g0801_0900.s0880_decoded_string_at_index.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0801_0900.s0880_decoded_string_at_index;

// #Medium #String #Stack #2022_03_28_Time_0_ms_(100.00%)_Space_42.1_MB_(34.90%)

/**
 * 880 - Decoded String at Index\.
 *
 * Medium
 *
 * 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:**
 *
 * **Input:** s = "leet2code3", k = 10
 *
 * **Output:** "o"
 *
 * **Explanation:** The decoded string is "leetleetcodeleetleetcodeleetleetcode". The 10^th letter in the string is "o".
 *
 * **Example 2:**
 *
 * **Input:** s = "ha22", k = 5
 *
 * **Output:** "h"
 *
 * **Explanation:** The decoded string is "hahahaha". The 5^th letter is "h".
 *
 * **Example 3:**
 *
 * **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 <= 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.
**/
@SuppressWarnings("java:S3518")
public class Solution {
    public String decodeAtIndex(String s, int k) {
        long length = 0;
        for (char c : s.toCharArray()) {
            if (c >= 48 && c <= 57) {
                length *= c - '0';
            } else {
                ++length;
            }
        }
        for (int i = s.length() - 1; i >= 0; i--) {
            char c = s.charAt(i);
            k %= length;
            if (c >= 48 && c <= 57) {
                length /= c - '0';
            } else if (k == 0) {
                return String.valueOf(c);
            } else {
                --length;
            }
        }
        return "";
    }
}