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leetcode/problems/src/math/DecodedStringAtIndex.java at master · KindleBooks66/leetcode · GitHub
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math
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DecodedStringAtIndex.java
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leetcode
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problems
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src
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math
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DecodedStringAtIndex.java
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package
math
;
/**
* Created by gouthamvidyapradhan on 21/07/2019 An encoded string S is given. To find and write the
* decoded string to a tape, the encoded string is read one character at a time and the following
* steps are taken:
*
* <p>If the character read is a letter, that letter is written onto the tape. If the character read
* is a digit (say d), the entire current tape is repeatedly written d-1 more times in total. Now
* for some encoded string S, and an index K, find and return the K-th letter (1 indexed) in the
* decoded string.
*
* <p>Example 1:
*
* <p>Input: S = "leet2code3", K = 10 Output: "o" Explanation: The decoded string is
* "leetleetcodeleetleetcodeleetleetcode". The 10th letter in the string is "o". Example 2:
*
* <p>Input: S = "ha22", K = 5 Output: "h" Explanation: The decoded string is "hahahaha". The 5th
* letter is "h". Example 3:
*
* <p>Input: S = "a2345678999999999999999", K = 1 Output: "a" Explanation: The decoded string is "a"
* repeated 8301530446056247680 times. The 1st letter is "a".
*
* <p>Note:
*
* <p>2 <= S.length <= 100 S will only contain lowercase letters and digits 2 through 9. S starts
* with a letter. 1 <= K <= 10^9 The decoded string is guaranteed to have less than 2^63 letters.
*
* <p>Solution: General idea is as shown below example: If S = "leet2" and K = 6 the answer is "e"
* which is same as finding answer for K = 2. As soon as the product exceeds the total value of K as
* in this case the product of 4 (leet) x 2 is 8 and 8 clearly exceeds 6 therefore we can reduce K
* to 8 - 6 = 2 and start from the beginning once again. Repeat the same process until we reach the
* answer.
*/
public
class
DecodedStringAtIndex
{
public
static
void
main
(
String
[]
args
) {
System
.
out
.
println
(
new
DecodedStringAtIndex
().
decodeAtIndex
(
"a2345678999999999999999"
,
1000000000
));
}
public
String
decodeAtIndex
(
String
S
,
int
K
) {
long
product
=
0
;
char
lastC
=
S
.
charAt
(
0
);
for
(
int
i
=
0
,
l
=
S
.
length
();
i
<
l
; ) {
char
c
=
S
.
charAt
(
i
);
if
(
Character
.
isLetter
(
c
)) {
lastC
=
c
;
product
++;
i
++;
if
(
K
==
product
)
break
;
}
else
{
long
temp
= (
product
*
Integer
.
parseInt
(
String
.
valueOf
(
c
)));
if
(
temp
==
K
)
break
;
else
{
if
(
temp
>
K
) {
long
x
= (
K
/
product
);
if
((
product
*
x
) ==
K
)
break
;
K
-= (
product
*
x
);
i
=
0
;
product
=
0
;
}
else
{
product
=
temp
;
i
++;
}
}
}
}
return
String
.
valueOf
(
lastC
);
}
}
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