给定一个带有以下符号的布尔表达式。
Symbols
'T' ---> true
'F' ---> false
并在符号之间填充以下运算符符
Operators
& ---> boolean AND
| ---> boolean OR
^ ---> boolean XOR
计算我们可以为表达式加上括号的方式的数量,以便表达式的值计算为真。
让输入采用两个数组的形式,一个包含按顺序的符号(T 和 F),另一个包含运算符(&、| 和 ^}
例子:
Input: symbol[] = {T, F, T}
operator[] = {^, &}
Output: 2
The given expression is "T ^ F & T", it evaluates true
in two ways "((T ^ F) & T)" and "(T ^ (F & T))"
Input: symbol[] = {T, F, F}
operator[] = {^, |}
Output: 2
The given expression is "T ^ F | F", it evaluates true
in two ways "( (T ^ F) | F )" and "( T ^ (F | F) )".
Input: symbol[] = {T, T, F, T}
operator[] = {|, &, ^}
Output: 4
The given expression is "T | T & F ^ T", it evaluates true
in 4 ways ((T|T)&(F^T)), (T|(T&(F^T))), (((T|T)&F)^T)
and (T|((T&F)^T)).
解决方案:
令T(i, j)表示将 i 和 j(两者都包括在内)之间的符号括起来的方法的数量,以便 i 和 j 之间的子表达式的计算结果为真。
–>
令F(i, j)表示将 i 和 j(两者都包括在内)之间的符号括起来以使 i 和 j 之间的子表达式计算为假的方法数。
–>
基本案例:
T(i, i) = 1 if symbol[i] = 'T'
T(i, i) = 0 if symbol[i] = 'F'
F(i, i) = 1 if symbol[i] = 'F'
F(i, i) = 0 if symbol[i] = 'T'
如果我们绘制上述递归解的递归树,我们可以观察到它有许多重叠的子问题。与其他动态规划问题一样,它可以通过自底向上填充表格来解决。以下是动态编程解决方案的 C++ 实现。
C++
#include
#include
using namespace std;
// Returns count of all possible
// parenthesizations that lead
// to result true for a boolean
// expression with symbols like
// true and false and operators
// like &, | and ^ filled
// between symbols
int countParenth(char symb[], char oper[], int n)
{
int F[n][n], T[n][n];
// Fill diaginal entries first
// All diagonal entries in
// T[i][i] are 1 if symbol[i]
// is T (true). Similarly,
// all F[i][i] entries are 1 if
// symbol[i] is F (False)
for (int i = 0; i < n; i++) {
F[i][i] = (symb[i] == 'F') ? 1 : 0;
T[i][i] = (symb[i] == 'T') ? 1 : 0;
}
// Now fill T[i][i+1],
// T[i][i+2], T[i][i+3]... in order
// And F[i][i+1], F[i][i+2],
// F[i][i+3]... in order
for (int gap = 1; gap < n; ++gap)
{
for (int i = 0, j = gap;
j < n; ++i, ++j)
{
T[i][j] = F[i][j] = 0;
for (int g = 0;
g < gap; g++)
{
// Find place of parenthesization using
// current value of gap
int k = i + g;
// Store Total[i][k]
// and Total[k+1][j]
int tik = T[i][k] + F[i][k];
int tkj = T[k + 1][j]
+ F[k + 1][j];
// Follow the recursive formulas
// according
// to the current operator
if (oper[k] == '&') {
T[i][j] += T[i][k]
* T[k + 1][j];
F[i][j] += (tik * tkj
- T[i][k]
* T[k + 1][j]);
}
if (oper[k] == '|') {
F[i][j] += F[i][k]
* F[k + 1][j];
T[i][j] += (tik * tkj
- F[i][k]
* F[k + 1][j]);
}
if (oper[k] == '^') {
T[i][j] += F[i][k]
* T[k + 1][j]
+ T[i][k]
* F[k + 1][j];
F[i][j] += T[i][k]
* T[k + 1][j]
+ F[i][k] * F[k + 1][j];
}
}
}
}
return T[0][n - 1];
}
// Driver code
int main()
{
char symbols[] = "TTFT";
char operators[] = "|&^";
int n = strlen(symbols);
// There are 4 ways
// ((T|T)&(F^T)), (T|(T&(F^T))), (((T|T)&F)^T) and
// (T|((T&F)^T))
cout << countParenth(symbols, operators, n);
return 0;
}
Java
class GFG {
// Returns count of all possible
// parenthesizations that lead to
// result true for a boolean
// expression with symbols like true
// and false and operators like &, |
// and ^ filled between symbols
static int countParenth(char symb[],
char oper[],
int n)
{
int F[][] = new int[n][n];
int T[][] = new int[n][n];
// Fill diaginal entries first
// All diagonal entries in T[i][i]
// are 1 if symbol[i] is T (true).
