20. Modular Exponentiation for large numbers

The problem can be found at the following link: Question Link

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++)

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;
		}
};

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