Groupon Interview Question
Software DevelopersCountry: United States
With a string of length n, you have 2 in power of (n-1) different possible coma delimited strings. You can represent state of coma positions as an array of 0s and 1s, where 1 means that comma is needed at i+1 position. In this example: 123, would have the following:
0,0 => 123
0,1 => 12,3
1,0 => 1,23
1,1 => 1,2,3
the way you do it is you need introduce a variable of type
int k;
and increment it by 1 at each cycle, until
k<input.length-1
. To check whether the coma is needed you need to check condition:
(k>>1)&1==1
The problem is numbers greater than 26 cannot be encoded. Thus 893 can only be encoded as 8,9,3 and no three digit numbers are allowed. I feel like recursion would be useful here as it would allow to not have to check "unallowed" permutations more than once at each level.
1,2,2,2
12,2,2
1,22,2
then
12,22
1,2,22
Adding a 5th is all of the above encodings plus
1,2,2,22
12,2,22
1,22,22
which happens to be the 3 digit encodings
6 gives all of the above plus
1,2,2,2,22
1,22,2,22
12,2,2,22
1,2,22,22
12,22,22
Which once again is all the 4 digit encodings
Thus the number of encodings for a N digit string(where every combination can encoded) is the number of encodings for an n-1 digit string + the number of encodings of a n-2 digit string. Note if every combination is not possible, such as 12,35 then make sure you check that each number is < 26 or else don't count it.
I would definitely use a dynamic programming approach or else this will become exponential in run time like fibonacci.
public static int findNumberOfDecodeSequences(String sequence) {
if(sequence.length() == 0) {
return 0;
}
int numStringsEndingWithSingleDigit = 1;
int numStringsEndingWithDoubleDigits = 0;
for(int i = 1; i < sequence.length(); i++) {
int doubleDigitVal = (sequence.charAt(i - 1) - '0') * 10 + (sequence.charAt(i) - '0');
if(doubleDigitVal <= 26) {
int temp = numStringsEndingWithDoubleDigits + numStringsEndingWithSingleDigit;
numStringsEndingWithDoubleDigits = numStringsEndingWithSingleDigit;
numStringsEndingWithSingleDigit = temp;
}
else {
numStringsEndingWithSingleDigit = numStringsEndingWithSingleDigit + numStringsEndingWithDoubleDigits;
numStringsEndingWithDoubleDigits = 0;
}
}
return (numStringsEndingWithDoubleDigits + numStringsEndingWithSingleDigit);
}
public static int findNumberOfDecodeSequences(String sequence) {
if(sequence.length() == 0) {
return 0;
}
int numStringsEndingWithSingleDigit = 1;
int numStringsEndingWithDoubleDigits = 0;
for(int i = 1; i < sequence.length(); i++) {
int doubleDigitVal = (sequence.charAt(i - 1) - '0') * 10 + (sequence.charAt(i) - '0');
if(doubleDigitVal <= 26) {
int temp = numStringsEndingWithDoubleDigits + numStringsEndingWithSingleDigit;
numStringsEndingWithDoubleDigits = numStringsEndingWithSingleDigit;
numStringsEndingWithSingleDigit = temp;
}
else {
numStringsEndingWithSingleDigit = numStringsEndingWithSingleDigit + numStringsEndingWithDoubleDigits;
numStringsEndingWithDoubleDigits = 0;
}
}
return (numStringsEndingWithDoubleDigits + numStringsEndingWithSingleDigit);
}
public static int decodeSeq(String num) {
if (num.length() == 0)
return 0;
int[] f = new int[num.length() + 1];
f[0] = 1;
f[1] = 1;
for (int i = 1; i < num.length(); i++) {
if ((num.charAt(i - 1) - '0') * 10 + (num.charAt(i) - '0') <= 26) {
f[i + 1] = f[i] + f[i - 1];
} else {
f[i + 1] = f[i];
}
}
return f[num.length()];
}
public static int decodeSeq(String num) {
if (num.length() == 0)
return 0;
int[] f = new int[num.length() + 1];
f[0] = 1;
f[1] = 1;
for (int i = 1; i < num.length(); i++) {
if ((num.charAt(i - 1) - '0') * 10 + (num.charAt(i) - '0') <= 26) {
f[i + 1] = f[i] + f[i - 1];
} else {
f[i + 1] = f[i];
}
}
return f[num.length()];
}
DP
package Java;
import java.util.Arrays;
/**
* Author: Nitin Gupta
* Date: 26/04/19
* Description:
*/
public class NumberOfWaysToDecodeDigitSeq {
public static void main(String args[]) {
String s = "1212";
System.out.println(ways(s));
}
private static int ways(String s) {
if (s == null || s.isEmpty())
return 0;
int n = s.length();
int count[] = new int[n + 1];
Arrays.fill(count, 0);
count[0] = 1; //every single character can be transform
count[1] = 1;
char digits[] = s.toCharArray();
for (int i = 2; i <= n; i++) {
if (digits[i - 1] > '0')
count[i] += count[i - 1];
if (digits[i - 2] == '1' || digits[i - 2] == '2' && digits[i - 1] < '7')
count[i] += count[i - 2];
}
return count[n];
}
}
- Anonymous June 08, 2015