// Similarly, all F[i][i] entries
// are 1 if symbol[i] is F (False)
for (int i = 0; i < n; i++) {
F[i][i] = (symb[i] == 'F') ? 1 : 0;
T[i][i] = (symb[i] == 'T') ? 1 : 0;
}
// Now fill T[i][i+1], T[i][i+2],
// T[i][i+3]... in order And F[i][i+1],
// F[i][i+2], F[i][i+3]... in order
for (int gap = 1; gap < n; ++gap)
{
for (int i = 0,
j = gap; j < n;
++i, ++j)
{
T[i][j] = F[i][j] = 0;
for (int g = 0; g < gap; g++)
{
// Find place of parenthesization
// using current value of gap
int k = i + g;
// Store Total[i][k]
// and Total[k+1][j]
int tik = T[i][k]
+ F[i][k];
int tkj = T[k + 1][j]
+ F[k + 1][j];
// Follow the recursive formulas
// according to the current operator
if (oper[k] == '&')
{
T[i][j] += T[i][k]
* T[k + 1][j];
F[i][j]
+= (tik * tkj
- T[i][k]
* T[k + 1][j]);
}
if (oper[k] == '|')
{
F[i][j] += F[i][k]
* F[k + 1][j];
T[i][j]
+= (tik * tkj
- F[i][k]
* F[k + 1][j]);
}
if (oper[k] == '^')
{
T[i][j] += F[i][k]
* T[k + 1][j]
+ T[i][k]
* F[k + 1][j];
F[i][j] += T[i][k]
* T[k + 1][j]
+ F[i][k]
* F[k + 1][j];
}
}
}
}
return T[0][n - 1];
}
// Driver code
public static void main(String[] args)
{
char symbols[] = "TTFT".toCharArray();
char operators[] = "|&^".toCharArray();
int n = symbols.length;
// There are 4 ways
// ((T|T)&(F^T)), (T|(T&(F^T))),
// (((T|T)&F)^T) and (T|((T&F)^T))
System.out.println(
countParenth(symbols, operators, n));
}
}
// This code has been contributed
// by 29AjayKumar
Python
# Returns count of all possible
# parenthesizations that lead to
# result true for a boolean
# expression with symbols like
# true and false and operators
# like &, | and ^ filled between symbols
def countParenth(symb, oper, n):
F = [[0 for i in range(n + 1)]
for i in range(n + 1)]
T = [[0 for i in range(n + 1)]
for i in range(n + 1)]
# Fill diaginal entries first
# All diagonal entries in
# T[i][i] are 1 if symbol[i]
# is T (true). Similarly, all
# F[i][i] entries are 1 if
# symbol[i] is F (False)
for i in range(n):
if symb[i] == 'F':
F[i][i] = 1
else:
F[i][i] = 0
if symb[i] == 'T':
T[i][i] = 1
else:
T[i][i] = 0
# Now fill T[i][i+1], T[i][i+2],
# T[i][i+3]... in order And
# F[i][i+1], F[i][i+2],
# F[i][i+3]... in order
for gap in range(1, n):
i = 0
for j in range(gap, n):
T[i][j] = F[i][j] = 0
for g in range(gap):
# Find place of parenthesization
# using current value of gap
k = i + g
# Store Total[i][k] and Total[k+1][j]
tik = T[i][k] + F[i][k]
tkj = T[k + 1][j] + F[k + 1][j]
# Follow the recursive formulas
# according to the current operator
if oper[k] == '&':
T[i][j] += T[i][k] * T[k + 1][j]
F[i][j] += (tik * tkj - T[i][k] *
T[k + 1][j])
if oper[k] == '|':
F[i][j] += F[i][k] * F[k + 1][j]
T[i][j] += (tik * tkj - F[i][k] *
F[k + 1][j])
if oper[k] == '^':
T[i][j] += (F[i][k] * T[k + 1][j] +
T[i][k] * F[k + 1][j])
F[i][j] += (T[i][k] * T[k + 1][j] +
F[i][k] * F[k + 1][j])
i += 1
return T[0][n - 1]
# Driver Code
symbols = "TTFT"
operators = "|&^"
n = len(symbols)
# There are 4 ways
# ((T|T)&(F^T)), (T|(T&(F^T))),
# (((T|T)&F)^T) and (T|((T&F)^T))
print(countParenth(symbols, operators, n))
# This code is contributed by
# sahil shelangia
C#
// C# program of above approach
using System;
class GFG
{
// Returns count of all possible
// parenthesizations that lead to
// result true for a boolean
// expression with symbols like true
// and false and operators like &, |
// and ^ filled between symbols
static int countParenth(char []symb,
char []oper, int n)
{
int [,]F = new int[n, n];
int [,]T = new int[n, n];
// Fill diaginal entries first
// All diagonal entries in T[i,i]
// are 1 if symbol[i] is T (true).
