Pow(x, n)

Identifying Problem Isomorphism

“Fast Powering Algorithm” is approximately isomorphic to “myPow”.

Reasoning:

Both problems require the implementation of an efficient algorithm for exponentiation, i.e., computing a number raised to an integer exponent. The approach to both problems is similar: applying the “divide and conquer” strategy to reduce the size of the problem at each recursive step.

In “Fast Powering Algorithm”, you are asked to compute the power of a number in the least number of multiplications, which is similar to “myPow” where you are also required to compute the power of a number. However, “Fast Powering Algorithm” can be more complex as it may ask for the minimum number of multiplications, thus requiring more intricate solution than “myPow”.

“Exponentiation by Squaring” is also a similar problem.

Reasoning:

It’s a technique used for quickly computing large positive integer powers of a number, and can be used to compute large Fibonacci numbers, amongst other things. This problem can be more complex than the “myPow” problem, as it requires a more profound understanding of number theory.

Remember, isomorphisms between problems don’t mean they are identical. They mean that the problems are structurally similar, and strategies or solutions for one problem can be adapted to solve the other. In this case, both problems are solvable by an efficient implementation of exponentiation, despite the different specifics and requirements.

From simple to complex, the order:

  1. myPow: This is the simplest problem of the three. It involves computing x raised to the power n. The main challenge is to handle large and negative exponents, which can be tackled using a straightforward divide-and-conquer approach.

  2. Fast Powering Algorithm: This problem is slightly more complex than the myPow problem. The task is similar – computing x to the power n – but the problem asks for the least number of multiplications. This adds an optimization layer to the problem.

  3. Exponentiation by Squaring: This is the most complex problem among the three. It involves the rapid computation of large positive integer powers of a number, and can be applied to calculate large Fibonacci numbers. Understanding this problem requires a deeper grasp of number theory.

10 Prerequisite LeetCode Problems

The “50. Pow(x, n)” problem involves calculating x raised to the power n (xn). It’s a problem about recursion and efficient power calculation. Here are 10 simpler problems to prepare for this problem:

  1. 371. Sum of Two Integers: This problem involves calculation without the use of + or -.

  2. 231. Power of Two: In this problem, you have to determine if a given integer is a power of 2.

  3. 326. Power of Three: A similar problem to the previous one but now you have to determine if a given integer is a power of 3.

  4. 342. Power of Four: Yet another similar problem but now you have to determine if a given integer is a power of 4.

  5. 168. Excel Sheet Column Title: This problem involves converting a given integer to its corresponding column title as it appears in an Excel sheet.

  6. 171. Excel Sheet Column Number: This is the reverse problem to the previous one, where you have to convert a column title to its corresponding integer number.

  7. 202. Happy Number: This problem involves manipulation of an integer’s digits and understanding of number properties.

  8. 9. Palindrome Number: Here, you are asked to determine if a number is a palindrome.

  9. 191. Number of 1 Bits: This problem involves bit manipulation and could give some useful insights into handling numbers at a lower level.

  10. 338. Counting Bits: Another problem involving bit manipulation where you need to count the number of 1s in the binary representation of numbers in a range.

These cover mathematical problem-solving and manipulation of numbers in programming. These concepts will be useful in tackling the “Pow(x, n)” problem.

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class Solution:
    def myPow(self, x: float, n: int) -> float:

        def function(base=x, exponent=abs(n)):
            if exponent == 0:
                return 1
            elif exponent % 2 == 0:
                return function(base * base, exponent // 2)
            else:
                return base * function(base * base, (exponent - 1) // 2)

        f = function()

        return float(f) if n >= 0 else 1/f

The code you’ve shared is an implementation of the power function, which calculates (x^n), where (x) is a float and (n) is an integer. This particular implementation uses a technique known as “fast exponentiation” or “exponentiation by squaring”. This technique reduces the time complexity from (O(n)) in a straightforward approach to (O(\log n)) in this approach.

