Multiply Large Numbers represented as Strings
Given two positive numbers as strings. The numbers may be very large (may not fit in long long int), the task is to find product of these two numbers.
Examples:
Input : num1 = 4154
num2 = 51454
Output : 213739916
Input : num1 = 654154154151454545415415454
num2 = 63516561563156316545145146514654
Output : 41549622603955309777243716069997997007620439937711509062916
The idea is based on school mathematics.
We start from last digit of second number multiply it with first number. Then we multiply second digit of second number with first number, and so on. We add all these multiplications. While adding, we put i-th multiplication shifted.
The approach used in below solution is to keep only one array for result. We traverse all digits first and second numbers in a loop and add the result at appropriate position.
C++
// 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 len1 = num1.size(); int len2 = num2.size(); if (len1 == 0 || len2 == 0) return "0"; // will keep the result number in vector // in reverse order vector<int> result(len1 + len2, 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=len1-1; 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=len2-1; 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 = "1235421415454545454545454544"; string str2 = "1714546546546545454544548544544545"; if((str1.at(0) == '-' || str2.at(0) == '-') && (str1.at(0) != '-' || str2.at(0) != '-' )) cout<<"-"; if(str1.at(0) == '-' && str2.at(0)!='-') { str1 = str1.substr(1); } else if(str1.at(0) != '-' && str2.at(0) == '-') { str2 = str2.substr(1); } else if(str1.at(0) == '-' && str2.at(0) == '-') { str1 = str1.substr(1); str2 = str2.substr(1); } cout << multiply(str1, str2); return 0; } |
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Java
// Java program to multiply two numbers // represented as Strings. class GFG { // Multiplies str1 and str2, and prints result. static String multiply(String num1, String num2) { int len1 = num1.length(); int len2 = num2.length(); if (len1 == 0 || len2 == 0) return "0"; // will keep the result number in vector // in reverse order int result[] = new int[len1 + len2]; // 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 = len1 - 1; i >= 0; i--) { int carry = 0; int n1 = num1.charAt(i) - '0'; // To shift position to left after every // multipliccharAtion of a digit in num2 i_n2 = 0; // Go from right to left in num2 for (int j = len2 - 1; j >= 0; j--) { // Take current digit of second number int n2 = num2.charAt(j) - '0'; // Multiply with current digit of first number // and add result to previously stored result // charAt current position. int sum = n1 * n2 + result[i_n1 + i_n2] + carry; // Carry for next itercharAtion 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 // multipliccharAtion 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"; // genercharAte the result String String s = ""; while (i >= 0) s += (result[i--]); return s; } // Driver code public static void main(String[] args) { String str1 = "1235421415454545454545454544"; String str2 = "1714546546546545454544548544544545"; if ((str1.charAt(0) == '-' || str2.charAt(0) == '-') && (str1.charAt(0) != '-' || str2.charAt(0) != '-')) System.out.print("-"); if (str1.charAt(0) == '-' && str2.charAt(0) != '-') { str1 = str1.substring(1); } else if (str1.charAt(0) != '-' && str2.charAt(0) == '-') { str2 = str2.substring(1); } else if (str1.charAt(0) == '-' && str2.charAt(0) == '-') { str1 = str1.substring(1); str2 = str2.substring(1); } System.out.println(multiply(str1, str2)); } } // This code is contributed by ankush_953 |
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Python3
# Python3 program to multiply two numbers # represented as strings. # Multiplies str1 and str2, and prints result. def multiply(num1, num2): len1 = len(num1) len2 = len(num2) if len1 == 0 or len2 == 0: return "0" # will keep the result number in vector # in reverse order result = [0] * (len1 + len2) # Below two indexes are used to # find positions in result. i_n1 = 0 i_n2 = 0 # Go from right to left in num1 for i in range(len1 - 1, -1, -1): carry = 0 n1 = ord(num1[i]) - 48 # To shift position to left after every # multiplication of a digit in num2 i_n2 = 0 # Go from right to left in num2 for j in range(len2 - 1, -1, -1): # Take current digit of second number n2 = ord(num2[j]) - 48 # Multiply with current digit of first number # and add result to previously stored result # at current position. summ = n1 * n2 + result[i_n1 + i_n2] + carry # Carry for next iteration carry = summ // 10 # Store result result[i_n1 + i_n2] = summ % 10 i_n2 += 1 # 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 += 1 # print(result) # ignore '0's from the right i = len(result) - 1 while (i >= 0 and result[i] == 0): i -= 1 # 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 s = "" while (i >= 0): s += chr(result[i] + 48) i -= 1 return s # Driver code str1 = "1235421415454545454545454544"str2 = "1714546546546545454544548544544545" if((str1[0] == '-' or str2[0] == '-') and (str1[0] != '-' or str2[0] != '-')): print("-", end = '') if(str1[0] == '-' and str2[0] != '-'): str1 = str1[1:] elif(str1[0] != '-' and str2[0] == '-'): str2 = str2[1:] elif(str1[0] == '-' and str2[0] == '-'): str1 = str1[1:] str2 = str2[1:] print(multiply(str1, str2)) # This code is contributed by ankush_953 |
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C#
