LeetCode-in-All

394. Decode String

Medium

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].

Solution

(define/contract (decode-string s)
  (-> string? string?)
  (let-values ([(result _) (decode-helper (string->list s) 0)])
    result))

(define (decode-helper chars pos)
  (let loop ([current-pos pos]
             [count 0]
             [result ""])
    (if (>= current-pos (length chars))
        (values result current-pos)
        (let ([char (list-ref chars current-pos)])
          (cond
            ; If it's a letter, append it to result
            [(char-alphabetic? char)
             (loop (add1 current-pos)
                   count
                   (string-append result (string char)))]
            
            ; If it's a digit, update count
            [(char-numeric? char)
             (loop (add1 current-pos)
                   (+ (* count 10) (- (char->integer char) (char->integer #\0)))
                   result)]
            
            ; If it's '[', handle nested string
            [(char=? char #\[)
             (let-values ([(nested-str new-pos) (decode-helper chars (add1 current-pos))])
               (loop new-pos
                     0
                     (string-append result 
                                  (string-join 
                                   (make-list count nested-str)
                                   ""))))]
            
            ; If it's ']', return current result and position
            [(char=? char #\])
             (values result (add1 current-pos))]
            
            ; Skip other characters
            [else (loop (add1 current-pos) count result)])))))

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