In C++, Write a function that receives a string of characters and return true if the string is in language, otherwise it returns false. You may use following function header (Assume a string is in language if it is read from the left side is the same as it is read from the right side. For example, a-b-d-c is not in the language, a-b-c-a-c-b-a is in language): bool isInLanguage_2(string aString)
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In C++,
Write a function that receives a string of characters and return true if the string is in language, otherwise it returns false. You may use following function header (Assume a string is in language if it is read from the left side is the same as it is read from the right side. For example, a-b-d-c is not in the language, a-b-c-a-c-b-a is in language):
bool isInLanguage_2(string aString)
And I also, want to know if my if statement is incorrect, and if it is how is it incorrect?
![ebool isInlanguage_2(string string_input){
// Checking for the length and the $ sign!
if (!(string_input.Length() % 2)){
return false;
Stack<char> charStack;
for (int i = 0; i< string_input.Length(); i++){
charStack.push(string_input[i]);
}
int temp_2 = 0;
while (string_input[temp_2] != '\@'){
if (string_input[temp_2]
charStack.peek()){
==
charStack.pop();
temp_2++;
else
return false;
return true;](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F346b36e0-a4d8-47d7-8bd1-1ad776cb1cf3%2F0f69a268-7b5a-4b85-9b7e-f89bd3339a42%2F2j54c9_processed.png&w=3840&q=75)
- A program is to be written in a C++ language.
- The program can be compiled on any one of the online compilers.
- According to the program if the string entered is in the language then true is to be returned else false is returned.
- In the below program 0 is returned if it is not in a language that is false.
- If it is language 1 is returned, which is true according to the boolean.
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- #include<bits/stdc++.h>using namespace std;bool isPalindrome(string &s){ int start=0; int end=s.length()-1; while(start<=end) { if(s[start]!=s[end]) { return false; } start++; end--; } return true;}int main(){ string s; cout<<"ENTER STRING:"; getline(cin,s); int n=s.length(); bool flag=true; for(int i=1;i<n;i++) { string lowerHalf=s.substr(0,i); string upperHalf=s.substr(i,n-i); if(isPalindrome(lowerHalf) && isPalindrome(upperHalf)) { flag=false; cout<<"String A is:"<<lowerHalf<<"\n"; cout<<"String B is:"<<upperHalf<<"\n"; break; } } if(flag) { cout<<"NO\n"; } return 0;} change to stdio.h string.himport java.lang.System; 3. public class FibonacciComparison { 4. // Fibonacci Sequence: 0, 1, 1, 2, 3, 5, 8 .... /* 7 input cases 8. 1) 0 9 2) 3 10 3) -1 11 4) 9 12 output cases 13 1) 0 14 2) 2 15 3) 0 16 4) 34 17 */ // Note that you need to return 0 if the input is negative. // Please pay close attention to the fact that the first index in our fib sequence is 0. 18 19 20 // Recursive Fibonacci public static int fib(int n) { // Code this func. 21 22 23 24 return -1; 25 26 // Iterative Fibonacci 27 28 public static int fiblinear(int n) { // Code this func. 29 30 return -1; 31 32 33 public static void main(String[] args) { 34 // list of fibonacci sequence numbers int[] nlist w { 5,10, 15, 20, 25, 30, 35, 40, 45}; 35 36 37 // Two arrays (one for fibLinear, other for fibRecursive) to store time for each run. // There are a total of nlist.length inputs that we will test double[] timingsEF = new double[nlist.Length]; double[] timingsLF = new double[nlist.length]; 38 39 40 41 42 // Every…7. The following function converts a postfix expression to an infix expression assuming the expression can only contain unsigned integer numbers and +' and -' operators. 1. public static String convert (String postfix) { 2. String operand1, operand2, infix; 3. Stack s = new AStack; 4. int i = 0; 5. while (i < postfix.length) { char nextc = postfix.charAt (i); if ((nextc 6. 7. '+')|| (nextc == '-') { == operand2 = s.pop (); operandl = s.pop () ; infix = '(' + operandl + nextc + operand2 + ')'; s.push (infix); i++; 8. 9. 10. 11. } 12. else if (Character.isDigit (nextc)){ int start = i; while (Character.isDigit (postfix.charAt (i)) i++; infix = postfix.substring (start, i); s.push (infix); 13. 14. 15. 16. 17. } else i++; 18. 19. } 20 return s.pop () ; 21. } (a) Rewrite only the lines of code that need to be changed so that the above function converts a postfix expression to a prefix expression. (b) Describe in words what changes are needed on the above given convert () function so that it…
