What is the time complexity of Pow(x, n)?

The optimal solution has a time complexity of O(log n).

What data structures are used to solve Pow(x, n)?

This problem uses Math, Recursion. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Pow(x, n)?

Insight 1: Directly multiplying x by itself n times would result in O(n) time complexity. The key is to use the property that x^n = (x^(n/2))^2 if n is even, and x^n = x * (x^(n/2))^2 if n is odd. This allows us to reduce the number of multiplications significantly. Insight 2: We can use bit manipulation to efficiently check if n is odd (n & 1) and divide n by 2 (n >> 1). This avoids using modulo and division operators, which can be slower.

Which companies ask LeetCode 50. Pow(x, n) in interviews?

This problem is frequently asked at Accenture, Adobe, Amazon, Apple, Arcesium and 21 more companies.

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Pow(x, n) - LeetCode 50 Solution

Q: How to solve Pow(x, n)?

The solution uses a binary exponentiation approach to calculate x^n efficiently. It iteratively squares the base x and halves the exponent n. If the current exponent n is odd, the current result is multiplied by the current value of x. This process continues until the exponent becomes 0.

Pow(x, n) - Complete Solution Guide

Pow(x, n) is LeetCode problem 50, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Implement pow(x, n) , which calculates x raised to the power n (i.e., x n ). 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 = 1/2 2 = 1/4 = 0.25 Constraints: -100.0 < x < 100.0 -2 31 <= n <= 2 31 -1 n is an integer. Either x is not zero or n > 0 . -10 4 <= x n <= 10 4

Detailed Explanation

The problem asks us to implement the `pow(x, n)` function, which calculates `x` raised to the power `n` (x^n). We are given a base `x` (a floating-point number) and an exponent `n` (an integer). The result should be a floating-point number. Constraints include bounds on x, n, and x^n, as well as handling the case where x is 0.

Solution Approach

The solution uses a binary exponentiation approach to calculate x^n efficiently. It iteratively squares the base x and halves the exponent n. If the current exponent n is odd, the current result is multiplied by the current value of x. This process continues until the exponent becomes 0.

Step-by-Step Algorithm

Step 1: Handle the case where n is negative. If n < 0, replace x with 1/x and n with -n.
Step 2: Initialize the result variable `res` to 1.0.
Step 3: Iterate while n > 0. In each iteration:
Step 4: Check if n is odd using the bitwise AND operator (n & 1). If it is, multiply the `res` by the current value of x.
Step 5: Square the value of x (x = x * x).
Step 6: Divide n by 2 using the bitwise right shift operator (n >>= 1).
Step 7: After the loop finishes, return the value of `res`.

Key Insights

Insight 1: Directly multiplying x by itself n times would result in O(n) time complexity. The key is to use the property that x^n = (x^(n/2))^2 if n is even, and x^n = x * (x^(n/2))^2 if n is odd. This allows us to reduce the number of multiplications significantly.
Insight 2: We can use bit manipulation to efficiently check if n is odd (n & 1) and divide n by 2 (n >> 1). This avoids using modulo and division operators, which can be slower.
Insight 3: Negative exponents can be handled by taking the reciprocal of the base (x) and negating the exponent (n). It's also important to consider integer overflow for the negative value of the exponent if the exponent is the minimum integer value (-2^31).
Insight 4: The iterative approach avoids recursion, leading to better performance and no call stack overhead.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

This problem involves: Math, Recursion.

Companies

Asked at: Accenture, Adobe, Amazon, Apple, Arcesium, Bloomberg, ByteDance, Citadel, EPAM Systems, Goldman Sachs, Infosys, LinkedIn, Meta, Microsoft, Millennium, Oracle, Qualcomm, Salesforce, ServiceNow, TikTok, Uber, Walmart Labs, Wix, Yahoo, eBay, tcs.

Pow(x, n) | LeetCode 50

Medium

Math Recursion

Time: O(log n)Space: O(1)

Solution Code

Available in:

class Solution {
    public double myPow(double x, int n) {
        if (n < 0) {
            x = 1 / x;
            n = -n;
        }
        
        double res = 1.0;
        while (n > 0) {
            if ((n & 1) == 1) {
                res *= x;
            }
            x *= x;
            n >>= 1;
        }
        
        return res;
    }
}

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Problem Info

Description

Problem 50: Pow(x, n) - Detailed explanation and solution approaches.

