To find out the characters in the palindrome we will follow this approach:

1. The max length will be the top right block.
2. Every time we will check the left and bottom adjacent blocks.
   1. If both the blocks contain the same value then we will go for any of them.
   2. If any block contains a value such that the difference between the current value and the value at the block is 2 then we will go for that block and make a note of that character present in the string whose index is the same as the column number.
   3. When we will get the value equals to 1 then stop the traversal.

C++ Implementation:

#include <bits/stdc++.h>
using namespace std;

int arr[101][101];

void print_palindrome(string str)
{
    int len = str.length();
    int i = 0;
    int j = len - 1;
    int num = arr[i][j];
    vector<char> v;
    int flag = 0;
    while (i < len && j >= 0) {
        if (num == 2) {
            flag = 1;
            break;
        }
        if (num == 1)
            break;
        //if the left block has the value by which
        //the difference will be 2
        if (arr[i][j - 1] + 2 == num) {
            v.push_back(str[j]);
            j--;
            num = arr[i][j];
            if (num == 1 || num == 2) {
                v.push_back(str[j]);
            }
        }
        //if the bottam block has the value by
        //which the difference will be 2
        if (arr[i + 1][j] + 2 == num) {
            v.push_back(str[i]);
            i++;
            num = arr[i][j];
            if (num == 1 || num == 2) {
                v.push_back(str[i]);
            }
        }
        if (arr[i][j - 1] == num) {
            j--;
        }
        else {
            i++;
        }
    }
    for (int i = 0; i < v.size(); i++) {
        cout << v[i];
    }
    //cout<<endl;
    for (int i = v.size() - 1; i >= 0; i--) {
        if (flag && i == v.size() - 1) {
            cout << v[i];
        }
        else if (i != v.size() - 1) {
            cout << v[i];
        }
    }
    cout << endl;
}

void length_of_palindrom(string str)
{
    int len = str.length();
    memset(arr, 0, sizeof(arr));
    for (int i = 0; i < len; i++) {
        arr[i][i] = 1;
    }
    for (int i = 0; i < len - 1; i++) {
        if (str[i] == str[i + 1])
            arr[i][i + 1] = 2;
        else
            arr[i][i + 1] = 1;
    }
    for (int l = 3; l <= len; l++) {
        for (int i = 0; i <= len - l; i++) {
            int j = i + l - 1;
            if (str[i] == str[j])
                arr[i][j] = arr[i + 1][j - 1] + 2;
            else
                arr[i][j] = max(arr[i + 1][j], arr[i][j - 1]);
        }
    }
    cout << "Palindrome : ";
    print_palindrome(str);
    cout << endl;
    return;
}

int main()
{
    int t;
    cout << "Test Case : ";
    cin >> t;
    while (t--) {
        cout << "Enter the string : ";
        string str;
        cin >> str;
        length_of_palindrom(str);
    }
}

Output

Test Case : 3
Enter the string : bghaufahgt
Palindrome : ghafahg

Enter the string : souabbuos
Palindrome : soubbuos

Enter the string : sahajkhahas
Palindrome :  sahajahas