The function function(base=x, exponent=abs(n)) is a recursive helper function used to calculate the power of a number. It uses the properties of exponentiation to reduce the number of multiplications needed.

Here’s how it works:

  1. If the exponent is zero, the result is always 1, regardless of the base (by definition of exponentiation).

  2. If the exponent is even, then (x^n) can be rewritten as ((x^2)^{n/2}). For example, (x^4 = (x^2)^2). This step essentially halves the exponent.

  3. If the exponent is odd, then (x^n) can be rewritten as (x \cdot x^{n-1}). The function calculates (x^{n-1}) as ((x^2)^{(n-1)/2}) to keep the exponent halving principle.

The result f from the helper function is then returned. If the original exponent n was negative, the function returns the reciprocal of the result, because (x^{-n} = 1/x^n).

This approach uses recursion to keep reducing the problem size until it reaches the base case (exponent equals 0). The calculated results are then combined to produce the final output.

Note that the base is squared at each recursive call, and the exponent is halved. This is how the function manages to perform fewer multiplications than a straightforward loop that would multiply the base (n) times.

This is a slightly different approach to solving the problem but the underlying principle is still divide-and-conquer. Here, an internal function function is defined inside the myPow method, which computes the power of base to the exponent.

There are two key differences in this version:

  1. Handling Negative Exponents: In the previous solution, negative exponents are handled within the recursive calls by transforming the problem to 1/x^-n. In this version, the absolute value of the exponent is used in the recursive calls, which ensures the exponent is always positive. The result is then inverted if the original n was negative. This could be a cleaner and more intuitive approach for some people.

  2. Recursive Call: This solution also uses the rule x^n = (x^2)^(n/2) for even n and x^n = x * (x^2)^((n-1)/2) for odd n, which is slightly different from the x^n = x * x^(n-1) for odd n in the previous solution. Both methods essentially reduce the problem size by about half in each recursive call.

Performance-wise, both solutions have the same time complexity: O(log n). They only differ in how they handle the special cases and how they reduce the problem size. Both approaches are valid and the choice between them could come down to personal preference or the specific requirements of your task.

Implement pow(x, n), which calculates x raised to the power n (i.e., xn).

Example 1:

Input: x = 2.00000, n = 10 Output: 1024.00000

Example 2:

Input: x = 2.10000, n = 3 Output: 9.26100

Example 3:

Input: x = 2.00000, n = -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 1/4 = 0.25

Constraints:

-100.0 < x < 100.0 -231 <= n <= 231-1 n is an integer. Either x is not zero or n > 0. -104 <= xn <= 104

We need to implement a function to compute x raised to the power n. Let’s call this function pow. The main challenges are handling large exponents and negative exponents.

Here is a simple solution using recursion and the principle of divide and conquer.

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class Solution:
    def myPow(self, x: float, n: int) -> float:
        if n == 0: 
            return 1.0
        if n < 0: 
            return 1.0 / self.myPow(x, -n)
        if n % 2: 
            return x * self.myPow(x, n - 1)
        return self.myPow(x*x, n // 2)

Let’s understand the function:

  1. If n is 0, the function returns 1. This is the base case for the recursion.

  2. If n is negative, the function converts the problem to a positive exponent problem by using the rule x^(-n) = 1/x^n.

  3. If n is odd (n%2 = 1), the function converts the problem to an even exponent problem. This is done by using the rule x^n = x * x^(n-1).

  4. If n is even (n%2 = 0), the function reduces the problem size by half. This is done by using the rule x^n = (x^2)^(n/2).

This function works in O(log n) time complexity. The reason is that the function reduces the problem size by half in each recursive call when n is even. This is much more efficient than a naive implementation that multiplies x by itself n times, which would take O(n) time.

Remember, Python has a built-in function pow(x, n) that performs the same calculation. But the purpose of this exercise is to learn how such a function could be implemented from scratch.

Why memoization is not used in this solution?

Memoization is a technique where you store the results of expensive function calls and reuse the cached result when the same inputs occur again. It’s typically useful in dynamic programming problems, where subproblems overlap and the same computations are performed multiple times.

