# Modular exponentiation (Recursive)

• Difficulty Level : Medium
• Last Updated : 31 May, 2022

Given three numbers a, b and c, we need to find (ab) % c
Now why do “% c” after exponentiation, because ab will be really large even for relatively small values of a, b and that is a problem because the data type of the language that we try to code the problem, will most probably not let us store such a large number.
Examples:

```Input : a = 2312 b = 3434 c = 6789
Output : 6343

Input : a = -3 b = 5 c = 89
Output : 24```

Auxiliary Space: O(1)

The idea is based on below properties.
Property 1:
(m * n) % p has a very interesting property:
(m * n) % p =((m % p) * (n % p)) % p
Property 2:
if b is even:
(a ^ b) % c = ((a ^ b/2) * (a ^ b/2))%c ? this suggests divide and conquer
if b is odd:
(a ^ b) % c = (a * (a ^( b-1))%c
Property 3:
If we have to return the mod of a negative number x whose absolute value is less than y:
then (x + y) % y will do the trick
Note:
Also as the product of (a ^ b/2) * (a ^ b/2) and a * (a ^( b-1) may cause overflow, hence we must be careful about those scenarios

## C++

 `// Recursive C++ program to compute modular power``#include ``using` `namespace` `std;` `int` `exponentMod(``int` `A, ``int` `B, ``int` `C)``{``    ``// Base cases``    ``if` `(A == 0)``        ``return` `0;``    ``if` `(B == 0)``        ``return` `1;` `    ``// If B is even``    ``long` `y;``    ``if` `(B % 2 == 0) {``        ``y = exponentMod(A, B / 2, C);``        ``y = (y * y) % C;``    ``}` `    ``// If B is odd``    ``else` `{``        ``y = A % C;``        ``y = (y * exponentMod(A, B - 1, C) % C) % C;``    ``}` `    ``return` `(``int``)((y + C) % C);``}` `// Driver code``int` `main()``{``    ``int` `A = 2, B = 5, C = 13;``    ``cout << ``"Power is "` `<< exponentMod(A, B, C);``    ``return` `0;``}` `// This code is contributed by SHUBHAMSINGH10`

## C

 `// Recursive C program to compute modular power``#include ` `int` `exponentMod(``int` `A, ``int` `B, ``int` `C)``{``    ``// Base cases``    ``if` `(A == 0)``        ``return` `0;``    ``if` `(B == 0)``        ``return` `1;` `    ``// If B is even``    ``long` `y;``    ``if` `(B % 2 == 0) {``        ``y = exponentMod(A, B / 2, C);``        ``y = (y * y) % C;``    ``}` `    ``// If B is odd``    ``else` `{``        ``y = A % C;``        ``y = (y * exponentMod(A, B - 1, C) % C) % C;``    ``}` `    ``return` `(``int``)((y + C) % C);``}` `// Driver program to test above functions``int` `main()``{``   ``int` `A = 2, B = 5, C = 13;``   ``printf``(``"Power is %d"``, exponentMod(A, B, C));``   ``return` `0;``}`

## Java

 `// Recursive Java program``// to compute modular power``import` `java.io.*;` `class` `GFG``{``static` `int` `exponentMod(``int` `A,``                       ``int` `B, ``int` `C)``{``        ` `    ``// Base cases``    ``if` `(A == ``0``)``        ``return` `0``;``    ``if` `(B == ``0``)``        ``return` `1``;``    ` `    ``// If B is even``    ``long` `y;``    ``if` `(B % ``2` `== ``0``)``    ``{``        ``y = exponentMod(A, B / ``2``, C);``        ``y = (y * y) % C;``    ``}``    ` `    ``// If B is odd``    ``else``    ``{``        ``y = A % C;``        ``y = (y * exponentMod(A, B - ``1``,``                             ``C) % C) % C;``    ``}``    ` `    ``return` `(``int``)((y + C) % C);``}` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``int` `A = ``2``, B = ``5``, C = ``13``;``    ``System.out.println(``"Power is "` `+``                        ``exponentMod(A, B, C));``}``}` `// This code is contributed``// by Swetank Modi.`

## Python3

 `# Recursive Python program``# to compute modular power``def` `exponentMod(A, B, C):``    ` `    ``# Base Cases``    ``if` `(A ``=``=` `0``):``        ``return` `0``    ``if` `(B ``=``=` `0``):``        ``return` `1``    ` `    ``# If B is Even``    ``y ``=` `0``    ``if` `(B ``%` `2` `=``=` `0``):``        ``y ``=` `exponentMod(A, B ``/` `2``, C)``        ``y ``=` `(y ``*` `y) ``%` `C``    ` `    ``# If B is Odd``    ``else``:``        ``y ``=` `A ``%` `C``        ``y ``=` `(y ``*` `exponentMod(A, B ``-` `1``,``                             ``C) ``%` `C) ``%` `C``    ``return` `((y ``+` `C) ``%` `C)` `# Driver Code``A ``=` `2``B ``=` `5``C ``=` `13``print``(``"Power is"``, exponentMod(A, B, C))``    ` `# This code is contributed``# by Swetank Modi.`

## C#

 `// Recursive C# program``// to compute modular power``class` `GFG``{``static` `int` `exponentMod(``int` `A, ``int` `B, ``int` `C)``{``        ` `    ``// Base cases``    ``if` `(A == 0)``        ``return` `0;``    ``if` `(B == 0)``        ``return` `1;``    ` `    ``// If B is even``    ``long` `y;``    ``if` `(B % 2 == 0)``    ``{``        ``y = exponentMod(A, B / 2, C);``        ``y = (y * y) % C;``    ``}``    ` `    ``// If B is odd``    ``else``    ``{``        ``y = A % C;``        ``y = (y * exponentMod(A, B - 1,``                             ``C) % C) % C;``    ``}``    ` `    ``return` `(``int``)((y + C) % C);``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `A = 2, B = 5, C = 13;``    ``System.Console.WriteLine(``"Power is "` `+``                    ``exponentMod(A, B, C));``}``}` `// This code is contributed``// by mits`

## PHP

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## Javascript

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Output:

`Power is 6`

Time Complexity : O(logn)

Auxiliary Space: O(logn)

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