Solution Approach
Approach 1: Brute Force Approach
The most straightforward idea is to go by the definition of exponential and multiply a to itself b times. We run a loop for b times and keep multiplying a to answer and taking modulus each time. Also, the answer should be initiated by 1.
Pseudocode
function exponentiation(int a, int b, int m){
int answer = 1;
for(int i=0;i<b;i++){
answer = (answer*a)%m;
}
return answer%m;
}
C++ implementation
#include<bits/stdc++.h>
using namespace std;
int exponentiation(int a, int b, int m){
int answer = 1;
for(int i=0;i<b;i++){
answer = (answer*a)%m;
}
return answer%m;
}
int main()
{
int a = 2;
int b = 5;
int m = 1000000;
cout<<exponentiation(a,b)<<endl;
}
Output
32
Complexity
Time complexity
O(b), where b is the exponent
Reason: Since we’re multiplying a to the answer b times using a loop, the time complexity will be O(b). Taking modulus takes constant time.
Space complexity
O(1)
Reason: All the spaces taken are constant.
Approach 2: Binary exponentiation
This is the most efficient approach to do modular exponentiation. We need to calculate ab%m, which can also be written as (a2)b/2%m. Notice that computing a2 takes only constant time, and the whole computation steps are reduced to b/2 steps from b steps. Thus, now we need to calculate xb/2, where x = a2. But, notice that if b is odd, then b/2 will be a decimal, and calculating that is not easy. Also, if b is odd, we can make it even. How? So, xb can be written as x*(xb-1). Thus, whenever we encounter an odd b, multiply x to the answer and reduce the value of b by 1. Then, keep dividing the power by 2 as long as it is even, and keep replacing the base by its square. Keep taking modulus at each step.
Pseudocode
function exponentiation(int a, int b, int m){
int answer = 1;
while(b>0){
if(b%2==1){
answer = (answer*a)%m;
b--;
}
a = (a*a)%m;
b = b/2;
}
return answer%m;
}
C++ implementation
#include<bits/stdc++.h>
using namespace std;
int exponentiation(int a, int b){
int answer = 1;
while(b>0){
if(b%2==1){
answer=(answer*a)%m;
b--;
}
a = (a*a)%m;
b = b/2;
}
return answer%m;
}
int main()
{
int a = 2;
int b = 5;
int m = 1000000;
cout<<exponentiation(a,b,m)<<endl;
}
Output
32
Complexities
Time complexity
O(log2b), where b is the exponent
Reason: Since we’re dividing b by 2 in each step, the total number of steps will be log2b. Taking modulus takes constant time.
Space complexity
O(1)
Reason: All the spaces taken are constant.
Check out this problem - Maximum Product Subarray
Frequently asked questions
What is modular exponentiation method?
Exponentiation is a mathematical operation written as ab, which means the value a is multiplied by itself b times.
What is the time complexity of the binary exponentiation method?
The time complexity of the binary exponentiation method is log(b), where b is the exponent.
What should we do if an exponentiation's value is too large to fit in the integer range?
If any answer's value is so large that it can’t fit in the integer range, take its modulo with a prime number. Generally, that prime number is 10^9 + 7.
What is modulo?
Modulo represents the remainder when a number a is divided by another number b.
Where can binary exponentiation be used?
Binary exponentiation can efficiently compute Fibonacci numbers and apply a permutation on an array k times.
Conclusion
This article discussed the approach to solving “Modular exponentiation.” This was a basic problem of number theory. You should solve some more number theory problems to have a good grasp of this topic. Some of these are: number of vehicles, Yogesh and primes, alien dictionary, nth Fibonacci number, jumping numbers, minimum jumps, and stocks are profitable.
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