Combinatorics problem with bit-strings [closed]

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When only three types of bit strings 0, 10,11 are available, how many valid $7$-bit strings can be represented? For example, the bit string 0101110, which is composed of 0, 10, 11, 10, is a valid bit string, but 1100101 is invalid.

edited Sep 2 at 4:26

tarit goswami

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asked Sep 2 at 4:17

Nursoltan Saipolda

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closed as off-topic by user21820, Did, amWhy, Holo, ccorn Sep 2 at 15:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â€“ user21820, Did, amWhy, Holo, ccorn

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edited Sep 2 at 4:26

tarit goswami

1,139219

asked Sep 2 at 4:17

Nursoltan Saipolda

122

closed as off-topic by user21820, Did, amWhy, Holo, ccorn Sep 2 at 15:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â€“ user21820, Did, amWhy, Holo, ccorn

If this question can be reworded to fit the rules in the help center, please edit the question.

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up vote
0
down vote

favorite

edited Sep 2 at 4:26

tarit goswami

1,139219

asked Sep 2 at 4:17

Nursoltan Saipolda

122

edited Sep 2 at 4:26

tarit goswami

1,139219

asked Sep 2 at 4:17

Nursoltan Saipolda

122

edited Sep 2 at 4:26

tarit goswami

1,139219

edited Sep 2 at 4:26

tarit goswami

1,139219

edited Sep 2 at 4:26

tarit goswami

1,139219

asked Sep 2 at 4:17

Nursoltan Saipolda

122

asked Sep 2 at 4:17

Nursoltan Saipolda

122

asked Sep 2 at 4:17

Nursoltan Saipolda

122

closed as off-topic by user21820, Did, amWhy, Holo, ccorn Sep 2 at 15:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â€“ user21820, Did, amWhy, Holo, ccorn

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by user21820, Did, amWhy, Holo, ccorn Sep 2 at 15:23

This question appears to be off-topic. The users who voted to close gave this specific reason:

"This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." â€“ user21820, Did, amWhy, Holo, ccorn

If this question can be reworded to fit the rules in the help center, please edit the question.

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3 Answers
3

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oldest

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up vote
6
down vote

accepted

Define $f(n)$ as the number of valid $n$-bit strings and define the empty string as valid. So $f(0)=f(1)=1$. We can create a valid string of length $n$ by taking a string of length $n-1$ and appending "$0$" or by taking a string of length $n-2$ and appending either "$10$" or "$11$". So, we have $f(n)=f(n-1)+2f(n-2)$. From here, we get

$$f(2)=f(1)+2f(0)=1+2=3$$
$$f(3)=f(2)+2f(1)=3+2=5$$
$$f(4)=f(3)+2f(2)=5+6=11$$
$$f(5)=f(4)+2f(3)=11+10=21$$
$$f(6)=f(5)+2f(4)=21+22=43$$
$$f(7)=f(6)+2f(5)=43+42=85$$

answered Sep 2 at 6:49

Mike

11.6k31642

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up vote
5
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The available strings form whats called a "prefix code", which means that no string is the prefix of another string. This allows us to create a simple algorithm that determines if a bit string is valid or not. Think of the bit string as a stream. Read in a character. If it is 0, discard it and move on to the next character. If it is a 1, then discard it and the next character.

In the first case, we are removing an instance of the 0 string. In the second, we are removing an instance of either 10 or 11.

This process only fails if you read in a 1 but there is no character following it.

Therefore, the only invalid strings are strings which end in a single 1. Can you finish the problem from here?

Edit: @YawarRaza7349 pointed out a potential misinterpretation; I'll complete this solution to avoid that.

Since all invalid strings end in 1, we know that all strings ending in 0 are valid. This already gives us $2^6=64$ valid strings.

Next, the only way for a valid string to end in a 1 is for the last two characters to be 11. If we remove these characters, we are left with $5$ characters that also form a valid string.

Thus we can apply the same argument again. If the last of these $5$ characters is 0, the string is valid. This gives $2^4=16$ valid strings.

Iterating again, we have $3$ characters which form a valid string and there are $2^2=4$ possibilities.

Finally, we have just a single character string that is valid, there is only $2^0=1$ option.

