Variable Sized Arrays | HackerRank

Goal

Create N number of variable-length arrays and find a element from quaries

Solution

If you know vector in C++, it would be easy problem.

you initialize vector of vector a and insert variable sized vector into a.

# Variable Sized Arrays

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int n = 0;
    int q = 0;
    
    cin >> n >> q;
    
     vector<vector<int>> a;
    
    for(int i = 0 ; i < n ; i++){
        int k = 0;
        
        cin >> k;
        
        vector<int> v;
        
        int value = 0;
        int index = 1;
        while(cin >> value){
            
            v.push_back(value);
            index++;
            
            if (index > k){
                break;
            }
            
        }
        
        a.push_back(v);
        
    }
    
    for(int i = 0; i < q ; i++){
        int index, j ;
        cin >> index >> j;
        cout << a[index][j] << endl;
    }
    
    
    return 0;
}

Arrays Introduction | HackerRank

Goal

Print the N integers of the array in the reverse order, space-separated on a single line.

Solution

1. Create array with N size.

2. Save N integers into array.

3. Print reverse array with space

// Arrays Introduction

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */
    int n;
    
    cin >> n;
    
    int a[n];
    int tmp;
    int i = 0;
    
    while (cin >> tmp)
    {
        a[i] = tmp;
        i++;
    }
    
    for(int index = n - 1; index >=0 ; index--){
        if(index !=0){
            cout << a[index] << " ";
        }
        else{
            cout << a[index] << endl;
        }
    }
    
    
    return 0;
}

Pointer | HackerRank

Goal

By using pointer in C++, save a + b value into a and | a - b | into b.

Solution

This problem makes us to pratice pointer. 

You can save sum and abs_min of a and b with * sign in a and b, respectively. 

// Pointer

#include <stdio.h>
#include <cstdlib>


void update(int *a,int *b) {
    // Complete this function

    int sum = *a + * b;
    
    int abs_min = (*a) - (*b);
    *a = sum;
    *b = abs(abs_min);
        
}

int main() {
    int a, b;
    int *pa = &a, *pb = &b;
    
    scanf("%d %d", &a, &b);
    update(pa, pb);
    printf("%d\n%d", a, b);

    return 0;
}

Functions | HackerRank

Goal

Get four arguments and return the greatest of the four integers

Solution

Because you just need to return the greatest integer.

You can compare a and b first, then c and d.

If you find already greateest integer, you can simply return the value.

// Functions

#include <iostream>
#include <cstdio>
using namespace std;

/*
Add `int max_of_four(int a, int b, int c, int d)` here.
*/

int max_of_four(int a, int b, int c, int d){
    
    
    if(a >= b){
        if (a >= c){
            if (a >= d){
                return a;
            }
            else{
                return d;
            }
        }
            
        else{
            
            if( c>=d){
                return c;
            }
            
            else{
                return d;
            }
        }
    }
    
    else{
        
        if (b >= c){
            if (b >= d){
                return b;
            }
            else {
                return d;
            }
        }
        
        else{
            
            if (c >= d){
                return c;
            }
            
            else {
                return d;
            }
            
        }
        
    }
    
    return 0;
}



int main() {
    int a, b, c, d;
    scanf("%d %d %d %d", &a, &b, &c, &d);
    int ans = max_of_four(a, b, c, d);
    printf("%d", ans);
    
    return 0;
}

For Loop | HackerRank

Goal

Print results according to two positive input values by using for loop statement.

Solution

To solve easily, I create vector string array included 1-9 numbers. If you use vector array, you can print numbers easily.

// For Loop
#include <iostream>
#include <cstdio>
#include <vector>
#include <string>
using namespace std;

int main() {
    // Complete the code.
    int a,b;
    cin >> a;
    cin >> b;
    
    vector<string> numbers {"", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
    
    for(int i = a; i <= b; i++){
        
        if(i>9){
            if( i % 2 == 0){
                cout << "even" << endl;
            }
            else{
                cout << "odd" << endl;
            }
        }
        
        else{
            
            cout << numbers[i] << endl;
        }
    }
    
    return 0;
}

https://www.hackerrank.com/challenges/c-tutorial-conditional-if-else/problem

Goal

Solution

You just need to create condtion statements for each condition like below.

// Conditional Statements

#include <bits/stdc++.h>

using namespace std;

string ltrim(const string &);
string rtrim(const string &);



int main()
{
    string n_temp;
    getline(cin, n_temp);

    int n = stoi(ltrim(rtrim(n_temp)));

    // Write your code here
    if(n > 9){
        printf("Greater than 9");
    }
    else if(n == 1){
        printf("one");
    }
    else if(n == 2){
        printf("two");
    }
    else if(n == 3){
        printf("three");
    }
    else if(n == 4){
        printf("four");
    }
    else if(n == 5){
        printf("five");
    }
    else if(n == 6){
        printf("six");
    }
    else if(n == 7){
        printf("seven");
    }
    else if(n == 8){
        printf("eight");
    }
    else { // n == 9
        printf("nine");
    }

    return 0;
}

string ltrim(const string &str) {
    string s(str);

    s.erase(
        s.begin(),
        find_if(s.begin(), s.end(), not1(ptr_fun<int, int>(isspace)))
    );

    return s;
}

string rtrim(const string &str) {
    string s(str);

    s.erase(
        find_if(s.rbegin(), s.rend(), not1(ptr_fun<int, int>(isspace))).base(),
        s.end()
    );

    return s;
}

Basic Data Types | HackerRank

Goal

1. Read space-seprated values : int, long, char, float and double, respectively

2. Print each element on a new line in the same order it was received as input. + correct decimal points of floating and double points.

Solution

Follow HackerRank explanation of c++ data types and their format specifiers.

// Basic Data Types

#include <iostream>
#include <cstdio>
using namespace std;

int main() {
    // Complete the code.
    int a;
    long b;
    char c;
    float d;
    double e;

    scanf("%d %ld %c %f %lf", &a, &b, &c, &d, &e);
    printf("%d\n", a);
    printf("%ld\n", b);
    printf("%c\n", c);
    printf("%f\n", d);
    printf("%lf\n", e);
    return 0;
}

Input and Output | HackerRank

Goal

Read 3 numbers from stdin and print their sum to stdout

Solution

This challenge is to practice reading input from stdin(standard input stream) and printing out to stdout(standard output stream).

// Input and Output

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;


int main() {
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */   
    int a, b,c ;
    int result = 0;
    cin >> a >> b >> c;
    result = a + b + c;
    cout << result << endl;
    return 0;
}

Say "Hello, World!" With C++ | HackerRank

Goal

Solution

This is simple challenge to print "Hello, World!" to stdout. We can use easily an object of class iostream. "std::cout."

// Say "Hello, World!" With C++

#include <iostream>
#include <cstdio>
using namespace std;

int main() {

    cout << "Hello, World!" << endl;
    return 0;
}