C++ Call by reference geeks for geeks

 

Rahul loves to play with numbers, he challenges his friend Ankush with a problem related to numbers in which he has to first reverse the order of two digits and then swap them. Let us take the two digits as a and b.

Example 1:

Input:
a = 1234 
b = 67890
Output: 
9876 4321
Explanation: Reversing the given two numbers
will give: 4321 and 9876. After swapping it
would come as: 9876 and 4321.

Example 2:

Input:
a = 10000 
b = 3254
Output: 
4523 1

 

Your Task: Complete the two functions reverse_dig() and swap() with arguments as a and b references. Don't return anything to the function.

Constraints:
0 <= a, b <= 105

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solution<1>:

// { Driver Code Starts
//Initial Template for C++

#include <iostream>
using namespace std;


// } Driver Code Ends
//User function Template for C++

void reverse_dig(int& a, int& b)
{
    //Add your code here.
    int num1 = 0, num2 = 0;
    while (a > 0)
    {
        int res = a % 10;
        num1 = num1 * 10 + res;
        a /= 10;
    }
    while (b > 0)
    {
        int res = b % 10;
        num2 = num2 * 10 + res;
        b /= 10;
    }
    a = num1;
    b = num2;
}
void swap(int& a, int& b)
{
    //Add your code here.
    int temp = a;
    a = b;
    b = temp;
}


// { Driver Code Starts.

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int a, b;
        cin >> a >> b;

        reverse_dig(a, b);
        swap(a, b);
        cout << a << " " << b << endl;
    }
    return 0;
}  // } Driver Code Ends

solution<2>:

void reverse_dig(int &a,int &b)
{
    //Add your code here.
    string a1,s1=to_string(a);   //to convert int to string.
    for(int i=s1.length()-1;i>=0;i--)
    {
        a1+=s1[i];
    }
    string a2,s2=to_string(b);
    for(int i=s2.length()-1;i>=0;i--)
    {
        a2+=s2[i];
    }
    a=stoi(a1);     //to convert string to int.
    b=stoi(a2);
    
}
void swap(int &a,int &b)
{
    //Add your code here.
    int temp=a;
    a=b;
    b=temp;
}