linear time and space complexity.

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