Binary exponentiation modulo
Implement modular exponentiation: return (a raised to the power b) modulo m, for b up to 10^18, WITHOUT using Python's built-in pow three-argument form. Use binary (square-and-multiply) exponentiation, reducing modulo m at every multiplication to keep numbers small. Return 0 when m == 1.
Implement
mod_pow(a: int, b: int, m: int) → intExamples
in
[2,10,1000]out24What a strong answer looks like
State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
0:00 of about 15 min
solution.py
InputExpectedGot
[2,10,1000]24not run yetsample