// Similarly, all F[i,i] entries
// are 1 if symbol[i] is F (False)
for (int i = 0; i < n; i++)
{
F[i,i] = (symb[i] == 'F') ? 1 : 0;
T[i,i] = (symb[i] == 'T') ? 1 : 0;
}
// Now fill T[i,i+1], T[i,i+2],
// T[i,i+3]... in order And F[i,i+1],
// F[i,i+2], F[i,i+3]... in order
for (int gap = 1; gap < n; ++gap)
{
for (int i = 0, j = gap; j < n; ++i, ++j)
{
T[i, j] = F[i, j] = 0;
for (int g = 0; g < gap; g++)
{
// Find place of parenthesization
// using current value of gap
int k = i + g;
// Store Total[i,k] and Total[k+1,j]
int tik = T[i, k] + F[i, k];
int tkj = T[k + 1, j] + F[k + 1, j];
// Follow the recursive formulas
// according to the current operator
if (oper[k] == '&')
{
T[i, j] += T[i, k]
* T[k + 1, j];
F[i, j] += (tik * tkj
- T[i, k] * T[k + 1, j]);
}
if (oper[k] == '|')
{
F[i,j] += F[i, k]
* F[k + 1, j];
T[i,j] += (tik
* tkj - F[i, k]
* F[k + 1, j]);
}
if (oper[k] == '^')
{
T[i, j] += F[i, k] * T[k + 1, j] +
T[i, k] * F[k + 1, j];
F[i, j] += T[i, k] * T[k + 1, j] +
F[i, k] * F[k + 1, j];
}
}
}
}
return T[0,n - 1];
}
// Driver code
public static void Main()
{
char []symbols = "TTFT".ToCharArray();
char []operators = "|&^".ToCharArray();
int n = symbols.Length;
// There are 4 ways
// ((T|T)&(F^T)), (T|(T&(F^T))),
// (((T|T)&F)^T) and (T|((T&F)^T))
Console.WriteLine(countParenth(symbols,
operators, n));
}
}
/* This code contributed by PrinciRaj1992 */
Javascript
C++
#include
using namespace std;
int dp[101][101][2];
int parenthesis_count(string s,
int i,
int j,
int isTrue)
{
// Base Condition
if (i > j)
return false;
if (i == j) {
if (isTrue == 1)
return s[i] == 'T';
else
return s[i] == 'F';
}
if (dp[i][j][isTrue] != -1)
return dp[i][j][isTrue];
int ans = 0;
for (int k = i + 1
; k <= j - 1; k = k + 2)
{
int leftF, leftT, rightT, rightF;
if (dp[i][k - 1][1] == -1)
{
leftT = parenthesis_count(s, i, k - 1, 1);
} // Count no. of T in left partition
else {
leftT = dp[i][k - 1][1];
}
if (dp[k + 1][j][1] == -1)
{
rightT = parenthesis_count(s, k + 1, j, 1);
} // Count no. of T in right partition
else
{
rightT = dp[k + 1][j][1];
}
if (dp[i][k - 1][0] == -1)
{
// Count no. of F in left partition
leftF = parenthesis_count(s, i, k - 1, 0);
}
else
{
leftF = dp[i][k - 1][0];
}
if (dp[k + 1][j][0] == -1)
{
// Count no. of F in right partition
rightF = parenthesis_count(s, k + 1, j, 0);
}
else
{
rightF = dp[k + 1][j][0];
}
if (s[k] == '&')
{
if (isTrue == 1)
ans += leftT * rightT;
else
ans += leftF * rightF + leftT * rightF
+ leftF * rightT;
}
else if (s[k] == '|')
{
if (isTrue == 1)
ans += leftT * rightT + leftT * rightF
+ leftF * rightT;
else
ans = ans + leftF * rightF;
}
else if (s[k] == '^')
{
if (isTrue == 1)
ans = ans + leftF * rightT + leftT * rightF;
else
ans = ans + leftT * rightT + leftF * rightF;
}
dp[i][j][isTrue] = ans;
}
return ans;
}
// Driver Code
int main()
{
string symbols = "TTFT";
string operators = "|&^";
string s;
int j = 0;
