# 20. Modular Exponentiation for large numbers

The problem can be found at the following link: [Question Link](https://www.geeksforgeeks.org/problems/modular-exponentiation-for-large-numbers5537/1)

## My Approach

* ans is initialized to 1 to store the final result.
* x is reduced modulo M to ensure it is within the range \[0, M-1].
* While n (the exponent) is greater than 0, repeat the following steps:
* If n is odd (i.e., n % 2 == 1):
* Multiply ans by x and take modulo M.
* This ensures that any intermediate results do not exceed the limits.
* Halve n by right-shifting it by 1 bit (n >>= 1), which is equivalent to n = n / 2.
* Square x and take modulo M (x = (x \* x) % M).
* This prepares x for the next iteration.
* After the loop completes (when n becomes 0), return ans as the result.

## Time and Auxiliary Space Complexity

* **Time Complexity**: The time complexity of this approach is `O(logN)`
* **Auxiliary Space Complexity**: The auxiliary space complexity is `O(1)`.

## Code (C++)

```cpp
class Solution
{
	public:
		long long int PowMod(long long int x,long long int n,long long int M)
		{
		    long long int ans=1;
            x%=M;
            while (n>0)
            {
                if (n%2==1)
                    ans=(ans*x)%M;
                n>>=1;
                x=(x*x)%M;
            }
            return ans;
		}
};
```

## Contribution and Support

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