Summing, we have $2^6+2^4+2^2=64+16+4+1=85$.

edited Sep 2 at 22:00

smci

354211

answered Sep 2 at 4:32

rikhavshah

1,072212

3

Your last paragraph could be read as implying that the answer is the number of 7-bit strings that end in 1, which isn't correct. Valid strings where 11 was consumed last also end in 1.
â€“Â YawarRaza7349
Sep 2 at 9:12

Sorry, I meant "the number of 7-bit strings that don't end in 1".
â€“Â YawarRaza7349
Sep 2 at 10:40

@YawarRaza7349 thanks, I updated my answer.
â€“Â rikhavshah
Sep 2 at 15:29

1

Interesting alternative to the one in Mike's answer. Another fix though: you need to take one more step and add $2^0$ (for the string 0111111). (I'm also not sure what you mean by "excluding the empty string"; it wasn't explicitly excluded, it wasn't counted like for the same reason all other strings that aren't 7 bits long weren't counted.)
â€“Â YawarRaza7349
Sep 2 at 15:50

2

I think this formulation uses a recurrence relation $g(n)$ defining the number of invalid n-bit strings, i.e. the complement of Mike's answer.
â€“Â smci
Sep 2 at 21:44

add a commentÂ |Â

up vote
-1
down vote

The numbers 11 and 10 have 2 digits(2 bits) and 0 has a single digit(1 bit).
The 7 bits can be formed in 4 cases: 1) three 2 bits and one 1 bit
2) two 2 bits and three 1 bits
3) one 2 bit and five 1 bits
4) zero 2 bits and seven 1 bits.

Case 1:
_ _ _ 0 the first three blanks can be filled in 8 ways(2 ways for 1st blankbecause it
can be filled by 10 or 11, 2 ways in 2nd and the same goes for the 3rd).

so by product rule 2*2*2=8 ways

Case 2:
_ _ 0 0 0 the first two blanks can be filled in 4 ways( 2 ways for 1st and in two for
2nd).

so by product rule we have 2*2 = 4.

Case 3:
_ 0 0 0 0 0 the first blank can be filled in 2 ways(with 10 or 11).

so this equals 2.

Case 4:
0 0 0 0 0 0 0 this is applicable in only one way.

(i didn't have the need to choose the one bit because it is 0 for sure)

now to obtain the final answer add all the cases.

which equals 8 + 4 + 2 + 1 = 15.

therefore there 15 strings.

I am happy to receive any corrections if this is wrong :)

answered Sep 2 at 5:01

Barath BR

2

I think the problem might be that the zeros don't have to all be at the end.
â€“Â Mike
Sep 2 at 6:55

but it wasn't mentioned right?
â€“Â Barath BR
Sep 2 at 12:54

1

The valid example the OP mentioned in the beginning, 0 10 11 10, starts with a 0, and thus is not counted in any of your cases.
â€“Â YawarRaza7349
Sep 2 at 14:13

add a commentÂ |Â

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

up vote
6
down vote

accepted

$$f(2)=f(1)+2f(0)=1+2=3$$
$$f(3)=f(2)+2f(1)=3+2=5$$
$$f(4)=f(3)+2f(2)=5+6=11$$
$$f(5)=f(4)+2f(3)=11+10=21$$
$$f(6)=f(5)+2f(4)=21+22=43$$
$$f(7)=f(6)+2f(5)=43+42=85$$

answered Sep 2 at 6:49

Mike

11.6k31642

add a commentÂ |Â

up vote
6
down vote

accepted

$$f(2)=f(1)+2f(0)=1+2=3$$
$$f(3)=f(2)+2f(1)=3+2=5$$
$$f(4)=f(3)+2f(2)=5+6=11$$
$$f(5)=f(4)+2f(3)=11+10=21$$
$$f(6)=f(5)+2f(4)=21+22=43$$
$$f(7)=f(6)+2f(5)=43+42=85$$

answered Sep 2 at 6:49

Mike

11.6k31642

add a commentÂ |Â

up vote
6
down vote

accepted

$$f(2)=f(1)+2f(0)=1+2=3$$
$$f(3)=f(2)+2f(1)=3+2=5$$
$$f(4)=f(3)+2f(2)=5+6=11$$
$$f(5)=f(4)+2f(3)=11+10=21$$
$$f(6)=f(5)+2f(4)=21+22=43$$
$$f(7)=f(6)+2f(5)=43+42=85$$

