grid47 Exploring patterns and algorithms
Sep 30, 2024
5 min read
Solution to LeetCode 372: Super Pow Problem
Given an integer a and an array b, where b represents a very large number, compute ( a^b % 1337 ). The array b stores the digits of b from most significant to least significant. You must handle extremely large values of b efficiently.
Problem
Approach
Steps
Complexity
Input: The input consists of an integer `a` and an array `b`, where `b[i]` is a digit of the large number `b`.
Example: Input: a = 3, b = [2]
Constraints:
• 1 <= a <= 231 - 1
• 1 <= b.length <= 2000
• 0 <= b[i] <= 9
• b does not contain leading zeros.
Output: Return the value of ( a^b % 1337 ).
Example: Output: 9
Constraints:
• The result should be computed modulo 1337.
Goal: The goal is to compute ( a^b % 1337 ) efficiently, even for extremely large values of `b`.
Steps:
• 1. Reduce `a` modulo 1337 to prevent overflow.
• 2. Use recursion to handle the large exponent, breaking the problem into smaller parts.
• 3. Use the principle of modular arithmetic to compute the result using smaller powers.
Goal: The constraints describe the valid range for `a` and the size of the array `b`.
Steps:
• 1 <= a <= 231 - 1
• 1 <= b.length <= 2000
• 0 <= b[i] <= 9
• b does not contain leading zeros.
Assumptions:
• The integer `a` is a positive integer.
• The array `b` represents a valid large integer without leading zeros.
• Input: Input: a = 3, b = [2]
Output: 9
• Explanation: The result is ( 3^2 % 1337 = 9 ), which is the expected output.
• Input: Input: a = 5, b = [1, 2]
Output: 25
• Explanation: The result is ( 5^{12} % 1337 = 25 ), which is computed using the recursive approach.
Approach: The approach uses recursion to break down the large exponent `b` and calculates the result using modular arithmetic.
Observations:
• Direct computation of large exponents is inefficient.
• Recursive solutions and modular arithmetic can break down the problem.
• By recursively calculating smaller powers, we can avoid overflow and handle the large size of `b`.
Steps:
• 1. Reduce `a` modulo 1337 to prevent overflow.
• 2. Recursively calculate the modular exponentiation of `a` raised to `b` using the divide and conquer method.
Empty Inputs:
• The array `b` will always have at least one digit.
Large Inputs:
• The array `b` can be up to 2000 digits long, so efficient handling of large numbers is necessary.
Special Values:
• If `a = 1`, the result is always 1, regardless of `b`.
Constraints:
• Ensure the solution handles the maximum and minimum values for `a` and handles large `b` values.
classSolution {
int base =1337;
intpowmod(int a, int k) {
a %= base;
int res =1;
for(int i =0; i < k; i++)
res = (res * a) % base;
return res;
}
public:int superPow(int a, vector<int>& b) {
if(b.empty()) return1;
int last_digit = b.back();
b.pop_back();
returnpowmod(superPow(a, b), 10) * powmod(a, last_digit) % base;
}
1 : Variable Initialization
int base =1337;
The `base` variable is initialized to 1337. This is the modulus used in the calculation to keep the result within manageable limits.
2 : Method Declaration
intpowmod(int a, int k) {
This line declares the helper method `powmod`, which calculates `a^k % base`. It takes two arguments: `a`, the base, and `k`, the exponent.
3 : Modulo Operation
a %= base;
This line reduces `a` modulo `base`, ensuring that the base remains within the modulus before performing exponentiation.
4 : Variable Initialization
int res =1;
The variable `res` is initialized to 1, which will hold the result of the power calculation.
5 : Loop Iteration
for(int i =0; i < k; i++)
This loop runs `k` times, each time multiplying `res` by `a` and reducing it modulo `base`.
6 : Multiplication
res = (res * a) % base;
In each iteration of the loop, `res` is updated to be the result of `res * a % base`, effectively calculating `a^k % base`.
7 : Return Statement
return res;
The final result of the modular exponentiation is returned as the output of the `powmod` method.
8 : Method Declaration
intsuperPow(int a, vector<int>& b) {
The method `superPow` is declared. This method calculates `a` raised to the power represented by the list of digits `b`.
9 : Base Case
if(b.empty()) return1;
If the vector `b` is empty, the result is `1`, as any number raised to the power of zero is 1.
10 : Pop Last Digit
int last_digit = b.back();
The last digit of the vector `b` is retrieved and stored in `last_digit`, representing the current power to apply.
11 : Pop Last Digit
b.pop_back();
The last digit is removed from the vector `b` to reduce the problem size for the recursive call.
The method uses recursion to calculate the result. It calls `superPow` recursively to calculate the power of `a` raised to the value represented by `b`. It then combines the result with the power of `a` raised to `last_digit`, applying the modulus operation at each step.
Best Case: O(log(b.length)) where `b.length` is the number of digits in `b`.
Average Case: O(log(b.length)) due to the recursive approach.
Worst Case: O(log(b.length)) as the recursion depth is determined by the length of `b`.
Description: The time complexity is logarithmic in relation to the length of `b`, as the problem is divided in each recursive step.
Best Case: O(1) if the recursion depth is minimal.
Worst Case: O(log(b.length)) due to the recursion stack.
Description: The space complexity is proportional to the recursion depth, which is logarithmic in relation to the number of digits in `b`.
