394. Decode String LeetCode Solution

In this guide, you will get 394. Decode String LeetCode Solution with the best time and space complexity. The solution to Decode String problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Decode String solution in C++
Decode String solution in Java
Decode String solution in Python
Additional Resources

394. Decode String LeetCode Solution image

Problem Statement of Decode String

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].

Complexity Analysis

Time Complexity: O(|\texttt{ans}|)
Space Complexity: O(|\texttt{ans}|)

394. Decode String LeetCode Solution in C++

class Solution {
 public:
  string decodeString(string s) {
    stack<pair<string, int>> stack;  // (prevStr, repeatCount)
    string currStr;
    int currNum = 0;

    for (const char c : s)
      if (isdigit(c)) {
        currNum = currNum * 10 + (c - '0');
      } else {
        if (c == '[') {
          stack.emplace(currStr, currNum);
          currStr = "";
          currNum = 0;
        } else if (c == ']') {
          const auto [prevStr, n] = stack.top();
          stack.pop();
          currStr = prevStr + getRepeatedStr(currStr, n);
        } else {
          currStr += c;
        }
      }

    return currStr;
  }

 private:
  // Returns s * n.
  string getRepeatedStr(const string& s, int n) {
    string repeat;
    while (n-- > 0)
      repeat += s;
    return repeat;
  }
};
/* code provided by PROGIEZ */

394. Decode String LeetCode Solution in Java

class Solution {
  public String decodeString(String s) {
    Stack<Pair<StringBuilder, Integer>> stack = new Stack<>(); // (prevStr, repeatCount)
    StringBuilder currStr = new StringBuilder();
    int currNum = 0;

    for (final char c : s.toCharArray())
      if (Character.isDigit(c)) {
        currNum = currNum * 10 + (c - '0');
      } else {
        if (c == '[') {
          stack.push(new Pair<>(currStr, currNum));
          currStr = new StringBuilder();
          currNum = 0;
        } else if (c == ']') {
          final Pair<StringBuilder, Integer> pair = stack.pop();
          final StringBuilder prevStr = pair.getKey();
          final int n = pair.getValue();
          currStr = prevStr.append(getRepeatedStr(currStr, n));
        } else {
          currStr.append(c);
        }
      }

    return currStr.toString();
  }

  // Returns s * n.
  private StringBuilder getRepeatedStr(StringBuilder s, int n) {
    StringBuilder sb = new StringBuilder();
    while (n-- > 0)
      sb.append(s);
    return sb;
  }
}
// code provided by PROGIEZ

394. Decode String LeetCode Solution in Python

class Solution:
  def decodeString(self, s: str) -> str:
    stack = []  # (prevStr, repeatCount)
    currStr = ''
    currNum = 0

    for c in s:
      if c.isdigit():
        currNum = currNum * 10 + int(c)
      else:
        if c == '[':
          stack.append((currStr, currNum))
          currStr = ''
          currNum = 0
        elif c == ']':
          prevStr, num = stack.pop()
          currStr = prevStr + num * currStr
        else:
          currStr += c

    return currStr
 # code by PROGIEZ