LeetCode 0880. Decoded String at Index Solution in Java, C++, Python & More | Explanation + Code

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).

PythonJavaC++GoTypeScript

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(code-box)

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; } } } }(code-box)

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; } } } };(code-box)

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-- } } }(code-box)

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; } } }(code-box)