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| 1 | +package math; |
| 2 | + |
| 3 | +/** |
| 4 | + * Created by gouthamvidyapradhan on 21/07/2019 An encoded string S is given. To find and write the |
| 5 | + * decoded string to a tape, the encoded string is read one character at a time and the following |
| 6 | + * steps are taken: |
| 7 | + * |
| 8 | + * <p>If the character read is a letter, that letter is written onto the tape. If the character read |
| 9 | + * is a digit (say d), the entire current tape is repeatedly written d-1 more times in total. Now |
| 10 | + * for some encoded string S, and an index K, find and return the K-th letter (1 indexed) in the |
| 11 | + * decoded string. |
| 12 | + * |
| 13 | + * <p>Example 1: |
| 14 | + * |
| 15 | + * <p>Input: S = "leet2code3", K = 10 Output: "o" Explanation: The decoded string is |
| 16 | + * "leetleetcodeleetleetcodeleetleetcode". The 10th letter in the string is "o". Example 2: |
| 17 | + * |
| 18 | + * <p>Input: S = "ha22", K = 5 Output: "h" Explanation: The decoded string is "hahahaha". The 5th |
| 19 | + * letter is "h". Example 3: |
| 20 | + * |
| 21 | + * <p>Input: S = "a2345678999999999999999", K = 1 Output: "a" Explanation: The decoded string is "a" |
| 22 | + * repeated 8301530446056247680 times. The 1st letter is "a". |
| 23 | + * |
| 24 | + * <p>Note: |
| 25 | + * |
| 26 | + * <p>2 <= S.length <= 100 S will only contain lowercase letters and digits 2 through 9. S starts |
| 27 | + * with a letter. 1 <= K <= 10^9 The decoded string is guaranteed to have less than 2^63 letters. |
| 28 | + * |
| 29 | + * <p>Solution: General idea is as shown below example: If S = "leet2" and K = 6 the answer is "e" |
| 30 | + * which is same as finding answer for K = 2. As soon as the product exceeds the total value of K as |
| 31 | + * in this case the product of 4 (leet) x 2 is 8 and 8 clearly exceeds 6 therefore we can reduce K |
| 32 | + * to 8 - 6 = 2 and start from the beginning once again. Repeat the same process until we reach the |
| 33 | + * answer. |
| 34 | + */ |
| 35 | +public class DecodedStringAtIndex { |
| 36 | + public static void main(String[] args) { |
| 37 | + System.out.println( |
| 38 | + new DecodedStringAtIndex().decodeAtIndex("a2345678999999999999999", 1000000000)); |
| 39 | + } |
| 40 | + |
| 41 | + public String decodeAtIndex(String S, int K) { |
| 42 | + long product = 0; |
| 43 | + char lastC = S.charAt(0); |
| 44 | + for (int i = 0, l = S.length(); i < l; ) { |
| 45 | + char c = S.charAt(i); |
| 46 | + if (Character.isLetter(c)) { |
| 47 | + lastC = c; |
| 48 | + product++; |
| 49 | + i++; |
| 50 | + if (K == product) break; |
| 51 | + } else { |
| 52 | + long temp = (product * Integer.parseInt(String.valueOf(c))); |
| 53 | + if (temp == K) break; |
| 54 | + else { |
| 55 | + if (temp > K) { |
| 56 | + long x = (K / product); |
| 57 | + if ((product * x) == K) break; |
| 58 | + K -= (product * x); |
| 59 | + i = 0; |
| 60 | + product = 0; |
| 61 | + } else { |
| 62 | + product = temp; |
| 63 | + i++; |
| 64 | + } |
| 65 | + } |
| 66 | + } |
| 67 | + } |
| 68 | + return String.valueOf(lastC); |
| 69 | + } |
| 70 | +} |
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