880. Decoded String at Index

Problem

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.

Solution

/**
 * @param {string} s
 * @param {number} k
 * @return {string}
 */
var decodeAtIndex = function(s, k) {
    var totalMap = Array(s.length);
    for (var i = 0; i < s.length; i++) {
        if (isDigit(s[i])) {
            totalMap[i] = Number(s[i]) * totalMap[i - 1];
        } else {
            totalMap[i] = (totalMap[i - 1] || 0) + 1;
        }
        if (totalMap[i] >= k) {
            return helper(s, k, totalMap, i);
        }
    }
};

var helper = function(s, k, totalMap, i) {
    if (isDigit(s[i])) {
        if (totalMap[i] === k || k % totalMap[i - 1] === 0) {
            var j = i - 1;
            while (isDigit(s[j])) j--;
            return s[j];
        } else {
            var n = k % totalMap[i - 1];
            return helper(s, n, totalMap, i - 1);
        }
    } else {
        if (totalMap[i] === k) {
            return s[i];
        } else {
            return helper(s, k, totalMap, i - 1);
        }
    }
};

var isDigit = function(char) {
    return char >= '2' && char <= '9';
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).