answered Sep 2 at 6:49

Mike

11.6k31642

$$f(2)=f(1)+2f(0)=1+2=3$$
$$f(3)=f(2)+2f(1)=3+2=5$$
$$f(4)=f(3)+2f(2)=5+6=11$$
$$f(5)=f(4)+2f(3)=11+10=21$$
$$f(6)=f(5)+2f(4)=21+22=43$$
$$f(7)=f(6)+2f(5)=43+42=85$$

answered Sep 2 at 6:49

Mike

11.6k31642

answered Sep 2 at 6:49

Mike

11.6k31642

answered Sep 2 at 6:49

Mike

11.6k31642

answered Sep 2 at 6:49

Mike

11.6k31642

add a commentÂ |Â

up vote
5
down vote

In the first case, we are removing an instance of the 0 string. In the second, we are removing an instance of either 10 or 11.

This process only fails if you read in a 1 but there is no character following it.

Therefore, the only invalid strings are strings which end in a single 1. Can you finish the problem from here?

Edit: @YawarRaza7349 pointed out a potential misinterpretation; I'll complete this solution to avoid that.

Since all invalid strings end in 1, we know that all strings ending in 0 are valid. This already gives us $2^6=64$ valid strings.

Next, the only way for a valid string to end in a 1 is for the last two characters to be 11. If we remove these characters, we are left with $5$ characters that also form a valid string.

Thus we can apply the same argument again. If the last of these $5$ characters is 0, the string is valid. This gives $2^4=16$ valid strings.

Iterating again, we have $3$ characters which form a valid string and there are $2^2=4$ possibilities.

Finally, we have just a single character string that is valid, there is only $2^0=1$ option.

Summing, we have $2^6+2^4+2^2=64+16+4+1=85$.

edited Sep 2 at 22:00

smci

354211

answered Sep 2 at 4:32

rikhavshah

1,072212

3

Your last paragraph could be read as implying that the answer is the number of 7-bit strings that end in 1, which isn't correct. Valid strings where 11 was consumed last also end in 1.
â€“Â YawarRaza7349
Sep 2 at 9:12

Sorry, I meant "the number of 7-bit strings that don't end in 1".
â€“Â YawarRaza7349
Sep 2 at 10:40

@YawarRaza7349 thanks, I updated my answer.
â€“Â rikhavshah
Sep 2 at 15:29

1

Interesting alternative to the one in Mike's answer. Another fix though: you need to take one more step and add $2^0$ (for the string 0111111). (I'm also not sure what you mean by "excluding the empty string"; it wasn't explicitly excluded, it wasn't counted like for the same reason all other strings that aren't 7 bits long weren't counted.)
â€“Â YawarRaza7349
Sep 2 at 15:50

2

I think this formulation uses a recurrence relation $g(n)$ defining the number of invalid n-bit strings, i.e. the complement of Mike's answer.
â€“Â smci
Sep 2 at 21:44

add a commentÂ |Â

up vote
5
down vote

In the first case, we are removing an instance of the 0 string. In the second, we are removing an instance of either 10 or 11.

This process only fails if you read in a 1 but there is no character following it.

Therefore, the only invalid strings are strings which end in a single 1. Can you finish the problem from here?

Edit: @YawarRaza7349 pointed out a potential misinterpretation; I'll complete this solution to avoid that.

Since all invalid strings end in 1, we know that all strings ending in 0 are valid. This already gives us $2^6=64$ valid strings.

Next, the only way for a valid string to end in a 1 is for the last two characters to be 11. If we remove these characters, we are left with $5$ characters that also form a valid string.

Thus we can apply the same argument again. If the last of these $5$ characters is 0, the string is valid. This gives $2^4=16$ valid strings.

Iterating again, we have $3$ characters which form a valid string and there are $2^2=4$ possibilities.

Finally, we have just a single character string that is valid, there is only $2^0=1$ option.

