## Algorithm

Problem Name: 880. Decoded String at Index

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.

## Code Examples

### #1 Code Example with Javascript Programming

```Code - Javascript Programming```

``````
const decodeAtIndex = function(S, K) {
let n = S.length;
let dp = Array(n + 1).fill(0);
for (let i = 0; i < n; i++) {
if (S[i] >= "2" && S[i] <= "9") {
dp[i + 1] = dp[i] * (S[i] - "0");
} else {
dp[i + 1] = dp[i] + 1;
}
}
K--;
for (let i = n - 1; i >= 0; i--) {
K %= dp[i + 1];
if (K + 1 == dp[i + 1] && !(S[i] >= "2" && S[i] <= "9")) {
return "" + S[i];
}
}
return null;
};
``````
Copy The Code &

Input

cmd
s = "leet2code3", k = 10

Output

cmd
"o"

### #2 Code Example with Python Programming

```Code - Python Programming```

``````
class Solution:
def decodeAtIndex(self, S, K):
stack, l = [], 0
for c in S:
l = l + 1 if c.isalpha() else l * int(c)
stack += c,
while l >= K:
while stack[-1].isdigit(): l //= int(stack.pop())
K = K % l
if not K: return stack[-1]
l -= 1
stack.pop()
``````
Copy The Code &

Input

cmd
s = "leet2code3", k = 10

Output

cmd
"o"