1. Vector X = {1, 2, 3, 7, 7, 100} Then what does the following return: adjacent_find (X.begin(), X.end()) Explain briefly. 2. Vector X = {101, 12, 3, 37, 7, 100} Then what can you say about the following application: result = binary_search(v.begin(), v.end(), 101); Explain briefly.
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- Suppose that two arraysx1,x2,…,xnandy1,y2,…,yn.( Need only handwritten solution please otherwise downvote).Write code for determining the index of a value in an OrderedVector. Be aware that if the value is not in the Vector, the routine returns the ideal location to insert the value. This may be a location that is outside the Vector(use java)#include <iostream> #include <fstream> using namespace std; struct Triple { int value, row, col; }; ostream& operator<<(ostream& os, Triple& t) { os << "(" << t.row << ", " << t.col << ", " << t.value << ")" <<endl; return os; } class Matrix; // forward declaration class MatrixNode { friend class Matrix; friend istream& operator>>(istream&, Matrix&); // for reading in a matrix friend ostream& operator<<(ostream&, Matrix&); private: MatrixNode *down, *right; bool head; union { // anonymous union MatrixNode *next; Triple triple; }; MatrixNode(bool, Triple*); // constructor }; MatrixNode::MatrixNode(bool b, Triple* t) { head = b; if (b) {right = next = this;} // row/column head node else triple = *t; // head node for list of headnodes OR element node } typedef MatrixNode* MatrixNodePtr; class Matrix { Matrix operator+ (const Matrix & b) const; Matrix operator*(const…
- Implement binary search algorithm as a template function.Part B: Given a vector of size S and a number N. Write a function that calculates if there is any pair of numbers in the vector whose sum is equal to N?Hint: you can use the binary search function you implemented in part A.Notes:You must use iterators in your implementation.Input Format: The input consists of 2 lines. The first line contains 2 numbers, S and N. The second line contains S numbers which represent the contents of the vector.Implement the following function without using any additional data structure, and without sorting the input vectors. /* Given two vectors of integers, check if the two vectors contain same set of values: e.g., V1=[3,4,10,4,10,11] and V2=[3,3,4, 11, 10] stores same set int: {3, 4, 10, 11}. Note that duplicates are removed when considering set */ @param list1, list2: two vectors of integers@pre: list1, list2 have been initialized@post: return true if list1 and list2 stores same values (in same or different order); return false, if not. */ bool SameSet (const vector<int> & list1, const vector<int> & list2) {//for each value in list1, check if it appears in list2, if not, return false //for each value in list2, check if the value appears in list1, if not return false //return true}Integer dataSize is read from input. Then, strings and integers are read and stored into string vector subjectList and integer vector pageList, respectively. Lastly, integer pageThreshold is read from input. Set matchCount with the number of element pairs with a page greater than or equal to pageThreshold. Output the subject of each pair found with a page greater than or equal to pageThreshold. End with a newline. Ex: If the input is: 3 Environment 416 Cooking 353 Science 306 349 Then the output is: Environment Cooking Total: 2 By using the following code: #include #include using namespace std; int main() { int numElements; int pageThreshold; int matchCount; unsigned int i; cin >> numElements; vector subjectList(numElements); vector pageList(numElements); for (i = 0; i < subjectList.size(); ++i) { cin >> subjectList.at(i); cin >> pageList.at(i); } cin >> pageThreshold; /* Your code goes here */ cout << "Total: " << matchCount;…
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- You are going to implement hashing with chaining with separate chaining. Please implement your hash function accordingly to distribute the data evenly in the array. Fill in the missing code! C only! ASAP!!! THE CODE: #include <stdio.h> // RecordTypestruct RecordType{int id;char name;int order;}; // Fill out this structurestruct HashType{ }; // Compute the hash functionint hash(int x){ } // parses input file to an integer arrayint parseData(char* inputFileName, struct RecordType** ppData){FILE* inFile = fopen(inputFileName, "r");int dataSz = 0;int i, n;char c;struct RecordType *pRecord;*ppData = NULL; if (inFile){fscanf(inFile, "%d\n", &dataSz);*ppData = (struct RecordType*) malloc(sizeof(struct RecordType) * dataSz);// Implement parse data blockif (*ppData == NULL){printf("Cannot allocate memory\n");exit(-1);}for (i = 0; i < dataSz; ++i){pRecord = *ppData + i;fscanf(inFile, "%d ", &n);pRecord->id = n;fscanf(inFile, "%c ", &c);pRecord->name =…Computer Science Done in Perl. Someone gave me a wrong/incomplete answer. Write a function, hashsum, which takes a hash reference as its first argument and an array reference as its second argument. The function should take the elements of the array and use them as keys in the hash. The values in the hash corresponding to the keys in the array should be totaled, and the sum should be returned to the user. If a key passed in the array does not exist in the hash, ignore it. Write a sample program to call this function.6) While there is a built-in pop_back() method for vectors, there is no built-in pop_front method. Suppose a program needs a pop_front() method that will remove the first element from the vector. For example if the original vector is [1, 2, 3, 4, 5], then after passing in this vector to pop_front() the new resulting vector will be [2, 3, 4, 5]. Which of the options below is the correct implementation of pop_front()? Group of answer choices D-) void pop_front(vector<int> &v){ for(int i=0; i<(v.size()-1); i++)v[i+1] = v[i]; v.pop_back(); } C-) void pop_front(vector<int> &v){ for(int i=0; i<(v.size()-1); i++)v[i] = v[i+1]; v.pop_back(); } B-) void pop_front(vector<int> &v){ for(int i=0; i<v.size(); i++)v[i+1] = v[i]; v.pop_back(); } A-) void pop_front(vector<int> &v){ for(int i=0; i<v.size(); i++)v[i] = v[i+1]; v.pop_back(); } 7) Suppose a program has the following vector: [99, 23, 55, 71, 87, 64, 35, 42] The goal is to…