Summing, we have $2^6+2^4+2^2=64+16+4+1=85$.

edited Sep 2 at 22:00

smci

354211

answered Sep 2 at 4:32

rikhavshah

1,072212

3

Your last paragraph could be read as implying that the answer is the number of 7-bit strings that end in 1, which isn't correct. Valid strings where 11 was consumed last also end in 1.
â€“Â YawarRaza7349
Sep 2 at 9:12

Sorry, I meant "the number of 7-bit strings that don't end in 1".
â€“Â YawarRaza7349
Sep 2 at 10:40

@YawarRaza7349 thanks, I updated my answer.
â€“Â rikhavshah
Sep 2 at 15:29

1

Interesting alternative to the one in Mike's answer. Another fix though: you need to take one more step and add $2^0$ (for the string 0111111). (I'm also not sure what you mean by "excluding the empty string"; it wasn't explicitly excluded, it wasn't counted like for the same reason all other strings that aren't 7 bits long weren't counted.)
â€“Â YawarRaza7349
Sep 2 at 15:50

2

I think this formulation uses a recurrence relation $g(n)$ defining the number of invalid n-bit strings, i.e. the complement of Mike's answer.
â€“Â smci
Sep 2 at 21:44

add a commentÂ |Â

up vote
5
down vote

In the first case, we are removing an instance of the 0 string. In the second, we are removing an instance of either 10 or 11.

This process only fails if you read in a 1 but there is no character following it.

Therefore, the only invalid strings are strings which end in a single 1. Can you finish the problem from here?

Edit: @YawarRaza7349 pointed out a potential misinterpretation; I'll complete this solution to avoid that.

Since all invalid strings end in 1, we know that all strings ending in 0 are valid. This already gives us $2^6=64$ valid strings.

Next, the only way for a valid string to end in a 1 is for the last two characters to be 11. If we remove these characters, we are left with $5$ characters that also form a valid string.

Thus we can apply the same argument again. If the last of these $5$ characters is 0, the string is valid. This gives $2^4=16$ valid strings.

Iterating again, we have $3$ characters which form a valid string and there are $2^2=4$ possibilities.

Finally, we have just a single character string that is valid, there is only $2^0=1$ option.

Summing, we have $2^6+2^4+2^2=64+16+4+1=85$.

edited Sep 2 at 22:00

smci

354211

answered Sep 2 at 4:32

rikhavshah

1,072212

In the first case, we are removing an instance of the 0 string. In the second, we are removing an instance of either 10 or 11.

This process only fails if you read in a 1 but there is no character following it.

Therefore, the only invalid strings are strings which end in a single 1. Can you finish the problem from here?

Edit: @YawarRaza7349 pointed out a potential misinterpretation; I'll complete this solution to avoid that.

Since all invalid strings end in 1, we know that all strings ending in 0 are valid. This already gives us $2^6=64$ valid strings.

Next, the only way for a valid string to end in a 1 is for the last two characters to be 11. If we remove these characters, we are left with $5$ characters that also form a valid string.

Thus we can apply the same argument again. If the last of these $5$ characters is 0, the string is valid. This gives $2^4=16$ valid strings.

Iterating again, we have $3$ characters which form a valid string and there are $2^2=4$ possibilities.

Finally, we have just a single character string that is valid, there is only $2^0=1$ option.

Summing, we have $2^6+2^4+2^2=64+16+4+1=85$.

edited Sep 2 at 22:00

smci

354211

answered Sep 2 at 4:32

rikhavshah

1,072212

edited Sep 2 at 22:00

smci

354211

edited Sep 2 at 22:00

smci

354211

edited Sep 2 at 22:00

smci

354211

answered Sep 2 at 4:32

rikhavshah

1,072212

answered Sep 2 at 4:32

rikhavshah

1,072212

answered Sep 2 at 4:32

rikhavshah

1,072212

3

Your last paragraph could be read as implying that the answer is the number of 7-bit strings that end in 1, which isn't correct. Valid strings where 11 was consumed last also end in 1.
â€“Â YawarRaza7349
Sep 2 at 9:12

Sorry, I meant "the number of 7-bit strings that don't end in 1".
â€“Â YawarRaza7349
Sep 2 at 10:40

@YawarRaza7349 thanks, I updated my answer.
â€“Â rikhavshah
Sep 2 at 15:29

1

Interesting alternative to the one in Mike's answer. Another fix though: you need to take one more step and add $2^0$ (for the string 0111111). (I'm also not sure what you mean by "excluding the empty string"; it wasn't explicitly excluded, it wasn't counted like for the same reason all other strings that aren't 7 bits long weren't counted.)
â€“Â YawarRaza7349
Sep 2 at 15:50

