# Multiply large integers under large modulo

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.

```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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

Efficient solution : Since a and b may be very large numbers, if we try to multiply directly then it will definitely overflow. Therefore we use the basic approach of multiplication i.e.,

a * b = a + a + … + a (b times)
So we can 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 we divide the above repeated steps of 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 C/C++ approach describing above explanation.

```// C program of finding modulo multiplication
#include<stdio.h>

// 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;
}
```
```Output:
4652135769797794
```

Time complexity: O(log b)
Auxiliary space: O(1)

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

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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