for (int i = 0; i < symbols.length(); i++)
{
s.push_back(symbols[i]);
if (j < operators.length())
s.push_back(operators[j++]);
}
// We obtain the string T|T&F^T
int n = s.length();
// There are 4 ways
// ((T|T)&(F^T)), (T|(T&(F^T))), (((T|T)&F)^T) and
// (T|((T&F)^T))
memset(dp, -1, sizeof(dp));
cout << parenthesis_count(s, 0, n - 1, 1);
return 0;
}
Java
import java.io.*;
import java.util.*;
class GFG {
public static int countWays(int N, String S)
{
int dp[][][] = new int[N + 1][N + 1][2];
for (int row[][] : dp)
for (int col[] : row)
Arrays.fill(col, -1);
return parenthesis_count(S, 0, N - 1, 1, dp);
}
public static int parenthesis_count(String str, int i,
int j, int isTrue,
int[][][] dp)
{
if (i > j)
return 0;
if (i == j)
{
if (isTrue == 1)
{
return (str.charAt(i) == 'T') ? 1 : 0;
}
else
{
return (str.charAt(i) == 'F') ? 1 : 0;
}
}
if (dp[i][j][isTrue] != -1)
return dp[i][j][isTrue];
int temp_ans = 0;
int leftTrue, rightTrue, leftFalse, rightFalse;
for (int k = i + 1; k <= j - 1; k = k + 2)
{
if (dp[i][k - 1][1] != -1)
leftTrue = dp[i][k - 1][1];
else
{
// Count number of True in left Partition
leftTrue = parenthesis_count(str, i, k - 1,
1, dp);
}
if (dp[i][k - 1][0] != -1)
leftFalse = dp[i][k - 1][0];
else
{
// Count number of False in left Partition
leftFalse = parenthesis_count(str, i, k - 1,
0, dp);
}
if (dp[k + 1][j][1] != -1)
rightTrue = dp[k + 1][j][1];
else
{
// Count number of True in right Partition
rightTrue = parenthesis_count(str, k + 1, j,
1, dp);
}
if (dp[k + 1][j][0] != -1)
rightFalse = dp[k + 1][j][0];
else
{
// Count number of False in right Partition
rightFalse = parenthesis_count(str, k + 1,
j, 0, dp);
}
// Evaluate AND operation
if (str.charAt(k) == '&')
{
if (isTrue == 1)
{
temp_ans
= temp_ans + leftTrue * rightTrue;
}
else
{
temp_ans = temp_ans
+ leftTrue * rightFalse
+ leftFalse * rightTrue
+ leftFalse * rightFalse;
}
}
// Evaluate OR operation
else if (str.charAt(k) == '|')
{
if (isTrue == 1)
{
temp_ans = temp_ans
+ leftTrue * rightTrue
+ leftTrue * rightFalse
+ leftFalse * rightTrue;
}
else
{
temp_ans
= temp_ans + leftFalse * rightFalse;
}
}
// Evaluate XOR operation
else if (str.charAt(k) == '^')
{
if (isTrue == 1)
{
temp_ans = temp_ans
+ leftTrue * rightFalse
+ leftFalse * rightTrue;
}
else
{
temp_ans = temp_ans
+ leftTrue * rightTrue
+ leftFalse * rightFalse;
}
}
dp[i][j][isTrue] = temp_ans;
}
return temp_ans;
}
// Driver code
public static void main(String[] args)
{
String symbols = "TTFT";
String operators = "|&^";
StringBuilder S = new StringBuilder();
int j = 0;
for (int i = 0; i < symbols.length(); i++)
{
S.append(symbols.charAt(i));
if (j < operators.length())
S.append(operators.charAt(j++));
}
// We obtain the string T|T&F^T
int N = S.length();
// There are 4 ways
// ((T|T)&(F^T)), (T|(T&(F^T))), (((T|T)&F)^T) and
// (T|((T&F)^T))
System.out.println(countWays(N, S.toString()));
}
}
// This code is contributed by farheenbano.
