Multiply large numbers as strings

This algorithm is very much like factorial. Multiplication of very large numbers represented as strings because they go out of the integer range.

#include <iostream>
#include <vector>
#include <string>
#include <sstream>

std::string multiply(std::string a, std::string b) {
  // take a vector of length 200 and initialize all of its elements to 0.
  std::vector<int> multiplication(200, 0); 

 int place = 0;
 for(int i = a.size() - 1 ; i >= 0; i--){
   int carry = 0;
   std::vector<int>::iterator it = multiplication.begin() + place;
   for(int j = b.size() - 1 ; j >= 0; j--){
     int a_element = a[i] - '0';
     int b_element = b[j] - '0';
     int prod = a_element*b_element + carry + *it;
     *it = prod % 10;
     carry = prod / 10;
     it++;
     while (carry){
       int current_carry = carry + *it;
       *it += current_carry % 10;
       it++;
       carry = current_carry / 10;
    }
  }
   place++;
 }

std::stringstream ss;
for(int i = multiplication.size(); i >= 0; i--){
 ss << multiplication[i];
}
 return ss.str();
}

Reverse a number

Algorithm for reversing the number is as follows:

#include <cmath> // for finding out the length of number
 
// find the length of the number to find out the highest power of 10
// then divide the number by 10 and take remainder to find digit 
// at the unit's place.
// multiply the digit with power of 10. repeat length times.
int reverse(int x) {
  int length = (int)log10(abs(x)); // magic formula to find length
  int reversed = 0;
  int remainder = 0;
  for(int i = length; i > 0 ; --i){
    remainder = x % 10;
    x = x / 10;
    reversed += remainder * pow(10, i);
  }
  reversed += x;
  return reversed;
 }

 

Factorial of a large number

Factorial of an integer is found by multiplying all the numbers starting from 1 up to the number. So factorial for 5 is 5! = 5 x 4 x 3 x 2 x 1 = 120.

A recursive algorithm for finding factorial is given as

int factorial(int num){
  return num > 1 ? : num * factorial(num - 1) : 1;
}

The issue is factorial quickly grows out of the range of int/double/float limit.

Example:

factorial of 100 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

factorial of 1000 402387260077093773543702433923003985719374864210714632543799910429938512398629020592044208486969404800479988610197196058631666872994808558901323829669944590997424504087073759918823627727188732519779505950995276120874975462497043601418278094646496291056393887437886487337119181045825783647849977012476632889835955735432513185323958463075557409114262417474349347553428646576611667797396668820291207379143853719588249808126867838374559731746136085379534524221586593201928090878297308431392844403281231558611036976801357304216168747609675871348312025478589320767169132448426236131412508780208000261683151027341827977704784635868170164365024153691398281264810213092761244896359928705114964975419909342221566832572080821333186116811553615836546984046708975602900950537616475847728421889679646244945160765353408198901385442487984959953319101723355556602139450399736280750137837615307127761926849034352625200015888535147331611702103968175921510907788019393178114194545257223865541461062892187960223838971476088506276862967146674697562911234082439208160153780889893964518263243671616762179168909779911903754031274622289988005195444414282012187361745992642956581746628302955570299024324153181617210465832036786906117260158783520751516284225540265170483304226143974286933061690897968482590125458327168226458066526769958652682272807075781391858178889652208164348344825993266043367660176999612831860788386150279465955131156552036093988180612138558600301435694527224206344631797460594682573103790084024432438465657245014402821885252470935190620929023136493273497565513958720559654228749774011413346962715422845862377387538230483865688976461927383814900140767310446640259899490222221765904339901886018566526485061799702356193897017860040811889729918311021171229845901641921068884387121855646124960798722908519296819372388642614839657382291123125024186649353143970137428531926649875337218940694281434118520158014123344828015051399694290153483077644569099073152433278288269864602789864321139083506217095002597389863554277196742822248757586765752344220207573630569498825087968928162753848863396909959826280956121450994871701244516461260379029309120889086942028510640182154399457156805941872748998094254742173582401063677404595741785160829230135358081840096996372524230560855903700624271243416909004153690105933983835777939410970027753472000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Following is program I wrote to find factorials and store them as strings instead of numbers.

#include <iostream> // for cout 
#include <vector> // used to store the numbers
#include <string> 
#include <sstream>

// method to multiply each number passed with each element 
// within the vector. If there is a carry, just add it to the next
// number. 
void multiply_vector(std::vector<int> & v, int multiply_num){
  int carry = 0;
  for(size_t i = 0; i < v.size(); i++){
    int prod = v[i] * multiply_num + carry;
    v[i] = prod % 10;
   carry = prod / 10;
 }

 while (carry)
 {
   int current_carry = carry % 10;
   v.push_back(current_carry);
   carry = carry / 10;
  }
}

// instead of recursion use loop and vector to store result 
// of multiplication
std::string factorial(int num){
   std::vector<int> factorial_digits;
   factorial_digits.push_back(1);
   for(int i = 2; i <= num; i++){
     multiply_vector(factorial_digits, i);
   }

// received the result in vector but need to reverse it..
  std::stringstream ss;
  for(int i = factorial_digits.size()-1; i >= 0; i--){
    ss << factorial_digits[i];
  }
  return ss.str();
}


int main(){
  // find factorial of 1000
  std::cout << factorial(1000) << '\n';
  return 0;
}

 

Pair in array with target sum

Given a number target and an array of numbers nums find the index of numbers which has sum equal to target.

Algorithm is:

  1. Iterate over the array
  2. subtract the first element from target, find the difference in the array
  3. Get index of current element and difference if found within the array.

c++ solution

#include <algorithm> // for std::distance and std::find
#include <vector>

std::vector<int> two_sum(std::vector<int>& nums, int target) {
  std::ptrdiff_t first_idx = 0, second_idx = 0;
  for(size_t i = 0; i <= nums.size(); i++){
    int difference = target - nums[i];
    second_idx = std::distance(nums.begin(), std::find(nums.begin()+i+1,
     nums.end(), difference));
     if(second_idx < nums.size()){
       first_idx = i;
        break;
      }
   }
   return {static_cast<int>(first_idx), static_cast<int>(second_idx)};
 }

Ruby solution

def two_sums(nums, target)
  first_idx, second_idx = 0, 0
  nums.each_with_index do |num, idx|
    difference = target - num
    second_idx = nums.rindex(difference)
    first_idx = idx unless second_idx.nil?
    break
  end
  [first_idx, second_idx]
end