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.
Input: s = "3[a]2[bc]"
Output: "aaabcbc"
Input: s = "3[a2[c]]"
Output: "accaccacc"
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].
Contents
- Solution 1 - Using a Stack
In this approach, we are going to loop through every character of input string s and store them into a Stack
and whenever we encounter a clsing bracket ']' that means we have reached to the end of the characters
which we need to decode using the integer before [, so we will construct a string using the characters from stack until we reach
'[', for example if the input string is 2[abc], after adding characters a, b, and c to the stack,
then when we see ], we will construct the string "abc".
Then, if the stack is not empty and the characters are digits 0-9, we construct the number before [ and repeat the string
constructed before and put back the resultant string into stack, since the brackets could be nested.
import java.util.Stack;
public class DecodeString {
static String decodeString(String s) {
Stack<String> stack = new Stack<>();
for(char c: s.toCharArray()) {
if( c != ']') {
stack.push(String.valueOf(c));
} else {
String current ="";
while (!"[".equals(stack.peek())) {
current = stack.pop() + current;
}
stack.pop(); // pop the '['
// now we need to get the K number before that
String k = "";
while(!stack.isEmpty() && Character.isDigit(stack.peek().toCharArray()[0])) {
k = stack.pop() + k;
}
String decoded = current.repeat(Integer.parseInt(k));
stack.push(decoded);
}
}
return String.join("", stack);
}
public static void main(String[] args) {
System.out.println(decodeString("2[abc]3[cd]ef"));
}
}
Complexity Analysis:
Time complexity: Time complexity really depends on how big is the k, and how many of them are present in the input string.
If there are n k's, the time complexity would be O (n * k)
Space complexity: O(n * k) for storing elements into stack.
Above implementations source code can be found at
GitHub link for Java code