Python3
def parenthesis_count(Str, i, j, isTrue, dp) :
if (i > j) :
return 0
if (i == j) :
if (isTrue == 1) :
return 1 if Str[i] == 'T' else 0
else :
return 1 if Str[i] == 'F' else 0
if (dp[i][j][isTrue] != -1) :
return dp[i][j][isTrue]
temp_ans = 0
for k in range(i + 1, j, 2) :
if (dp[i][k - 1][1] != -1) :
leftTrue = dp[i][k - 1][1]
else :
# Count number of True in left Partition
leftTrue = parenthesis_count(Str, i, k - 1, 1, dp)
if (dp[i][k - 1][0] != -1) :
leftFalse = dp[i][k - 1][0]
else :
# Count number of False in left Partition
leftFalse = parenthesis_count(Str, i, k - 1, 0, dp)
if (dp[k + 1][j][1] != -1) :
rightTrue = dp[k + 1][j][1]
else :
# Count number of True in right Partition
rightTrue = parenthesis_count(Str, k + 1, j, 1, dp)
if (dp[k + 1][j][0] != -1) :
rightFalse = dp[k + 1][j][0]
else :
# Count number of False in right Partition
rightFalse = parenthesis_count(Str, k + 1, j, 0, dp)
# Evaluate AND operation
if (Str[k] == '&') :
if (isTrue == 1) :
temp_ans = temp_ans + leftTrue * rightTrue
else :
temp_ans = temp_ans + leftTrue * rightFalse + leftFalse * rightTrue + leftFalse * rightFalse
# Evaluate OR operation
elif (Str[k] == '|') :
if (isTrue == 1) :
temp_ans = temp_ans + leftTrue * rightTrue + leftTrue * rightFalse + leftFalse * rightTrue
else :
temp_ans = temp_ans + leftFalse * rightFalse
# Evaluate XOR operation
elif (Str[k] == '^') :
if (isTrue == 1) :
temp_ans = temp_ans + leftTrue * rightFalse + leftFalse * rightTrue
else :
temp_ans = temp_ans + leftTrue * rightTrue + leftFalse * rightFalse
dp[i][j][isTrue] = temp_ans
return temp_ans
def countWays(N, S) :
dp = [[[-1 for k in range(2)] for i in range(N + 1)] for j in range(N + 1)]
return parenthesis_count(S, 0, N - 1, 1, dp)
symbols = "TTFT"
operators = "|&^"
S = ""
j = 0
for i in range(len(symbols)) :
S = S + symbols[i]
if (j < len(operators)) :
S = S + operators[j]
j += 1
# We obtain the string T|T&F^T
N = len(S)
# There are 4 ways
# ((T|T)&(F^T)), (T|(T&(F^T))), (((T|T)&F)^T) and
# (T|((T&F)^T))
print(countWays(N, S))
# This code is contributed by divyesh072019
C#
using System;
class GFG
{
static int parenthesis_count(string str, int i,
int j, int isTrue,
int[,,] dp)
{
if (i > j)
return 0;
if (i == j)
{
if (isTrue == 1)
{
return (str[i] == 'T') ? 1 : 0;
}
else
{
return (str[i] == 'F') ? 1 : 0;
}
}
if (dp[i, j, isTrue] != -1)
return dp[i, j, isTrue];
int temp_ans = 0;
int leftTrue, rightTrue, leftFalse, rightFalse;
for (int k = i + 1; k <= j - 1; k = k + 2)
{
if (dp[i, k - 1, 1] != -1)
leftTrue = dp[i, k - 1, 1];
else
{
// Count number of True in left Partition
leftTrue = parenthesis_count(str, i, k - 1, 1, dp);
}
if (dp[i, k - 1, 0] != -1)
leftFalse = dp[i, k - 1, 0];
else
{
// Count number of False in left Partition
leftFalse = parenthesis_count(str, i, k - 1, 0, dp);
}
if (dp[k + 1, j, 1] != -1)
rightTrue = dp[k + 1, j, 1];
else
{
// Count number of True in right Partition
rightTrue = parenthesis_count(str, k + 1, j,
1, dp);
}
if (dp[k + 1, j, 0] != -1)
rightFalse = dp[k + 1, j, 0];
else
{
// Count number of False in right Partition
rightFalse = parenthesis_count(str, k + 1,
j, 0, dp);
}
// Evaluate AND operation
if (str[k] == '&')
{
if (isTrue == 1)
{
temp_ans
= temp_ans + leftTrue * rightTrue;
}
else
{
temp_ans = temp_ans
+ leftTrue * rightFalse
+ leftFalse * rightTrue
+ leftFalse * rightFalse;
}
}
// Evaluate OR operation
else if (str[k] == '|')
{
if (isTrue == 1)
{
temp_ans = temp_ans
+ leftTrue * rightTrue
+ leftTrue * rightFalse
+ leftFalse * rightTrue;
}
else
{
temp_ans
= temp_ans + leftFalse * rightFalse;
}
}
// Evaluate XOR operation
else if (str[k] == '^')
{
if (isTrue == 1)
{
temp_ans = temp_ans
+ leftTrue * rightFalse
+ leftFalse * rightTrue;
}
else
{
temp_ans = temp_ans
+ leftTrue * rightTrue
+ leftFalse * rightFalse;
}
}
dp[i, j, isTrue] = temp_ans;
}
return temp_ans;
}
static int countWays(int N, string S)
{
int[,,] dp = new int[N + 1, N + 1, 2];
for(int i = 0; i < (N + 1); i++)
{
for(int j = 0; j < (N + 1); j++)
{
for(int k = 0; k < 2; k++)
{
dp[i, j, k] = -1;
}
}
}
return parenthesis_count(S, 0, N - 1, 1, dp);
}
// Driver code
static void Main()
{
string symbols = "TTFT";
string operators = "|&^";
string S = "";
int j = 0;
for (int i = 0; i < symbols.Length; i++)
{
S = S + symbols[i];
if (j < operators.Length)
S = S + operators[j++];
}
// We obtain the string T|T&F^T
int N = S.Length;
// There are 4 ways
// ((T|T)&(F^T)), (T|(T&(F^T))), (((T|T)&F)^T) and
// (T|((T&F)^T))
Console.WriteLine(countWays(N, S));
}
}
// This code is contributed by divyeshrabadiya07.
