c++ coding language   I need help with part B and C please. If you are unable to do both, then PLEASE prioritize part C. I am really stuck and really can use the help.       This is the code for c that was provided in order to guide me: const int N =31;    // N parking spaces bool parking[N];   // the garage   void EmptyTheLot(bool parking[], int N) {    for(int i=0; i<N; i++) p[i]=false;       // empty space }   // returns -1 if no space found,  //otherwise it returns   0<=i<N for a valid space.  int FindSpace(int PlateNumber, bool parking[], int N) {       // ?????                                    }    main() {  EmptyTheLot(parking, N);    // start with an empty parking garage. // get plate numbers and fill lot.  }

Programming language c++

#include <stdio.h>#include <string.h> int findRepeat(char* s){int p , i , j; p = -1; for (i = 0 ; i < strlen(s) ; i++){for (j = i + 1; j < strlen(s); j++){if (s[i] == s[j]){p = i;break;}}if (p != -1)break;}return p;} int main(){char str[] = "Hello World";int pos = findRepeat(str);if (pos == -1)printf("Not found");elseprintf("%c", str[pos]);return 0;}can you please explain this code for me

Q# Describe what a data structure is for a char**? (e.g. char** argv) Group of answer choices A. It is a pointer to a single c-string. B. It is an array of c-strings C. It is the syntax for dereferencing a c-string and returning the value.

#include <iostream>#include <string.h>using namespace std;int index1(char *T,char *P){int t=0,p=0,i,j,r;t = strlen(T);p = strlen(P);i=0;int max = t - p + 1;while(i<max){for(j=0;j<p;j++){if(P[j]==T[i+j]){r = i;break;}else r = -1;}i++;}return r;} int main(){ int l;char c[100];char d[100];cout<<"Enter First String\n"; cin>>c;cout<<"Enter Second String\n";cin>>d;l = index1(c,d);cout<<l; return 0;} Note: Remove Funcation

#include <iostream> #include <string.h> using namespace std; int index1(char *T,char *P) {     int t=0,p=0,i,j,r;     t = strlen(T);     p = strlen(P);     i=0;     int max = t - p + 1;     while(i<max)     {         for(j=0;j<p;j++)         {if(P[j]==T[i+j])          {r = i;           break;          }         else r = -1;         }         i++;     }     return r; }   int main() {       int l;     char c[50];     char d[50];     cout<<"Enter First String\n";     cin>>c;     cout<<"Enter Second String\n";     cin>>d;     l = index1(c,d);     cout<<"The returned index is:"<<l;       return 0; }Note:Remove Funcation

Language: JAVA Script Write a function 'mostFrequentLetter(string)' that takes a string as an argument and returns the character that appears the most often. In case of a tie, you may return any of the characters. The string will have at least one character.  Examples: console.log(most Frequent Letter ("apple")); // "p"  console.log(mostFrequentLétter("banana")); // "a" console.log(mostFrequentLetter("What about a longer string?")); //" " **************************************************************/ function mostFrequentLetter(string) { |// Your code here} /****DO NOT MODIFY ANYTHING UNDER THIS LINE****/ try{| module.exports = mostfrequentletter;}catch (e) {|module.exports=null;}

Static Length, Limited Dynamic Length, and Dynamic Length String implementations have what advantages and disadvantages?

#include <iostream> using namespace std; int BinSearch(int arr[],int beg, int end, int key){    if(beg > end){        return -1;    }    else{        int mid = beg + (end - beg) / 2;        if(arr[mid] == key){            return mid;        }        else if(key < arr[mid]){            return BinSearch(arr,beg,mid-1,key);        }        else{            return BinSearch(arr,mid+1,end,key);        }    }} int main(){    int arr[] = {1,2,3,4,5,6,7,8};    int n = sizeof(arr)/sizeof(arr[0]);        int key;    cout << "Enter the key to be searched: " << endl;    cin >> key;        int res = BinSearch(arr,0,n-1,key);    (res == -1) ? cout << "Element is not present in array"                   : cout << "Element is present at index " << res;         return 0;}Q: Remove Function in above algorithm