- Exercise Objectives Problem Description Write a program that reads a string and mirrors it around the middle character. Examples: abcd becomes cdab. abcde becomes deCab AhmadAlami becomes AlamiAhmad Page 1 of 2 Your program must: • Implement function void reflect (char* str) which receives a string (array of characters) and mirrors it. This function does not print anything. • Read from the user (in main()) a string and then print the string after calling function reflect(). • Use pointers and pointer arithmetic only. The use of array notation and/or functions from the string.h library is not allowed.Question 37 public static void main(String[] args) { Dog[] dogs = { new Dog(), new Dog()}; for(int i = 0; i >>"+decision()); } class Counter { private static int count; public static void inc() { count++;} public static int getCount() {return count;} } class Dog extends Counter{ public Dog(){} public void wo(){inc();} } class Cat extends Counter{ public Cat(){} public void me(){inc();} } The Correct answer: Nothing is output O 2 woofs and 5 mews O 2 woofs and 3 mews O 5 woofs and 5 mews O#include using namespace std; int find(int arr[], int value, int left, int right) { int midpt = (left+right)/2; if (left > right) return -1; if ( arr[midpt] return midpt; else if (arr[midpt] < value) == value) return else return find(arr,value, left,midpt-1); } void main(void) { int arr[] ={4,5,6,9,11}; cout<void funi (int iist , int s) { int sum=0; for (int i = 0; iData structure/ C language / Graph / Dijkstra’s algorithm implement a solution of a very common issue: howto get from one town to another using the shortest route.* design a solution that will let you find the shortest paths betweentwo input points in a graph, representing cities and towns, using Dijkstra’salgorithm. Your program should allow the user to enter the input filecontaining information of roads connecting cities/towns. The programshould then construct a graph based on the information provided from thefile. The user should then be able to enter pairs of cities/towns and thealgorithm should compute the shortest path between the two cities/townsentered.Attached a file containing a list of cities/towns with the following data:Field 1: Vertex ID of the 1st end of the segmentField 2: Vertex ID of the 2nd of the segmentField 3: Name of the townField 4: Distance in KilometerPlease note that all roads are two-ways. Meaning, a record may representboth the roads from feild1 to field2…void Path() { char[] userInput =" "; system("echo $PATH"); //append the path system("export PATH=$PATH:"); scanf("%c", userInput); printf("%c", "path + %c", userInput); } PLEASE FIX THIS FUNCTION!!! (C PROGRAMMING)The following function converts a postfix expression to an infix expression assuming the expression can only contain unsigned integer numbers and ‘+’ and ‘-’ operators. 1. public static String convert(String postfix) { 2. String operand1, operand2, infix; 3. Stack<String> s = new AStack<String>; 4. int i = 0; 5. while(i < postfix.length) { 6. char nextc = postfix.charAt(i); 7. if ((nextc == '+')||(nextc == '-') { 8. operand2 = s.pop(); operand1 = s.pop(); 9. infix = '(' + operand1 + nextc + operand2 + ')'; 10. s.push(infix); i++; 11. } 12. else if (Character.isDigit(nextc)){ 13. int start = i; 14. while(Character.isDigit(postfix.charAt(i)) i++; 15. infix = postfix.substring(start,i); 16. s.push(infix); 17. } 18. else i++; 19. } 20 return s.pop(); 21. } a) Rewrite only the lines of code that need to be changed so…Data structure/ C language / Graph / Dijkstra’s algorithm implement a solution of a very common issue: how to get from one town to another using the shortest route.* design a solution that will let you find the shortest paths between two input points in a graph, representing cities and towns, using Dijkstra’s algorithm. 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