C++ Program to convert number to spoken string

[Divide and Conquer Approach]
1). int varies from -2147483648 to +2147483647

2). e.g. 2,147,483,648 : 4 groups of three (out of which one group will contain only one digit)

3). Once we break the number in groups we can then represent it in, [w billion x million y thousand z], where billion, million and thousand are prepositions.

4). Each group will contain max upto 3 digits

5). Having basic knowledge of number to word will help us.
e.g. digit[0] = "zero";
digit[1] = "one";
digit[2] = "two"; and so on...

6). Preposition will help us merging back the solution.
e.g.
places[2] = "thousand";
places[3] = "million";
places[4] = "billion";

7). Go through the code for details :)

  
#include <string>
#include <vector>
#include <map>
using namespace std;

#include <math.h>

string numberToSpokenFormString(int n) {
 // assuming int is 4 byte
 
 map<int,string> digit;
 map<int,string> places;

 digit[0] = "zero";
 digit[1] = "one";
 digit[2] = "two";
 digit[3] = "three";
 digit[4] = "four";
 digit[5] = "five";
 digit[6] = "six";
 digit[7] = "seven";
 digit[8] = "eight";
 digit[9] = "nine";
 digit[10] = "ten";
 digit[11] = "eleven";
 digit[12] = "twelve";
 digit[13] = "thirteen";
 digit[14] = "fourteen";
 digit[15] = "fifteen";
 digit[16] = "sixteen";
 digit[17] = "seventeen";
 digit[18] = "eighteen";
 digit[19] = "nineteen";
 digit[20] = "twenty";
 digit[30] = "thirty";
 digit[40] = "fourty";
 digit[50] = "fifty";
 digit[60] = "sixty";
 digit[70] = "seventy";
 digit[80] = "eighty";
 digit[90] = "ninety";
 digit[100] = "hundred";
 
 places[2] = "thousand";
 places[3] = "million";
 places[4] = "billion";
 
 string str = "";
 int nNumOfDigits = 1;
 int nPairs = 1;
 
 if(n < 0)
 {
  n = abs(n);
  str += "minus ";
 }
 
 if(n == 0)
 {
  str = digit[0] ;
 }
 else
 {
  // [Number of digits in the integer]
  // For example log10 of number 34250000 is [ 7 + log10(3.425) ] = 7.xxxxxxx
  // take only integer part and you will have 7. Add 1 to it and number of digits = 8

  nNumOfDigits = (int)(log10((double)n))+1; 
  
  // [Groups of three]
  // int varies between -2147483648 to +2147483647
  // 2,147,483,648 : 4 groups of three (out of which one group will contain only one digit)
  // once we break the number in groups we can then represent it in,
  // w billion x million y thousand z, where billion, million and thousand are preposition.
  // [Divide and Conquer approch]
   
  nPairs = ceil(((double)nNumOfDigits/3));  
 
  int temp, num;
  for(int i = 0; i <= nPairs; i++) // Complexity O(n), n: number of digits.
  {
   num = n/(int)pow((double)1000, (double)nPairs-i);
   
   for(int j = 100; j >= 1; j=j/10)
   {
    temp = num / j;
    if(temp != 0)
    {
     if(j == 100)
     {
      str += digit[temp] + " " + digit[j] + " " ;
     }
     else if(j == 10)
     {
      if(num <= 20) // Check if number is less than or equal to 20
      {
       str += digit[num]+ " " ;
       break;
      }
      else
      {
       str += digit[temp*j] + " " ;
      }
     }
     else
     {
      str += digit[num] + " ";
     }
     num %= j;
    }
   }
   str += places[nPairs - i + 1] + " " ;
   
   if(nPairs > 1)
   {
    n %= (int)pow((double)1000, (double)nPairs-i);
   }
  }
 }
 
 return str;
  }