0394. Decode String

Medium | String + Stack | 20 ms (99.12%), 14.2 MB (81.02%)

❓ Problem Statement

Source: LeetCode - Decode String GitHub: Solution / Performance

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; 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 won't be input like 3a or 2[4].

✍🏻 Constraints & Example

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].

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

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

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

Input: s = "abc3[cd]xyz"
Output: "abccdcdcdxyz"

💡 Ideas

Use stack to record the following information in a tuple before the left square bracket [:

• Decoded string aspre_DecodedStr

• n times asnTimes

When meeting the right square bracket ], pop the tuple from the stack. We then concatenate the returned string by pre_DecodedStr + nTimes * cur_DecodedStr

🤖 Python3

class Solution:
    def decodeString(self, s: str) -> str:
        
        # ==================================================
        #  String + Stack                                  =
        # ==================================================
        # time  : O(n)
        # space : O(m), m is the number of pairs of square brackets
        
        ans, stack, curNum = '', [], 0
        
        for char in s:
            if char.isdigit():
                curNum = curNum * 10 + int(char)
                
            elif char == '[':
                stack.append((ans, curNum))
                ans, curNum = '', 0
                
            elif char == ']':
                prev, num = stack.pop()
                ans = prev + num * ans
                
            else:
                ans += char
        
        return ans

🤖 Java

class Solution {
    /**
     * @time  : O(n)
     * @space : O(m), m is the number of pairs of square brackets
     */
    
    public String decodeString(String s) {
        int curNum = 0;
        String ans = "";
        Stack<Integer> numStack = new Stack<>();
        Stack<String> strStack = new Stack<>();
        
        for(int i=0 ; i<s.length() ; i++) {
            char c = s.charAt(i);
            
            if(Character.isDigit(c)) {
                curNum = curNum * 10 + (c - '0');
                
            } else if(c == '[') {
                numStack.push(curNum);
                strStack.push(ans);
                curNum = 0;
                ans = "";
                
            } else if(c == ']') {
                String prev = strStack.pop();
                int num = numStack.pop();
                ans = prev + ans.repeat(num);
                
            } else {
                ans += c;
            }
        }
        
        return ans;
    }
}