public class Main{public static int greenTicket(int a, int b, int c){ if(a == b) { if(b == c) return 20; return 10; } if(a == c || b == c) return 10; return 0;}public static void main (String[] args) { System.out.println(greenTicket(2,2,2));}}
public class Main{
public static int greenTicket(int a, int b, int c)
// CPP program to convert infix to prefix #include <bits/stdc++.h> using namespace std; bool isOperator(char c) { return (!isalpha(c) && !isdigit(c)); } int getPriority(char C) { if (C == '-' || C == '+') return 1; else if (C == '*' || C == '/') return 2; else if (C == '^') rRead more
// CPP program to convert infix to prefix
#include <bits/stdc++.h>
using namespace std;
bool isOperator(char c)
{
return (!isalpha(c) && !isdigit(c));
}
int getPriority(char C)
{
if (C == '-' || C == '+')
return 1;
else if (C == '*' || C == '/')
return 2;
else if (C == '^')
return 3;
return 0;
}
string infixToPostfix(string infix)
{
infix = '(' + infix + ')';
int l = infix.size();
stack<char> char_stack;
string output;
for (int i = 0; i < l; i++) {
// If the scanned character is an
// operand, add it to output.
if (isalpha(infix[i]) || isdigit(infix[i]))
output += infix[i];
// If the scanned character is an
// ‘(‘, push it to the stack.
else if (infix[i] == '(')
char_stack.push('(');
// If the scanned character is an
// ‘)’, pop and output from the stack
// until an ‘(‘ is encountered.
else if (infix[i] == ')') {
while (char_stack.top() != '(') {
output += char_stack.top();
char_stack.pop();
}
// Remove '(' from the stack
char_stack.pop();
}
// Operator found
else
{
if (isOperator(char_stack.top()))
{
if(infix[i] == '^')
{
while (getPriority(infix[i]) <= getPriority(char_stack.top()))
{
output += char_stack.top();
char_stack.pop();
}
}
else
{
while (getPriority(infix[i]) < getPriority(char_stack.top()))
{
output += char_stack.top();
char_stack.pop();
}
}
// Push current Operator on stack
char_stack.push(infix[i]);
}
}
}
while(!char_stack.empty()){
output += char_stack.top();
char_stack.pop();
}
return output;
}
string infixToPrefix(string infix)
{
/* Reverse String
* Replace ( with ) and vice versa
* Get Postfix
* Reverse Postfix * */
int l = infix.size();
// Reverse infix
reverse(infix.begin(), infix.end());
// Replace ( with ) and vice versa
for (int i = 0; i < l; i++) {
if (infix[i] == '(') {
infix[i] = ')';
i++;
}
else if (infix[i] == ')') {
infix[i] = '(';
i++;
}
}
string prefix = infixToPostfix(infix);
// Reverse postfix
reverse(prefix.begin(), prefix.end());
return prefix;
}
// Driver code
int main()
{
string s = ("a+b");
cout << infixToPrefix(s) << std::endl;
return 0;
}
C Program to Traverse the Tree Non-Recursively.
#include <stdio.h> #include <stdlib.h> struct node { int a; struct node *left; struct node *right; }; void generate(struct node **, int); int search(struct node *, int); void delete(struct node **); int main() { struct node *head = NULL; int choice = 0, num, flag = 0, key; do { printf("\Read more
You have a green lottery ticket, with ints a, b, and c on it. If the numbers are all different …
i have updated the whole code now run by your own
i have updated the whole code now run by your own
See lessYou have a green lottery ticket, with ints a, b, and c on it. If the numbers are all different …
run the code by your own on compiler java change greenTickect(2,2,2) with your questions input to get proper output
run the code by your own on compiler java change greenTickect(2,2,2) with your questions input to get proper output
See lessYou have a green lottery ticket, with ints a, b, and c on it. If the numbers are all different …
public class Main{public static int greenTicket(int a, int b, int c){ if(a == b) { if(b == c) return 20; return 10; } if(a == c || b == c) return 10; return 0;}public static void main (String[] args) { System.out.println(greenTicket(2,2,2));}}
Give the infix expression of the following prefix expressions. (A) * – + A B C D (B) + – a * B C D
Give the infix expression of the following prefix expressions. (A) * – + A B C D (B) + – a * B C D TAP ON ATTACHMENT FOR ANSWER OF THIS QUESTION:
Give the infix expression of the following prefix expressions. (A) * – + A B C D (B) + – a * B C D
TAP ON ATTACHMENT FOR ANSWER OF THIS QUESTION:
See lessInfix to prefix
// CPP program to convert infix to prefix #include <bits/stdc++.h> using namespace std; bool isOperator(char c) { return (!isalpha(c) && !isdigit(c)); } int getPriority(char C) { if (C == '-' || C == '+') return 1; else if (C == '*' || C == '/') return 2; else if (C == '^') rRead more