Given a number N, our task is to find the closest Palindrome number whose absolute difference with given number is minimum. If 2 Palindome numbers have same absolute difference from the given number, then output the smaller one.

Given a number N, our task is to find the closest Palindrome number whose absolute difference with given number is minimum. If 2 Palindome numbers have same absolute difference from the given number, then output the smaller one.

For each test case, the print the closest palindrome number.
Note:  If the difference of two closest palindromes numbers is equal then we print smaller number as output.

For Input:
5
848
1001
500999
1756183123
815901158
Your Output is:
848
1001
501105
1756226571
815898518

#include <bits/stdc++.h>
using namespace std;
long long int palindrome(long long int n)
{
    long long int pre=n;
    vector<int> v;
//just spliting number into digits
    while(n>0)
    {
        v.push_back(n%10);
        n/=10;
    }
//vector stored digits in reverse order
    int i=0,j=v.size()-1;
//so we have taking mirror of leftside
     while(i<=j)
     {
         if(v[i]!=v[j])
          v[i]=v[j];
          ++i,--j;
     }
   
    long int a=0,b=0,c=0;
//
    for(i=0;i<v.size();++i)
      a=a*10+v[i];
// if given is palindrome return itself
     if(pre==a)
     return pre;
// checking length of size of vector
     if(v.size()%2==1)
     {
//if it is odd
//compare with actual value
         if(a<pre)
         {
            
// if a less than actual then just add 1 to middle if it is not 9
// if it is 9 then on adding 1 become 10 so add to left of middle
//making that mirror to right of middle
             if(v[v.size()/2]==9)
             {
             v[v.size()/2-1]+=1;
             v[v.size()/2]=0;
             v[v.size()/2+1]=v[v.size()/2-1];
             }else{
              v[v.size()/2]+=1;
             }
            
         }
         else
         {  
// if a greater than actual then just substruct 1 to middle
// if middle is greater than 0
// if middle is 0 then substruct to left of middle
// and in middle place 9 and right of middle is mirror of left
             if(v[v.size()/2]==0)
             {
             v[v.size()/2-1]-=1;
             v[v.size()/2]=9;
             v[v.size()/2+1]=v[v.size()/2-1];
             }else{
              v[v.size()/2]-=1;
             }
         }
//calculating the possible nearest palindrome number
         for(i=0;i<v.size();++i)
            b=b*10+v[i];

 // if equal distance from actual number then return smallest number
// else return small distance palindrome 
          if(abs(a-pre)==abs(b-pre))
            {
                return min(a,b);
           }else if(abs(a-pre)<abs(b-pre))
              {
               return a;
           }else{
               return b;
           }
           

        
     }else{
  //if length is even
//we have add 1 and substract to middle element
//to genarate two palindrome
// one left and one right of 'a' number

          for(i=0;i<v.size();++i)
          {

             if((i==v.size()/2)||(i==(v.size()/2-1)))
              {

                    if(v[i]<=9)
                    c=c*10+v[i]-1;
                    if(v[i]>=0)
                    b=b*10+v[i]+1;

                   
              }else{
                  b=b*10+v[i];
                  c=c*10+v[i];
              }

          }
//compare and finding small distance palindrome number
         
       if(abs(a-pre)<abs(b-pre)&&abs(c-pre)>abs(a-pre))
       {
        return a;
       }else
        if(abs(a-pre)>abs(b-pre)&&abs(c-pre)>abs(b-pre))

    {
     return b;
    }else if(abs(a-pre)>abs(c-pre)&&abs(c-pre)<abs(b-pre))
        {

    return c;
        }

     }
  
    return a;
}

int main() {
    //code
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
          if(n<10)
          {
              cout<<n<<endl;
          }else{
             cout<<palindrome(n)<<endl;
          }
    }
    return 0;
}