// C# program to multiply two numbers // represented as Strings. using System; class GFG { // Multiplies str1 and str2, and prints result. static String multiply(String num1, String num2) { int len1 = num1.Length; int len2 = num2.Length; if (len1 == 0 || len2 == 0) return "0"; // will keep the result number in vector // in reverse order int []result = new int[len1 + len2]; // Below two indexes are used to // find positions in result. int i_n1 = 0; int i_n2 = 0; int i; // Go from right to left in num1 for (i = len1 - 1; i >= 0; i--) { int carry = 0; int n1 = num1[i] - '0'; // To shift position to left after every // multipliccharAtion of a digit in num2 i_n2 = 0; // Go from right to left in num2 for (int j = len2 - 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 // charAt current position. int sum = n1 * n2 + result[i_n1 + i_n2] + carry; // Carry for next itercharAtion 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 // multipliccharAtion 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"; // genercharAte the result String String s = ""; while (i >= 0) s += (result[i--]); return s; } // Driver code public static void Main(String[] args) { String str1 = "1235421415454545454545454544"; String str2 = "1714546546546545454544548544544545"; if ((str1[0] == '-' || str2[0] == '-') && (str1[0] != '-' || str2[0] != '-')) Console.Write("-"); if (str1[0] == '-' && str2[0] != '-') { str1 = str1.Substring(1); } else if (str1[0] != '-' && str2[0] == '-') { str2 = str2.Substring(1); } else if (str1[0] == '-' && str2[0] == '-') { str1 = str1.Substring(1); str2 = str2.Substring(1); } Console.WriteLine(multiply(str1, str2)); } } // This code is contributed by Rajput-Ji |
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Output:
2118187521397235888154583183918321221520083884298838480662480
The above code is adapted from the code provided by Gaurav.
Time Complexity: O(m*n), where m and n are length of two number that need to be multiplied.
Another method:
// Java program to multiply two numbers represented // as strings. import java.util.Scanner; public class StringMultiplication { // Driver code public static void main(String[] args) { String num1 = "1235421415454545454545454544"; String tempnum1 = num1; String num2 = "1714546546546545454544548544544545"; String tempnum2 = num2; // Check condition if one string is negative if(num1.charAt(0) == '-' && num2.charAt(0)!='-') { num1 = num1.substring(1); } else if(num1.charAt(0) != '-' && num2.charAt(0) == '-') { num2 = num2.substring(1); } else if(num1.charAt(0) == '-' && num2.charAt(0) == '-') { num1 = num1.substring(1); num2 = num2.substring(1); } String s1 = new StringBuffer(num1).reverse().toString(); String s2 = new StringBuffer(num2).reverse().toString(); int[] m = new int[s1.length()+s2.length()]; // Go from right to left in num1 for (int i = 0; i < s1.length(); i++) { // Go from right to left in num2 for (int j = 0; j < s2.length(); j++) { m[i+j] = m[i+j]+(s1.charAt(i)-'0')*(s2.charAt(j)-'0'); } } String product = new String(); // Multiply with current digit of first number // and add result to previously stored product // at current position. for (int i = 0; i < m.length; i++) { int digit = m[i]%10; int carry = m[i]/10; if(i+1<m.length) { m[i+1] = m[i+1] + carry; } product = digit+product; } // ignore '0's from the right while(product.length()>1 && product.charAt(0) == '0') { product = product.substring(1); } // Check condition if one string is negative if(tempnum1.charAt(0) == '-' && tempnum2.charAt(0)!='-') { product = new StringBuffer(product).insert(0,'-').toString(); } else if(tempnum1.charAt(0) != '-' && tempnum2.charAt(0) == '-') { product = new StringBuffer(product).insert(0,'-').toString(); } else if(tempnum1.charAt(0) == '-' && tempnum2.charAt(0) == '-') { product = product; } System.out.println("Product of the two numbers is :"+"\n"+product); } } |
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Output:
2118187521397235888154583183918321221520083884298838480662480
Related Article :
Karatsuba algorithm for fast multiplication
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