/** * Author : ~swim~ * Date : 14-June-2018 * Description : [ASSIGNME:c NT]: Next Palindromic Number TASK: You will be given a positive integer, num. Create the function nextPal( ) that will output the smallest palindrome number higher than num. Input >> Output Examples:: nextPal(11) --> return 22 nextPal(188) --> return 191 nextPal(191) --> return 202 nextPal(2541) --> return 2552 NOTE: You will be receiving values higher than 10, all valid! HappyCodings!:) https://code.sololearn.com/WklnpoXLWJFH/?ref=app Input: Enter an integer >= 10 and submit */ #include #include #include #include #include using namespace std; using ULL = unsigned long long int; auto Len = [](auto num) -> ULL { return floor(log10(num)) + 1; }; auto Halfn = [](auto num, auto olen) { for (unsigned i = 0; i < olen / 2; num /= 10, ++i); return num; }; int main() { ULL num; cin >> num; // some basic checking if (num < 10) { cerr << "Error, enter a number >= 10"; return 0; } if (numeric_limits::max() < num) { cerr << "Error, number too large\n"; return 0; } cout << ([](auto num) { auto olen = Len(num); auto lhalf = Halfn(num, olen); // the whole program is written as // single cout statement using // lambda function. The function // object is used to capture the // inner lambda which allows inner // lambda to be used again. // I could have been written named // inner lambda function, but i am // experimenting // with function :) function palin = [olen](auto lhalf) { auto half = to_string(lhalf); // this is the mirror string tha // needs to be appended to left // half to form the final // palindrome auto tmp = half.substr(0, olen / 2); // take care of a corner case // which happens when the // number of digits in the left half // of number decrease, while // trying to find a lower palindrome // e.g. // num = 100000 // left half = 100 // the value passed to the // function palin, is 99. if all digits // are 9 then we need to adjust the // count of 9, otherwise if i simply // replicate it, i'll get 9999 but one // more 9 is required for correct output // // Without this adjustment, you // will still get the nearest // palindrome but it will // always be on the higher side // of the number if (olen / 2 > tmp.size() && all_of(tmp.begin(), tmp.end(), [](auto c) { return '9' == c; })) { tmp += '9'; } // append, convert snd return half = half + string(tmp.crbegin(), tmp.crend()); return stoull(half); }; auto bpalin = palin(lhalf); ULL hpalin = palin(lhalf + 1); ULL lpalin = palin(lhalf - 1); stringstream ss; ss << "base palindrome = " << bpalin << endl; ss << "higher palindrome = " << hpalin << endl; ss << "lower palindrome = " << lpalin << endl; auto diffb = labs(bpalin - num); auto diffh = labs(hpalin - num); auto diffl = labs(lpalin - num); auto nearest = (diffb < diffh) ? (diffb < diffl) ? bpalin : lpalin : (diffh < diffl) ? hpalin : lpalin; ss << "nearest palindrome = " << nearest << endl; return move(ss.str()); }(num)); }

next_palindrome.cpp