Algorithm
Problem Name: 880. Decoded String at Index
Problem Link: https://leetcode.com/problems/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 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.
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;
};
Input
s = "leet2code3", k = 10
Output
"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()
Input
s = "leet2code3", k = 10
Output
"o"