880. Decoded String at Index
Đề Bài
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 writtend - 1more 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 <= 100sconsists of lowercase English letters and digits2through9.sstarts with a letter.1 <= k <= 109- It is guaranteed that
kis less than or equal to the length of the decoded string. - The decoded string is guaranteed to have less than
263letters.
Thuật Toán & Kỹ Thuật
⏱️ Thời gian
O(n²)
💾 Không gian
O(1)
Lời Giải
C++
0880-decoded-string-at-index.cpp
class Solution {
public:
string decodeAtIndex(string s, int k) {
const int n = s.size();
long long real_size = 0;
for (char c : s) {
if (isdigit(c)) {
real_size = real_size * (c - '0');
}
else {
real_size++;
}
}
for (int i = n - 1; i >= 0; --i) {
char c = s[i];
if (isdigit(c)) {
real_size /= (c - '0');
k %= real_size;
}
else {
if (k % real_size == 0) {
return string(1, c);
}
real_size--;
}
}
return "";
}
};