Hello Programmers/Coders, Today we are going to share solutions of Programming problems of HackerRank of Programming Language C++ . At Each Problem with Successful submission with all Test Cases Passed, you will get an score or marks. And after solving maximum problems, you will be getting stars. This will highlight you profile to the recruiters.
In this post, you will find the solution for Operator Overloading in C++-HackerRank Problem. We are providing the correct and tested solutions of coding problems present on HackerRank. If you are not able to solve any problem, then you can take help from our Blog/website.
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Introduction To C++
C++ is a general-purpose programming language that was developed as an enhancement of the C language to include object-oriented paradigm. It is an imperative and a compiled language.
C++ is a middle-level language rendering it the advantage of programming low-level (drivers, kernels) and even higher-level applications (games, GUI, desktop apps etc.). The basic syntax and code structure of both C and C++ are the same.
C++ programming language was developed in 1980 by Bjarne Stroustrup at bell laboratories of AT&T (American Telephone & Telegraph), located in U.S.A. Bjarne Stroustrup is known as the founder of C++ language.
Operator Overloading in C++ - Hackerrank Solution
Objective
Classes define new types in C++. Types in C++ not only interact by means of constructions and assignments but also via operators. For example:
int a=2, b=1, c; c = b + a;
The result of variable c will be 3.
Similarly, classes can also perform operations using operator overloading. Operators are overloaded by means of operator functions, which are regular functions with special names. Their name begins with the operator keyword followed by the operator sign that is overloaded. The syntax is:
type operator sign (parameters) { /*... body ...*/ }
You are given a main() function which takes a set of inputs to create two matrices and prints the result of their addition. You need to write the class Matrix which has a member a of type vector<vector<int> >. You also need to write a member function to overload the operator +. The function’s job will be to add two objects of Matrix type and return the resultant Matrix.
Input Format :
First line will contain the number of test cases T. For each test case, there are three lines of input. The first line of each test case will contain two integers N and M which denote the number of the rows and columns respectively of the two matrices that will follow on the next two lines. These next two lines will each contain N*M elements describing the two matrices in row-wise format i.e. first M elements belong to the first row,next M elements belong to the second row and so on.
Constraints :
- 1 <= T <= 1000
- 1 <= N <= 100
- 1 <= M <= 100
- 1 <= Aij <= 10, where Aij is the element in the ith row and jth Column of the Matrix.
Output Format :
The code provided in the editor will use your class Matrix and overloaded operator function to add the two matrices and give the output.
Sample Input :
1 2 2 2 2 2 2 1 2 3 4
Sample Output :
3 4 5 6
Explanation :
The sum of first matrix and the second matrix is the matrix given in the output.
Operator Overloading in C++ – Hacker Rank Solution
#include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; /* Operator Overloading in C++ - Hacker Rank Solution START */ class Matrix { public: vector<vector<int>> a; Matrix() {} Matrix operator+(const Matrix &o) { Matrix sum(*this); for (size_t i = 0; i < sum.a.size(); i++) { for (size_t j = 0; j < sum.a[i].size(); j++) { sum.a[i][j] += o.a[i][j]; } } return sum; } }; /* Operator Overloading in C++ - Hacker Rank Solution END */ int main () { int cases,k; cin >> cases; for(k=0;k<cases;k++) { Matrix x; Matrix y; Matrix result; int n,m,i,j; cin >> n >> m; for(i=0;i<n;i++) { vector<int> b; int num; for(j=0;j<m;j++) { cin >> num; b.push_back(num); } x.a.push_back(b); } for(i=0;i<n;i++) { vector<int> b; int num; for(j=0;j<m;j++) { cin >> num; b.push_back(num); } y.a.push_back(b); } result = x+y; for(i=0;i<n;i++) { for(j=0;j<m;j++) { cout << result.a[i][j] << " "; } cout << endl; } } return 0; }
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