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Multiply Large Numbers represented as Strings
• Difficulty Level : Hard
• Last Updated : 24 Aug, 2020

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
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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 ` `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; ` `} `

## 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 `

## 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 `

## 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 == ``'-'` `|| str2 == ``'-'``) &&  ` `        ``(str1 != ``'-'` `|| str2 != ``'-'``))  ` `        ``Console.Write(``"-"``);  ` ` `  `    ``if` `(str1 == ``'-'` `&& str2 != ``'-'``)  ` `    ``{  ` `        ``str1 = str1.Substring(1);  ` `    ``}  ` `    ``else` `if` `(str1 != ``'-'` `&& str2 == ``'-'``)  ` `    ``{  ` `        ``str2 = str2.Substring(1);  ` `    ``}  ` `    ``else` `if` `(str1 == ``'-'` `&& str2 == ``'-'``)  ` `    ``{  ` `        ``str1 = str1.Substring(1);  ` `        ``str2 = str2.Substring(1);  ` `    ``}  ` `    ``Console.WriteLine(multiply(str1, str2));  ` `}  ` `} ` ` `  `// This code is contributed by Rajput-Ji `

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````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); ` `    ``} ` `} `

Output:

```2118187521397235888154583183918321221520083884298838480662480
```

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