2

I think this formulation uses a recurrence relation $g(n)$ defining the number of invalid n-bit strings, i.e. the complement of Mike's answer.
â€“Â smci
Sep 2 at 21:44

add a commentÂ |Â

3

Your last paragraph could be read as implying that the answer is the number of 7-bit strings that end in 1, which isn't correct. Valid strings where 11 was consumed last also end in 1.
â€“Â YawarRaza7349
Sep 2 at 9:12

Sorry, I meant "the number of 7-bit strings that don't end in 1".
â€“Â YawarRaza7349
Sep 2 at 10:40

@YawarRaza7349 thanks, I updated my answer.
â€“Â rikhavshah
Sep 2 at 15:29

1

Interesting alternative to the one in Mike's answer. Another fix though: you need to take one more step and add $2^0$ (for the string 0111111). (I'm also not sure what you mean by "excluding the empty string"; it wasn't explicitly excluded, it wasn't counted like for the same reason all other strings that aren't 7 bits long weren't counted.)
â€“Â YawarRaza7349
Sep 2 at 15:50

2

I think this formulation uses a recurrence relation $g(n)$ defining the number of invalid n-bit strings, i.e. the complement of Mike's answer.
â€“Â smci
Sep 2 at 21:44

Your last paragraph could be read as implying that the answer is the number of 7-bit strings that end in 1, which isn't correct. Valid strings where 11 was consumed last also end in 1.
â€“Â YawarRaza7349
Sep 2 at 9:12

Sorry, I meant "the number of 7-bit strings that don't end in 1".
â€“Â YawarRaza7349
Sep 2 at 10:40

@YawarRaza7349 thanks, I updated my answer.
â€“Â rikhavshah
Sep 2 at 15:29

Interesting alternative to the one in Mike's answer. Another fix though: you need to take one more step and add $2^0$ (for the string 0111111). (I'm also not sure what you mean by "excluding the empty string"; it wasn't explicitly excluded, it wasn't counted like for the same reason all other strings that aren't 7 bits long weren't counted.)
â€“Â YawarRaza7349
Sep 2 at 15:50

I think this formulation uses a recurrence relation $g(n)$ defining the number of invalid n-bit strings, i.e. the complement of Mike's answer.
â€“Â smci
Sep 2 at 21:44

add a commentÂ |Â

up vote
-1
down vote

Case 1:
_ _ _ 0 the first three blanks can be filled in 8 ways(2 ways for 1st blankbecause it
can be filled by 10 or 11, 2 ways in 2nd and the same goes for the 3rd).

so by product rule 2*2*2=8 ways

Case 2:
_ _ 0 0 0 the first two blanks can be filled in 4 ways( 2 ways for 1st and in two for
2nd).

so by product rule we have 2*2 = 4.

Case 3:
_ 0 0 0 0 0 the first blank can be filled in 2 ways(with 10 or 11).

so this equals 2.

Case 4:
0 0 0 0 0 0 0 this is applicable in only one way.

(i didn't have the need to choose the one bit because it is 0 for sure)

now to obtain the final answer add all the cases.

which equals 8 + 4 + 2 + 1 = 15.

therefore there 15 strings.

I am happy to receive any corrections if this is wrong :)

answered Sep 2 at 5:01

Barath BR

2

I think the problem might be that the zeros don't have to all be at the end.
â€“Â Mike
Sep 2 at 6:55

but it wasn't mentioned right?
â€“Â Barath BR
Sep 2 at 12:54

1

The valid example the OP mentioned in the beginning, 0 10 11 10, starts with a 0, and thus is not counted in any of your cases.
â€“Â YawarRaza7349
Sep 2 at 14:13

add a commentÂ |Â

up vote
-1
down vote

Case 1:
_ _ _ 0 the first three blanks can be filled in 8 ways(2 ways for 1st blankbecause it
can be filled by 10 or 11, 2 ways in 2nd and the same goes for the 3rd).

so by product rule 2*2*2=8 ways

Case 2:
_ _ 0 0 0 the first two blanks can be filled in 4 ways( 2 ways for 1st and in two for
2nd).

so by product rule we have 2*2 = 4.

