Super Pow
#993 Most AskedMedium
Super Pow #993 Most Asked
Medium
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Topics:
ArraysRecursionMath
Your task is to calculate ab mod 1337 where a is a positive integer and b is an extremely large positive integer given in the form of an array.
Example 1:
Input: a = 2, b = [3] Output: 8
Example 2:
Input: a = 2, b = [1,0] Output: 1024
Example 3:
Input: a = 1, b = [4,3,3,8,5,2] Output: 1
Constraints:
1 <= a <= 231 - 11 <= b.length <= 20000 <= b[i] <= 9bdoes not contain leading zeros.
Solution
Clarifying Questions
When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
- What is the range of values for 'a' and for each element in the 'b' array? Are we dealing with strictly positive integers?
- Can 'b' be an empty array or contain zero?
- What is the modulo value that should be used to compute the result? Is it a constant, and what is its value?
- Should I handle potential integer overflow issues when calculating intermediate power values?
- If 'a' or elements within 'b' are zero, how should the function behave?
Brute Force Solution
Approach
The goal is to compute a very large power of a number, but breaking it down into smaller, manageable steps. The brute force approach directly applies the definition of exponentiation: repeated multiplication.
Here's how the algorithm would work step-by-step:
- Start with the initial number.
- Multiply this number by itself repeatedly, where the number of repetitions is determined by the exponent.
- After each multiplication, take the remainder of the result when divided by a specified value. This keeps the numbers from getting too large.
- Repeat the multiplication and remainder operation until you have multiplied the initial number by itself the correct number of times, based on the exponent.
- The final result is the answer.
Code Implementation
def super_pow_brute_force(base, exponent_array):
modulo = 1337
result = 1
# Need to iterate through the exponent digits
for digit in exponent_array:
exponent_value = digit
# Simulate exponentiation by repeated multiplication
for _ in range(exponent_value):
result = (result * base) % modulo
return resultBig(O) Analysis
Time Complexity
O(n) – The algorithm performs repeated multiplication a number of times determined by the exponent, which is represented as an array of digits. The number of iterations is directly proportional to the magnitude of the exponent, which we can approximate by the number of digits 'n' in the exponent array. Each multiplication and modulo operation takes constant time. Therefore, the overall time complexity is O(n).
Space Complexity
O(1) – The provided explanation describes an iterative process of repeated multiplication and modulo operations. It does not mention the creation of any auxiliary data structures like lists, arrays, or hash maps. The algorithm only needs to store a few variables to keep track of the intermediate result and the number of repetitions. Since the space used is independent of the exponent's size (N), the space complexity is constant.
Optimal Solution
Approach
Calculating very large powers directly is inefficient and can lead to overflows. The trick is to break down the exponent into smaller parts and apply modular arithmetic properties to keep the intermediate results manageable.
Here's how the algorithm would work step-by-step:
- Recognize that you don't need to compute the entire power first and then take the modulus. You can take the modulus at each step.
- Break down the exponent into its individual digits. Think of the exponent as a series of powers of 10.
- Apply the property that (a*b) mod m is the same as ((a mod m) * (b mod m)) mod m. This allows us to work with smaller numbers.
- Repeatedly square the base 'a', taking the modulus at each step. This allows us to calculate powers of 2 very quickly.
- Combine these results to calculate the final power, again taking the modulus at each step. This avoids large intermediate values.
- Essentially, we are using the properties of modular exponentiation and a divide-and-conquer approach to handle the large exponent.
Code Implementation
def super_pow(base, exponent_array):
modulo = 1337
result = 1
for exponent in exponent_array:
# Apply modular exponentiation property.
result = pow(result, 10, modulo) * pow(base, exponent, modulo) % modulo
return resultBig(O) Analysis
Time Complexity
O(n) – The dominant operation is iterating through the digits of the exponent represented as an array (denoted as 'b' with size 'n'). Within the loop, modular exponentiation (specifically repeated squaring) takes place, which has a logarithmic time complexity relative to the base, but the number of squaring operations inside the loop is bounded by the digits of a single element in the array 'b' (0-9). Therefore, the loop iterates 'n' times, with each iteration performing a constant number of mathematical operations related to modular exponentiation. Thus, the overall time complexity is O(n), where n is the number of digits in the exponent.
Space Complexity
O(1) – The algorithm primarily uses modular arithmetic and repeated squaring. It breaks down the exponent into individual digits, but it doesn't store all digits simultaneously in an auxiliary data structure. Intermediate results are calculated and immediately used in modular operations. Therefore, the auxiliary space used remains constant, regardless of the size of the exponent array. Thus the space complexity is O(1).
Edge Cases
| Case | How to Handle |
|---|---|
| a is 0 and b is not an empty array | Return 0 since 0 to any positive power is 0. |
| a is 1 | Return 1 since 1 to any power is 1. |
| b is an empty array | Return 1 since any number to the power of 0 is 1. |
| Large a leading to potential overflow during intermediate calculations | Apply modulo operation after each multiplication to prevent overflow. |
| Large b array representing a huge exponent | The properties of modular arithmetic and Euler's theorem allows us to significantly reduce the effective exponent. |
| The base 'a' itself is equal to the modulo value. | If a is equal to the modulo value, the result is always 0 regardless of exponent. |
| b contains leading zeros (e.g., [0, 0, 1, 2]) | Ignore the leading zeros since they don't affect the resulting exponent. |
| a is negative | Since the exponent 'b' is treated as a whole number, ignore the sign of 'a' and return the positive power mod the base. |
a is 0 and b is not an empty array
How to Handle:
Return 0 since 0 to any positive power is 0.
a is 1
How to Handle:
Return 1 since 1 to any power is 1.
b is an empty array
How to Handle:
Return 1 since any number to the power of 0 is 1.
Large a leading to potential overflow during intermediate calculations
How to Handle:
Apply modulo operation after each multiplication to prevent overflow.
Large b array representing a huge exponent
How to Handle:
The properties of modular arithmetic and Euler's theorem allows us to significantly reduce the effective exponent.
The base 'a' itself is equal to the modulo value.
How to Handle:
If a is equal to the modulo value, the result is always 0 regardless of exponent.
b contains leading zeros (e.g., [0, 0, 1, 2])
How to Handle:
Ignore the leading zeros since they don't affect the resulting exponent.
a is negative
How to Handle:
Since the exponent 'b' is treated as a whole number, ignore the sign of 'a' and return the positive power mod the base.
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