In this power function, we’re using a divide-and-conquer approach. Each recursive call works on a unique, non-overlapping subproblem: either n/2 or n-1. This means we don’t repeat the same computations with the same inputs, which is when memoization would provide a benefit.

Due to this lack of overlapping subproblems, memoization wouldn’t bring a noticeable improvement to the efficiency of this specific algorithm. On the contrary, it could potentially increase the memory usage of the algorithm due to storing unnecessary computation results. Therefore, we haven’t used memoization in this solution.

However, in problems where you see that you’re computing the same function with the same arguments multiple times, memoization could be a good strategy to optimize the function and save computation time.

Language Agnostic Coding Drills

  1. Variable Declaration: Understanding how to declare variables is the most basic requirement for any programming language.

  2. Conditional Statements (if, elif, else): Conditional statements are used to perform different actions based on different conditions. These are fundamental in controlling the flow of the program.

  3. Recursion: In this code, a recursive function function is implemented. Understanding recursion, the process where a function calls itself as a subroutine, is a key concept in many algorithmic solutions.

  4. Mathematical operations (Multiplication, Division, Modulus): The code uses multiplication (*), division (//), and modulus (%) operators. Familiarity with these operators is important in performing mathematical calculations in programming.

  5. Function Definition and Call: Functions are reusable pieces of code. They allow code to be modular and more readable. In this code, the function function is defined and then called.

  6. Default arguments in Functions: The function function uses default arguments base and exponent which are initialized with x and abs(n). Understanding how to define and use default arguments is necessary for flexible function design.

  7. Object-Oriented Programming (Defining methods within a class): This code is written in a class Solution and involves the concept of object-oriented programming.

  8. Return statement: The return statement ends the execution of a function and returns the result to the caller. This code uses return statements to return the result of the computation.

  9. Type Conversion: The float function is used to convert the output to a float. Understanding type conversion functions is useful for ensuring that variables are of the correct type for a given operation.

  10. Working with numbers (Exponentiation, Absolute value): The code uses abs(n) to get the absolute value of n. Understanding these functions and their uses is important for working with numbers.

In Python, these concepts can be practiced with simple exercises like creating a basic function, implementing a recursive function, performing mathematical operations, defining classes and methods, and so on.

Targeted Drills in Python

Sure, here are coding drills for each of the concepts in Python:

  1. Variable Declaration: Declare variables and print their values.
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x = 10
n = 2
print(x, n)
  1. Conditional Statements (if, elif, else): Write a conditional statement to check if a number is positive, negative, or zero.
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num = -5
if num > 0:
    print("Positive number")
elif num == 0:
    print("Zero")
else:
    print("Negative number")
  1. Recursion: Write a recursive function to calculate factorial of a number.
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def factorial(n):
    if n == 1:
        return 1
    else:
        return n * factorial(n-1)
print(factorial(5))  # 120
  1. Mathematical operations (Multiplication, Division, Modulus): Perform various mathematical operations.
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print(10 * 5)  # 50
print(10 // 3)  # 3
print(10 % 3)  # 1
  1. Function Definition and Call: Define a function to calculate square of a number and call it.
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def square(n):
    return n ** 2
print(square(4))  # 16
  1. Default arguments in Functions: Define a function with default arguments.
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def power(base, exponent=2):
    return base ** exponent
print(power(4))  # 16
print(power(2, 3))  # 8
  1. Object-Oriented Programming (Defining methods within a class): Define a class with a method.
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class Solution:
    def square(self, n):
        return n ** 2
s = Solution()
print(s.square(4))  # 16
  1. Return statement: Use the return statement in a function.
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def add(a, b):
    return a + b
print(add(3, 2))  # 5
  1. Type Conversion: Convert an integer to a float.
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num = 10
print(float(num))  # 10.0
  1. Working with numbers (Exponentiation, Absolute value): Calculate the absolute value and exponentiation of a number.
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print(abs(-10))  # 10
print(2 ** 3)  # 8