Javascript
输出:
4
时间复杂度: O(n 3 )
辅助空间: O(n 2 )
方法二:
我们也可以使用递归方法(Top Down dp),这种方法使用记忆化。
C++
#include
using namespace std;
int dp[101][101][2];
int parenthesis_count(string s,
int i,
int j,
int isTrue)
{
// Base Condition
if (i > j)
return false;
if (i == j) {
if (isTrue == 1)
return s[i] == 'T';
else
return s[i] == 'F';
}
if (dp[i][j][isTrue] != -1)
return dp[i][j][isTrue];
int ans = 0;
for (int k = i + 1
; k <= j - 1; k = k + 2)
{
int leftF, leftT, rightT, rightF;
if (dp[i][k - 1][1] == -1)
{
leftT = parenthesis_count(s, i, k - 1, 1);
} // Count no. of T in left partition
else {
leftT = dp[i][k - 1][1];
}
if (dp[k + 1][j][1] == -1)
{
rightT = parenthesis_count(s, k + 1, j, 1);
} // Count no. of T in right partition
else
{
rightT = dp[k + 1][j][1];
}
if (dp[i][k - 1][0] == -1)
{
// Count no. of F in left partition
leftF = parenthesis_count(s, i, k - 1, 0);
}
else
{
leftF = dp[i][k - 1][0];
}
if (dp[k + 1][j][0] == -1)
{
// Count no. of F in right partition
rightF = parenthesis_count(s, k + 1, j, 0);
}
else
{
rightF = dp[k + 1][j][0];
}
if (s[k] == '&')
{
if (isTrue == 1)
ans += leftT * rightT;
else
ans += leftF * rightF + leftT * rightF
+ leftF * rightT;
}
else if (s[k] == '|')
{
if (isTrue == 1)
ans += leftT * rightT + leftT * rightF
+ leftF * rightT;
else
ans = ans + leftF * rightF;
}
else if (s[k] == '^')
{
if (isTrue == 1)
ans = ans + leftF * rightT + leftT * rightF;
else
ans = ans + leftT * rightT + leftF * rightF;
}
dp[i][j][isTrue] = ans;
}
return ans;
}
// Driver Code
int main()
{
string symbols = "TTFT";
string operators = "|&^";
string s;
int j = 0;
for (int i = 0; i < symbols.length(); i++)
{
s.push_back(symbols[i]);
if (j < operators.length())
s.push_back(operators[j++]);
}
// We obtain the string T|T&F^T
int n = s.length();
// There are 4 ways
// ((T|T)&(F^T)), (T|(T&(F^T))), (((T|T)&F)^T) and
// (T|((T&F)^T))
memset(dp, -1, sizeof(dp));
cout << parenthesis_count(s, 0, n - 1, 1);
return 0;
}
Java
import java.io.*;
import java.util.*;
class GFG {
public static int countWays(int N, String S)
{
int dp[][][] = new int[N + 1][N + 1][2];
for (int row[][] : dp)
for (int col[] : row)
Arrays.fill(col, -1);
return parenthesis_count(S, 0, N - 1, 1, dp);
}
public static int parenthesis_count(String str, int i,
int j, int isTrue,
int[][][] dp)
{
if (i > j)
return 0;
if (i == j)
{
if (isTrue == 1)
{
return (str.charAt(i) == 'T') ? 1 : 0;
}
else
{
return (str.charAt(i) == 'F') ? 1 : 0;
}
}
if (dp[i][j][isTrue] != -1)
return dp[i][j][isTrue];
int temp_ans = 0;
int leftTrue, rightTrue, leftFalse, rightFalse;
for (int k = i + 1; k <= j - 1; k = k + 2)
{
if (dp[i][k - 1][1] != -1)
leftTrue = dp[i][k - 1][1];
else
{
// Count number of True in left Partition
leftTrue = parenthesis_count(str, i, k - 1,
1, dp);
}
if (dp[i][k - 1][0] != -1)
leftFalse = dp[i][k - 1][0];
else
{
// Count number of False in left Partition
leftFalse = parenthesis_count(str, i, k - 1,
0, dp);
}
if (dp[k + 1][j][1] != -1)
rightTrue = dp[k + 1][j][1];
else
{
// Count number of True in right Partition
