# Multiply large integers under large modulo

• Difficulty Level : Medium
• Last Updated : 20 Dec, 2022

Given an integer a, b, m. Find (a * b ) mod m, where a, b may be large and their direct multiplication may cause overflow. However, they are smaller than half of the maximum allowed long long int value.

Examples:

Input: a = 426, b = 964, m = 235
Output: 119
Explanation: (426 * 964) % 235  = 410664 % 235 = 119

Input: a = 10123465234878998,
b = 65746311545646431
m = 10005412336548794
Output: 4652135769797794

Naive Approach: A naive approach is to use arbitrary precision data types such as int in python or Biginteger class in Java. But that approach will not be fruitful because the internal conversion of string to int and then perform operation will lead to slow down the calculations of addition and multiplications in the binary number system.

Efficient Approach: Since a and b may be very large numbers, if we try to multiply directly, they will definitely overflow. Therefore we use the basic approach of multiplication i.e., a * b = a + a + … + a (b times). Now easily compute the value of addition (under modulo m) without any overflow in the calculation. But if we try to add the value of a repeatedly up to b times then it will definitely timeout for the large value of b, since the time complexity of this approach would become O(b).

So, divide the above-repeated steps for a in simpler way i.e.,

If b is even then
a * b = 2 * a * (b / 2),

otherwise
a * b = a + a * (b – 1)

Below is the approach describing the above explanation :

## C++

 `// C++ program of finding modulo multiplication``#include ` `using` `namespace` `std;` `// Returns (a * b) % mod``long` `long` `moduloMultiplication(``long` `long` `a, ``long` `long` `b,``                               ``long` `long` `mod)``{``    ``long` `long` `res = 0; ``// Initialize result` `    ``// Update a if it is more than``    ``// or equal to mod``    ``a %= mod;` `    ``while` `(b) {``        ``// If b is odd, add a with result``        ``if` `(b & 1)``            ``res = (res + a) % mod;` `        ``// Here we assume that doing 2*a``        ``// doesn't cause overflow``        ``a = (2 * a) % mod;` `        ``b >>= 1; ``// b = b / 2``    ``}` `    ``return` `res;``}` `// Driver program``int` `main()``{``    ``long` `long` `a = 426;``    ``long` `long` `b = 964;``    ``long` `long` `m = 235;``    ``cout << moduloMultiplication(a, b, m);``    ``return` `0;``}` `// This code is contributed``// by Akanksha Rai`

## C

 `// C program of finding modulo multiplication``#include` `// Returns (a * b) % mod``long` `long` `moduloMultiplication(``long` `long` `a,``                               ``long` `long` `b,``                               ``long` `long` `mod)``{``    ``long` `long` `res = 0;  ``// Initialize result` `    ``// Update a if it is more than``    ``// or equal to mod``    ``a %= mod;` `    ``while` `(b)``    ``{``        ``// If b is odd, add a with result``        ``if` `(b & 1)``            ``res = (res + a) % mod;` `        ``// Here we assume that doing 2*a``        ``// doesn't cause overflow``        ``a = (2 * a) % mod;` `        ``b >>= 1;  ``// b = b / 2``    ``}` `    ``return` `res;``}` `// Driver program``int` `main()``{``    ``long` `long` `a = 10123465234878998;``    ``long` `long` `b = 65746311545646431;``    ``long` `long` `m = 10005412336548794;``    ``printf``(``"%lld"``, moduloMultiplication(a, b, m));``    ``return` `0;``}`

## Java

 `// Java program of finding modulo multiplication``import` `java.util.*;``import` `java.io.*;``class` `GFG``{` `    ``// Returns (a * b) % mod``    ``static` `long` `moduloMultiplication(``long` `a,``                            ``long` `b, ``long` `mod)``    ``{``        ` `        ``// Initialize result``        ``long` `res = ``0``; ` `        ``// Update a if it is more than``        ``// or equal to mod``        ``a %= mod;` `        ``while` `(b > ``0``)``        ``{``            ` `            ``// If b is odd, add a with result``            ``if` `((b & ``1``) > ``0``)``            ``{``                ``res = (res + a) % mod;``            ``}` `            ``// Here we assume that doing 2*a``            ``// doesn't cause overflow``            ``a = (``2` `* a) % mod;` `            ``b >>= ``1``; ``// b = b / 2``        ``}``        ``return` `res;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``long` `a = 10123465234878998L;``        ``long` `b = 65746311545646431L;``        ``long` `m = 10005412336548794L;``        ``System.out.print(moduloMultiplication(a, b, m));``    ``}``}` `// This code is contributed by Rajput-JI`

## Python3

 `# Python 3 program of finding``# modulo multiplication` `# Returns (a * b) % mod``def` `moduloMultiplication(a, b, mod):` `    ``res ``=` `0``; ``# Initialize result` `    ``# Update a if it is more than``    ``# or equal to mod``    ``a ``=` `a ``%` `mod;` `    ``while` `(b):``    ` `        ``# If b is odd, add a with result``        ``if` `(b & ``1``):``            ``res ``=` `(res ``+` `a) ``%` `mod;``            ` `        ``# Here we assume that doing 2*a``        ``# doesn't cause overflow``        ``a ``=` `(``2` `*` `a) ``%` `mod;` `        ``b >>``=` `1``; ``# b = b / 2``    ` `    ``return` `res;` `# Driver Code``a ``=` `10123465234878998``;``b ``=` `65746311545646431``;``m ``=` `10005412336548794``;``print``(moduloMultiplication(a, b, m));``    ` `# This code is contributed``# by Shivi_Aggarwal`

## C#

 `// C# program of finding modulo multiplication``using` `System;` `class` `GFG``{``    ` `// Returns (a * b) % mod``static` `long` `moduloMultiplication(``long` `a,``                            ``long` `b,``                            ``long` `mod)``{``    ``long` `res = 0; ``// Initialize result` `    ``// Update a if it is more than``    ``// or equal to mod``    ``a %= mod;` `    ``while` `(b > 0)``    ``{``        ``// If b is odd, add a with result``        ``if` `((b & 1) > 0)``            ``res = (res + a) % mod;` `        ``// Here we assume that doing 2*a``        ``// doesn't cause overflow``        ``a = (2 * a) % mod;` `        ``b >>= 1; ``// b = b / 2``    ``}` `    ``return` `res;``}` `// Driver code``static` `void` `Main()``{``    ``long` `a = 10123465234878998;``    ``long` `b = 65746311545646431;``    ``long` `m = 10005412336548794;``    ``Console.WriteLine(moduloMultiplication(a, b, m));``}``}` `// This code is contributed``// by chandan_jnu`

## PHP

 `>= 1; ``// b = b / 2``    ``}` `    ``return` `\$res``;``}` `    ``// Driver Code``    ``\$a` `= 10123465234878998;``    ``\$b` `= 65746311545646431;``    ``\$m` `= 10005412336548794;``    ``echo` `moduloMultiplication(``\$a``, ``\$b``, ``\$m``);` `// This oce is contributed by ajit``?>`

## Javascript

 ``

Output

`119`

Time complexity: O(log b), A number n has log(n) bits therefore the loop will run log(b) times.
Auxiliary space: O(1)

Note: Above approach will only work if 2 * m can be represented in standard data type otherwise it will lead to overflow.

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