Given two positive integers a and n, calculate an modulo 1337.
Example 1:
Input: a = 2, n = 3 Output: 8
Example 2:
Input: a = 2, n = 10 Output: 1024
Example 3:
Input: a = 1, n = 1000 Output: 1
Example 4:
Input: a = 2147483647, n = 200 Output: 1198
Constraints:
1 <= a, n <= 231 - 1
The core challenge of this problem is to compute large powers efficiently. Direct computation of an
can result in extremely large numbers, which are impractical to handle directly due to memory and time constraints. The significance of this problem lies in its applications in cryptography, computer graphics, and numerical methods.
Potential pitfalls include integer overflow and inefficient algorithms that do not scale well with large inputs.
To solve this problem, we can use the method of Exponentiation by Squaring, which is an efficient way to compute large powers modulo a number. This method reduces the time complexity significantly compared to the naive approach.
The naive solution involves multiplying the base a
by itself n
times. This approach has a time complexity of O(n)
, which is not feasible for large values of n
.
Exponentiation by Squaring reduces the time complexity to O(log n)
. The idea is to break down the power calculation into smaller parts using the properties of exponents:
n
is even, an = (a2)n/2
n
is odd, an = a * an-1
Here is a step-by-step breakdown of the Exponentiation by Squaring algorithm:
n
is greater than 0:
n
is odd, multiply the result by a
and reduce n
by 1.a
and halve n
.
/**
* Function to calculate a^n % 1337 using Exponentiation by Squaring
* @param {number} a - base
* @param {number} n - exponent
* @return {number} - result of a^n % 1337
*/
function powerMod(a, n) {
const MOD = 1337;
let result = 1;
a = a % MOD; // Reduce a modulo 1337 initially
while (n > 0) {
// If n is odd, multiply the result by a
if (n % 2 === 1) {
result = (result * a) % MOD;
}
// Square a and halve n
a = (a * a) % MOD;
n = Math.floor(n / 2);
}
return result;
}
// Example usage:
console.log(powerMod(2, 3)); // Output: 8
console.log(powerMod(2, 10)); // Output: 1024
console.log(powerMod(1, 1000)); // Output: 1
console.log(powerMod(2147483647, 200)); // Output: 1198
The time complexity of the Exponentiation by Squaring algorithm is O(log n)
because we halve the exponent in each step. The space complexity is O(1)
as we use a constant amount of extra space.
Potential edge cases include:
a = 1
and any n
: The result is always 1.n = 0
: The result is 1 for any a
(by definition of exponentiation).a
and n
: The algorithm handles these efficiently due to the modulo operation and logarithmic time complexity.To test the solution comprehensively, consider the following test cases:
a
and n
(e.g., a = 2, n = 3
).a
and n
(e.g., a = 2147483647, n = 200
).a = 1
and n = 0
.Use a testing framework like Jest or Mocha to automate and validate these test cases.
When approaching such problems:
In this blog post, we discussed how to efficiently compute large powers modulo a number using Exponentiation by Squaring. This method significantly reduces the time complexity and handles large inputs effectively. Understanding and applying such algorithms is crucial in fields like cryptography and numerical computing.
Practice and explore further to deepen your understanding and improve your problem-solving skills.