rightTrue = parenthesis_count(str, k + 1, j,
1, dp);
}
if (dp[k + 1][j][0] != -1)
rightFalse = dp[k + 1][j][0];
else
{
// Count number of False in right Partition
rightFalse = parenthesis_count(str, k + 1,
j, 0, dp);
}
// Evaluate AND operation
if (str.charAt(k) == '&')
{
if (isTrue == 1)
{
temp_ans
= temp_ans + leftTrue * rightTrue;
}
else
{
temp_ans = temp_ans
+ leftTrue * rightFalse
+ leftFalse * rightTrue
+ leftFalse * rightFalse;
}
}
// Evaluate OR operation
else if (str.charAt(k) == '|')
{
if (isTrue == 1)
{
temp_ans = temp_ans
+ leftTrue * rightTrue
+ leftTrue * rightFalse
+ leftFalse * rightTrue;
}
else
{
temp_ans
= temp_ans + leftFalse * rightFalse;
}
}
// Evaluate XOR operation
else if (str.charAt(k) == '^')
{
if (isTrue == 1)
{
temp_ans = temp_ans
+ leftTrue * rightFalse
+ leftFalse * rightTrue;
}
else
{
temp_ans = temp_ans
+ leftTrue * rightTrue
+ leftFalse * rightFalse;
}
}
dp[i][j][isTrue] = temp_ans;
}
return temp_ans;
}
// Driver code
public static void main(String[] args)
{
String symbols = "TTFT";
String operators = "|&^";
StringBuilder S = new StringBuilder();
int j = 0;
for (int i = 0; i < symbols.length(); i++)
{
S.append(symbols.charAt(i));
if (j < operators.length())
S.append(operators.charAt(j++));
}
// We obtain the string T|T&F^T
int N = S.length();
// There are 4 ways
// ((T|T)&(F^T)), (T|(T&(F^T))), (((T|T)&F)^T) and
// (T|((T&F)^T))
System.out.println(countWays(N, S.toString()));
}
}
// This code is contributed by farheenbano.
蟒蛇3
def parenthesis_count(Str, i, j, isTrue, dp) :
if (i > j) :
return 0
if (i == j) :
if (isTrue == 1) :
return 1 if Str[i] == 'T' else 0
else :
return 1 if Str[i] == 'F' else 0
if (dp[i][j][isTrue] != -1) :
return dp[i][j][isTrue]
temp_ans = 0
for k in range(i + 1, j, 2) :
if (dp[i][k - 1][1] != -1) :
leftTrue = dp[i][k - 1][1]
else :
# Count number of True in left Partition
leftTrue = parenthesis_count(Str, i, k - 1, 1, dp)
if (dp[i][k - 1][0] != -1) :
leftFalse = dp[i][k - 1][0]
else :
# Count number of False in left Partition
leftFalse = parenthesis_count(Str, i, k - 1, 0, dp)
if (dp[k + 1][j][1] != -1) :
rightTrue = dp[k + 1][j][1]
else :
# Count number of True in right Partition
rightTrue = parenthesis_count(Str, k + 1, j, 1, dp)
if (dp[k + 1][j][0] != -1) :
rightFalse = dp[k + 1][j][0]
else :
# Count number of False in right Partition
rightFalse = parenthesis_count(Str, k + 1, j, 0, dp)
# Evaluate AND operation
if (Str[k] == '&') :
if (isTrue == 1) :
temp_ans = temp_ans + leftTrue * rightTrue
else :
temp_ans = temp_ans + leftTrue * rightFalse + leftFalse * rightTrue + leftFalse * rightFalse
# Evaluate OR operation
elif (Str[k] == '|') :
if (isTrue == 1) :
temp_ans = temp_ans + leftTrue * rightTrue + leftTrue * rightFalse + leftFalse * rightTrue
else :
temp_ans = temp_ans + leftFalse * rightFalse
# Evaluate XOR operation
elif (Str[k] == '^') :
if (isTrue == 1) :
temp_ans = temp_ans + leftTrue * rightFalse + leftFalse * rightTrue
else :
temp_ans = temp_ans + leftTrue * rightTrue + leftFalse * rightFalse
dp[i][j][isTrue] = temp_ans
