Question

880. Decoded String at Index

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Description

You are given an encoded string s. To decode the string to a tape, the encoded string is read one character at a time and the following steps are taken:

• If the character read is a letter, that letter is written onto the tape.
• If the character read is a digit d, the entire current tape is repeatedly written d - 1 more times in total.

Given an integer k, return _the _kth_ letter (1-indexed) in the decoded string_.

 

Example 1:

Input: s = "leet2code3", k = 10
Output: "o"
Explanation: The decoded string is "leetleetcodeleetleetcodeleetleetcode".
The 10th letter in the string is "o".

Example 2:

Input: s = "ha22", k = 5
Output: "h"
Explanation: The decoded string is "hahahaha".
The 5th letter is "h".

Example 3:

Input: s = "a2345678999999999999999", k = 1
Output: "a"
Explanation: The decoded string is "a" repeated 8301530446056247680 times.
The 1st letter is "a".

 

Constraints:

• 2 <= s.length <= 100
• s consists of lowercase English letters and digits 2 through 9.
• s starts with a letter.
• 1 <= k <= 109
• It is guaranteed that k is less than or equal to the length of the decoded string.
• The decoded string is guaranteed to have less than 263 letters.

Solutions

Solution 1: Reverse Thinking

We can first calculate the total length $m$ of the decoded string, then traverse the string from back to front. Each time, we update $k$ to be $k \bmod m$, until $k$ is $0$ and the current character is a letter, then we return the current character. Otherwise, if the current character is a number, we divide $m$ by this number. If the current character is a letter, we subtract $1$ from $m$.

The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(1)$.

class Solution:
  def decodeAtIndex(self, s: str, k: int) -> str:
    m = 0
    for c in s:
      if c.isdigit():
        m *= int(c)
      else:
        m += 1
    for c in s[::-1]:
      k %= m
      if k == 0 and c.isalpha():
        return c
      if c.isdigit():
        m //= int(c)
      else:
        m -= 1

class Solution {
  public String decodeAtIndex(String s, int k) {
    long m = 0;
    for (int i = 0; i < s.length(); ++i) {
      if (Character.isDigit(s.charAt(i))) {
        m *= (s.charAt(i) - '0');
      } else {
        ++m;
      }
    }
    for (int i = s.length() - 1;; --i) {
      k %= m;
      if (k == 0 && !Character.isDigit(s.charAt(i))) {
        return String.valueOf(s.charAt(i));
      }
      if (Character.isDigit(s.charAt(i))) {
        m /= (s.charAt(i) - '0');
      } else {
        --m;
      }
    }
  }
}

class Solution {
public:
  string decodeAtIndex(string s, int k) {
    long long m = 0;
    for (char& c : s) {
      if (isdigit(c)) {
        m *= (c - '0');
      } else {
        ++m;
      }
    }
    for (int i = s.size() - 1;; --i) {
      k %= m;
      if (k == 0 && isalpha(s[i])) {
        return string(1, s[i]);
      }
      if (isdigit(s[i])) {
        m /= (s[i] - '0');
      } else {
        --m;
      }
    }
  }
};

func decodeAtIndex(s string, k int) string {
  m := 0
  for _, c := range s {
    if c >= '0' && c <= '9' {
      m *= int(c - '0')
    } else {
      m++
    }
  }
  for i := len(s) - 1; ; i-- {
    k %= m
    if k == 0 && s[i] >= 'a' && s[i] <= 'z' {
      return string(s[i])
    }
    if s[i] >= '0' && s[i] <= '9' {
      m /= int(s[i] - '0')
    } else {
      m--
    }
  }
}

function decodeAtIndex(s: string, k: number): string {
  let m = 0n;
  for (const c of s) {
    if (c >= '1' && c <= '9') {
      m *= BigInt(c);
    } else {
      ++m;
    }
  }
  for (let i = s.length - 1; ; --i) {
    if (k >= m) {
      k %= Number(m);
    }
    if (k === 0 && s[i] >= 'a' && s[i] <= 'z') {
      return s[i];
    }
    if (s[i] >= '1' && s[i] <= '9') {
      m = (m / BigInt(s[i])) | 0n;
    } else {
      --m;
    }
  }
}