Topics

Math Recursion

Companies

AccentureAdobeAmazonAppleArcesiumBloombergByteDanceCitadel

Stats

Difficulty:Medium

Languages:4

Companies:26

Problem #:50

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Handle the case where n is negative. If n < 0, replace x with 1/x and n with -n.
2
Step 2: Initialize the result variable `res` to 1.0.
3
Step 3: Iterate while n > 0. In each iteration:
4
Step 4: Check if n is odd using the bitwise AND operator (n & 1). If it is, multiply the `res` by the current value of x.
5
Step 5: Square the value of x (x = x * x).
6
Step 6: Divide n by 2 using the bitwise right shift operator (n >>= 1).
7
Step 7: After the loop finishes, return the value of `res`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log n)

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is O(log n) because the exponent n is being divided by 2 in each iteration of the while loop until it reaches 0. The space complexity is O(1) because we are only using a constant amount of extra space, irrespective of the input.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach

A recursive implementation of binary exponentiation. The base case is when n is 0, returning 1. Otherwise, recursively calculate pow(x, n/2) and square the result. If n is odd, multiply by x.

Time: O(log n)Space: O(log n) due to the call stack in recursion.

Trade-offs: The recursive approach can be more readable but might have a higher space complexity due to the call stack. The iterative approach is generally preferred for its better space efficiency.

✨

Key Insights

1Insight 1: Directly multiplying x by itself n times would result in O(n) time complexity. The key is to use the property that x^n = (x^(n/2))^2 if n is even, and x^n = x * (x^(n/2))^2 if n is odd. This allows us to reduce the number of multiplications significantly.
2Insight 2: We can use bit manipulation to efficiently check if n is odd (n & 1) and divide n by 2 (n >> 1). This avoids using modulo and division operators, which can be slower.
3Insight 3: Negative exponents can be handled by taking the reciprocal of the base (x) and negating the exponent (n). It's also important to consider integer overflow for the negative value of the exponent if the exponent is the minimum integer value (-2^31).
4Insight 4: The iterative approach avoids recursion, leading to better performance and no call stack overhead.

📚

Key Terms Explained

For Students

Recursion

A function that calls itself to solve smaller versions of the same problem.

Edge Cases

Edge case 1: n = 0. The solution correctly returns 1.0, as any number raised to the power of 0 is 1.
Edge case 2: x = 0 and n < 0. This is explicitly disallowed by the problem constraints (Either `x` is not zero or `n > 0`).
Edge case 3: n is a large positive integer. Squaring x repeatedly might lead to overflow in some cases depending on the value of 'x'. However, the constraints state -104 <= xn <= 104, so overflow issues are minimized within the permitted range.
Edge case 4: n is Integer.MIN_VALUE (-2147483648). Taking the negation of n could result in integer overflow in some languages because the absolute value of Integer.MIN_VALUE exceeds Integer.MAX_VALUE. This issue is typically circumvented using long data types internally or modifying the while loop condition to include logic for n being negative one.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using a naive approach of multiplying x by itself n times. This results in O(n) time complexity, which is inefficient for large values of n.

✗Common mistake 2: Not handling negative exponents correctly. Forgetting to take the reciprocal of x when n is negative will lead to incorrect results.

✗Common mistake 3: Integer overflow when negating Integer.MIN_VALUE. Care must be taken to prevent overflow issues. In some languages using long for intermediate calculations can help.

✗Common mistake 4: Assuming the constraints prevent large intermediate values that result from squaring 'x' when they actually do not. Reviewing constraints is imperative.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Cryptography. Exponentiation is a fundamental operation in many cryptographic algorithms, such as RSA. Efficient exponentiation algorithms are crucial for the performance of these algorithms.

🏦Real-world scenario 2: Scientific computing. Many scientific and engineering calculations involve raising numbers to large powers. For example, calculating compound interest or modeling exponential growth/decay.

🎮Real-world scenario 3: Computer graphics. Calculations involving lighting, shading, and transformations often involve exponentiation.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 21, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Pow(x, n)"?

For this medium-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether recursion techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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