You are given a string and 2 operators ( & , | ). Return the total number of ways in which you can evaluate to true using the given string and operator set.
Example : Input : TF Output : 1 Input : TFF Output : 2 ( T | F & F , T | F | F )
# | User | Rating |
---|---|---|
1 | tourist | 3985 |
2 | jiangly | 3814 |
3 | jqdai0815 | 3682 |
4 | Benq | 3529 |
5 | orzdevinwang | 3526 |
6 | ksun48 | 3517 |
7 | Radewoosh | 3410 |
8 | hos.lyric | 3399 |
9 | ecnerwala | 3392 |
9 | Um_nik | 3392 |
# | User | Contrib. |
---|---|---|
1 | cry | 169 |
2 | maomao90 | 162 |
2 | Um_nik | 162 |
4 | atcoder_official | 161 |
5 | djm03178 | 158 |
6 | -is-this-fft- | 157 |
7 | adamant | 155 |
8 | awoo | 154 |
8 | Dominater069 | 154 |
10 | luogu_official | 150 |
You are given a string and 2 operators ( & , | ). Return the total number of ways in which you can evaluate to true using the given string and operator set.
Example : Input : TF Output : 1 Input : TFF Output : 2 ( T | F & F , T | F | F )
Name |
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I have come across a problem like this before. But i dont know how to solve it using dp. Someone explain please
I will discuss it using recursion + memoization which can be easily converted to an iterative DP solution.
Let your state be where you are in the given string
i
, and what is the cumulative (Prefix) answer till nowlast
.Each time you can either AND this bit
s[i]
or OR it withlast
.After finishing all the string, if
last
is truereturn 1
, elsereturn 0
.So the code may be something like: Assuming that
s
is the given string,mem
is the memoization array filled initially by-1
.