Square of large number represented as String

Given a very large number, the task is to write a program to compute its square.

Examples:

Input: 9999
Output: 99980001
9999*9999 = 99980001



Input: 45454545
Output: 2066115661157025
45454545*45454545 = 2066115661157025

Naive Approach: A naive approach is to calculate the squares my multiplying the number with itself. But in C++, if the input is a large number, the resultant square will overflow.

Efficient Approach: An efficient approach is to store the number as strings, and perform multiplication of two large numbers.

Below is the implementation of the above approach:

C++

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// C++ program to multiply two numbers
// represented as strings.
#include <bits/stdc++.h>
using namespace std;
  
// Multiplies str1 and str2, and prints result.
string multiply(string num1, string num2)
{
    int n1 = num1.size();
    int n2 = num2.size();
    if (n1 == 0 || n2 == 0)
        return "0";
  
    // will keep the result number in vector
    // in reverse order
    vector<int> result(n1 + n2, 0);
  
    // Below two indexes are used to find positions
    // in result.
    int i_n1 = 0;
    int i_n2 = 0;
  
    // Go from right to left in num1
    for (int i = n1 - 1; i >= 0; i--) {
        int carry = 0;
        int n1 = num1[i] - '0';
  
        // To shift position to left after every
        // multiplication of a digit in num2
        i_n2 = 0;
  
        // Go from right to left in num2
        for (int j = n2 - 1; j >= 0; j--) {
            // Take current digit of second number
            int n2 = num2[j] - '0';
  
            // Multiply with current digit of first number
            // and add result to previously stored result
            // at current position.
            int sum = n1 * n2 + result[i_n1 + i_n2] + carry;
  
            // Carry for next iteration
            carry = sum / 10;
  
            // Store result
            result[i_n1 + i_n2] = sum % 10;
  
            i_n2++;
        }
  
        // store carry in next cell
        if (carry > 0)
            result[i_n1 + i_n2] += carry;
  
        // To shift position to left after every
        // multiplication of a digit in num1.
        i_n1++;
    }
  
    // ignore '0's from the right
    int i = result.size() - 1;
    while (i >= 0 && result[i] == 0)
        i--;
  
    // If all were '0's - means either both or
    // one of num1 or num2 were '0'
    if (i == -1)
        return "0";
  
    // generate the result string
    string s = "";
    while (i >= 0)
        s += std::to_string(result[i--]);
  
    return s;
}
  
// Driver code
int main()
{
    string str1 = "454545454545454545";
  
    cout << multiply(str1, str1);
  
    return 0;
}

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// Java program to multiply two numbers  


// represented as strings. 


  


class GFG  


{ 


  


    // Multiplies str1 and str2, and prints result. 


    public static String multiply(String num1, String num2) 


    { 


        int n1 = num1.length(); 


        int n2 = num2.length(); 


        if (n1 == 0 || n2 == 0) 


            return "0"; 


  


        // will keep the result number in vector 


        // in reverse order 


        int[] result = new int[n1 + n2]; 


  


        // Below two indexes are used to find positions 


        // in result. 


        int i_n1 = 0; 


        int i_n2 = 0; 


  


        // Go from right to left in num1 


        for (int i = n1 - 1; i >= 0; i--)  


        { 


            int carry = 0; 


            int n_1 = num1.charAt(i) - '0'; 


  


            // To shift position to left after every 


            // multiplication of a digit in num2 


            i_n2 = 0; 


  


            // Go from right to left in num2 


            for (int j = n2 - 1; j >= 0; j--) 


            { 


  


                // Take current digit of second number 


                int n_2 = num2.charAt(j) - '0'; 


  


                // Multiply with current digit of first number 


                // and add result to previously stored result 


                // at current position. 


                int sum = n_1 * n_2 + result[i_n1 + i_n2] + carry; 


  


                // Carry for next iteration 


                carry = sum / 10; 


  


                // Store result 


                result[i_n1 + i_n2] = sum % 10; 


  


                i_n2++; 


            } 


  


            // store carry in next cell 


            if (carry > 0) 


                result[i_n1 + i_n2] += carry; 


  


            // To shift position to left after every 


            // multiplication of a digit in num1. 


            i_n1++; 


        } 


  


        // ignore '0's from the right 


        int i = result.length - 1; 


        while (i >= 0 && result[i] == 0) 


            i--; 


  


        // If all were '0's - means either both or 


        // one of num1 or num2 were '0' 


        if (i == -1) 


            return "0"; 


  


        // generate the result string 


        String s = ""; 


        while (i >= 0) 


            s += Integer.toString(result[i--]); 


  


        return s; 


    } 


  


    // Driver code 


    public static void main(String[] args)  


    { 


        String str1 = "454545454545454545"; 


        System.out.println(multiply(str1, str1)); 


  


    } 


} 


  


// This code is contributed by 


// sanjeev2552 



















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C#














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// C# program to multiply two numbers  


// represented as strings. 


using System; 


using System.Collections.Generic; 


      


class GFG  


{ 


  


    // Multiplies str1 and str2,  


    // and prints result. 


    public static String multiply(String num1,  


                                  String num2) 


    { 


        int n1 = num1.Length; 


        int n2 = num2.Length; 


        if (n1 == 0 || n2 == 0) 


            return "0"; 


  


        // will keep the result number in vector 


        // in reverse order 


        int[] result = new int[n1 + n2]; 


  


        // Below two indexes are used to  


        // find positions in result. 


        int i_n1 = 0; 


        int i_n2 = 0; 


        int i = 0; 


          


        // Go from right to left in num1 


        for (i = n1 - 1; i >= 0; i--)  


        { 


            int carry = 0; 


            int n_1 = num1[i] - '0'; 


  


            // To shift position to left after every 


            // multiplication of a digit in num2 


            i_n2 = 0; 


  


            // Go from right to left in num2 


            for (int j = n2 - 1; j >= 0; j--) 


            { 


  


                // Take current digit of second number 


                int n_2 = num2[j] - '0'; 


  


                // Multiply with current digit of first number 


                // and add result to previously stored result 


                // at current position. 


                int sum = n_1 * n_2 +  


                          result[i_n1 + i_n2] + carry; 


  


                // Carry for next iteration 


                carry = sum / 10; 


  


                // Store result 


                result[i_n1 + i_n2] = sum % 10; 


  


                i_n2++; 


            } 


  


            // store carry in next cell 


            if (carry > 0) 


                result[i_n1 + i_n2] += carry; 


  


            // To shift position to left after every 


            // multiplication of a digit in num1. 


            i_n1++; 


        } 


  


        // ignore '0's from the right 


        i = result.Length - 1; 


        while (i >= 0 && result[i] == 0) 


            i--; 


  


        // If all were '0's - means either both or 


        // one of num1 or num2 were '0' 


        if (i == -1) 


            return "0"; 


  


        // generate the result string 


        String s = ""; 


        while (i >= 0) 


            s += (result[i--]).ToString(); 


  


        return s; 


    } 


  


    // Driver code 


    public static void Main(String[] args)  


    { 


        String str1 = "454545454545454545"; 


        Console.WriteLine(multiply(str1, str1)); 


    } 


} 


  


// This code is contributed by Princi Singh 



















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Output:


206611570247933883884297520661157025






          
          
          
          

            

    

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Improved By :  sanjeev2552, princi singh


      
      
      
      

      

      
					
		

    Article Tags : 
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Strings

           	
Practice Tags : 
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