# Decoded String At Index Problem

## Description

LeetCode Problem 880.

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 k^th letter (1-indexed) in the decoded string.

Example 1:

``````1
2
3
4
Input: s = "leet2code3", k = 10
Output: "o"
Explanation: The decoded string is "leetleetcodeleetleetcodeleetleetcode".
The 10^th letter in the string is "o".
``````

Example 2:

``````1
2
3
4
Input: s = "ha22", k = 5
Output: "h"
Explanation: The decoded string is "hahahaha".
The 5^th letter is "h".
``````

Example 3:

``````1
2
3
4
Input: s = "a2345678999999999999999", k = 1
Output: "a"
Explanation: The decoded string is "a" repeated 8301530446056247680 times.
The 1^st 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 <= 10^9
• 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 2^63 letters.

## Sample C++ Code

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
class Solution {
public:
string decodeAtIndex(string S, int K) {
long size = 0;
int N = S.size();

for (int i = 0; i < N; ++i) {
if (isdigit(S[i]))
size *= S[i] - '0';
else
size++;
}

for (int i = N-1; i >=0; --i) {
K %= size;
if (K == 0 && isalpha(S[i]))
return (string) "" + S[i];

if (isdigit(S[i]))
size /= S[i] - '0';
else
size--;
}
return S.substr(K, 1);
}
};
``````