return temp_ans
def countWays(N, S) :
dp = [[[-1 for k in range(2)] for i in range(N + 1)] for j in range(N + 1)]
return parenthesis_count(S, 0, N - 1, 1, dp)
symbols = "TTFT"
operators = "|&^"
S = ""
j = 0
for i in range(len(symbols)) :
S = S + symbols[i]
if (j < len(operators)) :
S = S + operators[j]
j += 1
# We obtain the string T|T&F^T
N = len(S)
# There are 4 ways
# ((T|T)&(F^T)), (T|(T&(F^T))), (((T|T)&F)^T) and
# (T|((T&F)^T))
print(countWays(N, S))
# This code is contributed by divyesh072019
C#
using System;
class GFG
{
static int parenthesis_count(string str, int i,
int j, int isTrue,
int[,,] dp)
{
if (i > j)
return 0;
if (i == j)
{
if (isTrue == 1)
{
return (str[i] == 'T') ? 1 : 0;
}
else
{
return (str[i] == 'F') ? 1 : 0;
}
}
if (dp[i, j, isTrue] != -1)
return dp[i, j, isTrue];
int temp_ans = 0;
int leftTrue, rightTrue, leftFalse, rightFalse;
for (int k = i + 1; k <= j - 1; k = k + 2)
{
if (dp[i, k - 1, 1] != -1)
leftTrue = dp[i, k - 1, 1];
else
{
// Count number of True in left Partition
leftTrue = parenthesis_count(str, i, k - 1, 1, dp);
}
if (dp[i, k - 1, 0] != -1)
leftFalse = dp[i, k - 1, 0];
else
{
// Count number of False in left Partition
leftFalse = parenthesis_count(str, i, k - 1, 0, dp);
}
if (dp[k + 1, j, 1] != -1)
rightTrue = dp[k + 1, j, 1];
else
{
// Count number of True in right Partition
rightTrue = parenthesis_count(str, k + 1, j,
1, dp);
}
if (dp[k + 1, j, 0] != -1)
rightFalse = dp[k + 1, j, 0];
else
{
// Count number of False in right Partition
rightFalse = parenthesis_count(str, k + 1,
j, 0, dp);
}
// Evaluate AND operation
if (str[k] == '&')
{
if (isTrue == 1)
{
temp_ans
= temp_ans + leftTrue * rightTrue;
}
else
{
temp_ans = temp_ans
+ leftTrue * rightFalse
+ leftFalse * rightTrue
+ leftFalse * rightFalse;
}
}
// Evaluate OR operation
else if (str[k] == '|')
{
if (isTrue == 1)
{
temp_ans = temp_ans
+ leftTrue * rightTrue
+ leftTrue * rightFalse
+ leftFalse * rightTrue;
}
else
{
temp_ans
= temp_ans + leftFalse * rightFalse;
}
}
// Evaluate XOR operation
else if (str[k] == '^')
{
if (isTrue == 1)
{
temp_ans = temp_ans
+ leftTrue * rightFalse
+ leftFalse * rightTrue;
}
else
{
temp_ans = temp_ans
+ leftTrue * rightTrue
+ leftFalse * rightFalse;
}
}
dp[i, j, isTrue] = temp_ans;
}
return temp_ans;
}
static int countWays(int N, string S)
{
int[,,] dp = new int[N + 1, N + 1, 2];
for(int i = 0; i < (N + 1); i++)
{
for(int j = 0; j < (N + 1); j++)
{
for(int k = 0; k < 2; k++)
{
dp[i, j, k] = -1;
}
}
}
return parenthesis_count(S, 0, N - 1, 1, dp);
}
// Driver code
static void Main()
{
string symbols = "TTFT";
string operators = "|&^";
string S = "";
int j = 0;
for (int i = 0; i < symbols.Length; i++)
{
S = S + symbols[i];
if (j < operators.Length)
S = S + operators[j++];
}
// We obtain the string T|T&F^T
int N = S.Length;
// There are 4 ways
// ((T|T)&(F^T)), (T|(T&(F^T))), (((T|T)&F)^T) and
// (T|((T&F)^T))
Console.WriteLine(countWays(N, S));
}
}
// This code is contributed by divyeshrabadiya07.
Javascript
输出
4
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