LeetCode 394 - Decode String

Table of Contents

• Problem Statement
• Using Two Stacks
• Conclusion

Problem Statement

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.

Using Two Stacks

Explanation

We use two stacks:

• countStack to store repeat counts (k).
• resultStack to store intermediate decoded strings.

1. Digit Handling (0-9)
   • Build k by processing consecutive digits.
     
     
2. Opening Bracket ([)
   • Push k onto countStack and the current string onto resultStack.
   • Reset k and currentResult for the next segment.
     
     
3. Closing Bracket (])
   • Pop k from countStack and the last decoded segment from resultStack.
   • Repeat currentResult k times and append that to the previous result.
     
     
4. Character Handling(a-z)
   • Append normal characters to currentResult.
     
     

class Solution {
public:
    string decodeString(string s) {
        stack<int> countStack;
        stack<string> resultStack;
        string currentResult = "";
        int k = 0;

        for (char c : s) {
            if (isdigit(c)) {
                k = k * 10 + (c - '0');

            } else if (c == '[') {
                countStack.push(k);
                resultStack.push(currentResult);
                k = 0;
                currentResult = "";

            } else if (c == ']') {
                int repeatCount = countStack.top();
                countStack.pop();
                string decodedSegment = resultStack.top();
                resultStack.pop();

                while (repeatCount--) {
                    decodedSegment += currentResult;
                }
                currentResult = decodedSegment;

            } else {
                currentResult += c;
            }
        }
        return currentResult;
    }
};

Conclusion

• Time Complexity: 𝑂(𝑛)
  • Each character is processed exactly once.
  • Stack operations (push/pop) are 𝑂(1).
• Space Complexity: 𝑂(𝑛)
  • Worst-case: deeply nested structures require storing intermediate results in stacks.

Source:

394. Decode String