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880. Decoded String at Index

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 {
    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;
            }
        }
    }
};

  • 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


  • 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;
        }
    }
}

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