Case 3:
_ 0 0 0 0 0 the first blank can be filled in 2 ways(with 10 or 11).

so this equals 2.

Case 4:
0 0 0 0 0 0 0 this is applicable in only one way.

(i didn't have the need to choose the one bit because it is 0 for sure)

now to obtain the final answer add all the cases.

which equals 8 + 4 + 2 + 1 = 15.

therefore there 15 strings.

I am happy to receive any corrections if this is wrong :)

answered Sep 2 at 5:01

Barath BR

2

I think the problem might be that the zeros don't have to all be at the end.
â€“Â Mike
Sep 2 at 6:55

but it wasn't mentioned right?
â€“Â Barath BR
Sep 2 at 12:54

1

The valid example the OP mentioned in the beginning, 0 10 11 10, starts with a 0, and thus is not counted in any of your cases.
â€“Â YawarRaza7349
Sep 2 at 14:13

add a commentÂ |Â

up vote
-1
down vote

Case 1:
_ _ _ 0 the first three blanks can be filled in 8 ways(2 ways for 1st blankbecause it
can be filled by 10 or 11, 2 ways in 2nd and the same goes for the 3rd).

so by product rule 2*2*2=8 ways

Case 2:
_ _ 0 0 0 the first two blanks can be filled in 4 ways( 2 ways for 1st and in two for
2nd).

so by product rule we have 2*2 = 4.

Case 3:
_ 0 0 0 0 0 the first blank can be filled in 2 ways(with 10 or 11).

so this equals 2.

Case 4:
0 0 0 0 0 0 0 this is applicable in only one way.

(i didn't have the need to choose the one bit because it is 0 for sure)

now to obtain the final answer add all the cases.

which equals 8 + 4 + 2 + 1 = 15.

therefore there 15 strings.

I am happy to receive any corrections if this is wrong :)

answered Sep 2 at 5:01

Barath BR

Case 1:
_ _ _ 0 the first three blanks can be filled in 8 ways(2 ways for 1st blankbecause it
can be filled by 10 or 11, 2 ways in 2nd and the same goes for the 3rd).

so by product rule 2*2*2=8 ways

Case 2:
_ _ 0 0 0 the first two blanks can be filled in 4 ways( 2 ways for 1st and in two for
2nd).

so by product rule we have 2*2 = 4.

Case 3:
_ 0 0 0 0 0 the first blank can be filled in 2 ways(with 10 or 11).

so this equals 2.

Case 4:
0 0 0 0 0 0 0 this is applicable in only one way.

(i didn't have the need to choose the one bit because it is 0 for sure)

now to obtain the final answer add all the cases.

which equals 8 + 4 + 2 + 1 = 15.

therefore there 15 strings.

I am happy to receive any corrections if this is wrong :)

answered Sep 2 at 5:01

Barath BR

answered Sep 2 at 5:01

Barath BR

answered Sep 2 at 5:01

Barath BR

answered Sep 2 at 5:01

Barath BR

2

I think the problem might be that the zeros don't have to all be at the end.
â€“Â Mike
Sep 2 at 6:55

but it wasn't mentioned right?
â€“Â Barath BR
Sep 2 at 12:54

1

The valid example the OP mentioned in the beginning, 0 10 11 10, starts with a 0, and thus is not counted in any of your cases.
â€“Â YawarRaza7349
Sep 2 at 14:13

add a commentÂ |Â

2

I think the problem might be that the zeros don't have to all be at the end.
â€“Â Mike
Sep 2 at 6:55

but it wasn't mentioned right?
â€“Â Barath BR
Sep 2 at 12:54

1

The valid example the OP mentioned in the beginning, 0 10 11 10, starts with a 0, and thus is not counted in any of your cases.
â€“Â YawarRaza7349
Sep 2 at 14:13

I think the problem might be that the zeros don't have to all be at the end.
â€“Â Mike
Sep 2 at 6:55

but it wasn't mentioned right?
â€“Â Barath BR
Sep 2 at 12:54

The valid example the OP mentioned in the beginning, 0 10 11 10, starts with a 0, and thus is not counted in any of your cases.
â€“Â YawarRaza7349
Sep 2 at 14:13

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