What is the time complexity of Sqrt(x)?

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

What data structures are used to solve Sqrt(x)?

This problem uses Math, Binary Search. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Sqrt(x)?

Reframing the square root problem as a search for the largest integer `k` such that `k^2 0`). Employing `mid <= x // mid` instead of `mid * mid <= x` is a crucial defensive programming choice. It elegantly prevents potential integer overflow issues when `mid` is large, especially in languages with fixed-size integers, ensuring the solution remains robust for very large input values of `x` without sacrificing correctness.

Which companies ask LeetCode 69. Sqrt(x) in interviews?

This problem is frequently asked at Accenture, Adobe, Amazon, Apple, Bloomberg and 12 more companies.

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Sqrt(x) - LeetCode 69 Solution

Q: How to solve Sqrt(x)?

The provided solution cleverly employs binary search to navigate the problem's constraints and achieve an efficient result. Instead of trying to calculate the square root directly, we reframe the problem as 'find the largest integer `k` such that `k * k 0`, we set our search range from `left = 1` to `right = x`.

Q: Which companies ask LeetCode 69. Sqrt(x) in interviews?

This problem is frequently asked at Accenture, Adobe, Amazon, Apple, Bloomberg and 12 more companies.

Sqrt(x) - Complete Solution Guide

Sqrt(x) is LeetCode problem 69, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a non-negative integer x , return the square root of x rounded down to the nearest integer . The returned integer should be non-negative as well. You must not use any built-in exponent function or operator. For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python. Example 1: Input: x = 4 Output: 2 Explanation: The square root of 4 is 2, so we return 2. Example 2: Input: x = 8 Output: 2 Explanation: The square root of 8 is 2.82842..., and since we round it down to the nearest intege

Detailed Explanation

The core task here is to calculate the integer square root of a given non-negative integer `x`, always rounding down to the nearest whole number. For instance, if `x` is 4, we expect 2. If `x` is 8, where the true square root is approximately 2.828, we still return 2 because the problem explicitly states we must round down. The returned value also needs to be non-negative.

Solution Approach

The provided solution cleverly employs binary search to navigate the problem's constraints and achieve an efficient result. Instead of trying to calculate the square root directly, we reframe the problem as 'find the largest integer `k` such that `k * k <= x`.' Our search space for `k` is naturally bounded: the smallest possible non-negative square root is 0, and the largest it could possibly be is `x` itself (e.g., `sqrt(1) = 1`, `sqrt(0) = 0`). So, for `x > 0`, we set our search range from `left = 1` to `right = x`.

Step-by-Step Algorithm

Step 1: Initialize `left` to 1 and `right` to `x` (or `x/2` for optimization). Initialize `ans` to 0.
Step 2: While `left` is less than or equal to `right`, calculate the middle element `mid`.
Step 3: Check if `mid * mid <= x` (or equivalently, `mid <= x / mid` to prevent overflow). If true, update `ans` to `mid` and move `left` to `mid + 1`. Otherwise, move `right` to `mid - 1`.
Step 4: Repeat Step 2 and 3 until the loop terminates. The final value of `ans` is the integer square root.

Key Insights

Reframing the square root problem as a search for the largest integer `k` such that `k^2 <= x` immediately highlights binary search as an optimal strategy, transforming a direct computation into an efficient search over a naturally ordered range (`[1, x]` for `x > 0`).
Employing `mid <= x // mid` instead of `mid * mid <= x` is a crucial defensive programming choice. It elegantly prevents potential integer overflow issues when `mid` is large, especially in languages with fixed-size integers, ensuring the solution remains robust for very large input values of `x` without sacrificing correctness.
The binary search's update logic, where `ans` is stored when `mid^2 <= x` and then `left` is pushed higher (`left = mid + 1`), intrinsically satisfies the 'round down' requirement. This ensures that `ans` consistently tracks the largest integer whose square does not exceed `x`, converging directly to the problem's desired output.

Complexity Analysis

Time Complexity: O(log n)

Space Complexity: O(1)

Topics

This problem involves: Math, Binary Search.

Companies

Asked at: Accenture, Adobe, Amazon, Apple, Bloomberg, Citadel, Goldman Sachs, Grammarly, Infosys, Meta, Microsoft, SAP, Samsung, TikTok, Uber, Yahoo, tcs.

Sqrt(x) | LeetCode 69

Easy

Math Binary Search

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

Solution Code

Available in:

class Solution {
    public int mySqrt(int x) {
        if (x == 0) {
            return 0;
        }

        int left = 1;
        int right = x;
        int ans = 0;

        while (left <= right) {
            int mid = left + (right - left) / 2;

            if (mid <= x / mid) {
                ans = mid;
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }

        return ans;
    }
}

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Problem Info

Description

Problem 69: Sqrt(x) - Detailed explanation and solution approaches.

Topics

Math Binary Search

Companies

AccentureAdobeAmazonAppleBloombergCitadelGoldman SachsGrammarly

Stats

Difficulty:Easy

Languages:4

Companies:17

Problem #:69

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

4 steps

1
Step 1: Initialize `left` to 1 and `right` to `x` (or `x/2` for optimization). Initialize `ans` to 0.
2
Step 2: While `left` is less than or equal to `right`, calculate the middle element `mid`.
3
Step 3: Check if `mid * mid <= x` (or equivalently, `mid <= x / mid` to prevent overflow). If true, update `ans` to `mid` and move `left` to `mid + 1`. Otherwise, move `right` to `mid - 1`.
4
Step 4: Repeat Step 2 and 3 until the loop terminates. The final value of `ans` is the integer square root.

💡

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 binary search algorithm halves the search space in each iteration. Therefore, the number of iterations is logarithmic with respect to the input `x`. The algorithm uses a constant amount of extra space regardless of the input size.

📖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

Newton's Method

An iterative approach that refines an initial guess using the derivative of the function f(x) = x^2 - n (where n is the input).

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

Trade-offs: Newton's method can converge faster than binary search in some cases, but it involves more complex calculations and might require handling potential floating-point errors.

Linear Scan (Inefficient)

Iterate from 1 up to x, checking if `i*i <= x` in each iteration. Stop when `i*i > x` and return `i-1`.

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

Trade-offs: This approach is much less efficient than binary search and should be avoided for large input values.

✨

Key Insights

1Reframing the square root problem as a search for the largest integer `k` such that `k^2 <= x` immediately highlights binary search as an optimal strategy, transforming a direct computation into an efficient search over a naturally ordered range (`[1, x]` for `x > 0`).
2Employing `mid <= x // mid` instead of `mid * mid <= x` is a crucial defensive programming choice. It elegantly prevents potential integer overflow issues when `mid` is large, especially in languages with fixed-size integers, ensuring the solution remains robust for very large input values of `x` without sacrificing correctness.
3The binary search's update logic, where `ans` is stored when `mid^2 <= x` and then `left` is pushed higher (`left = mid + 1`), intrinsically satisfies the 'round down' requirement. This ensures that `ans` consistently tracks the largest integer whose square does not exceed `x`, converging directly to the problem's desired output.

📚

Key Terms Explained

For Students

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: x = 0. The solution correctly handles this by returning 0.
Edge case 2: x = 1. The solution handles this by returning 1.
Edge case 3: Large values of x. The use of `mid <= x / mid` prevents potential integer overflow issues that would occur if `mid * mid` were used directly for very large input values.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Integer overflow: Not handling the multiplication `mid * mid` carefully can lead to integer overflow for large input values of `x`. Using `mid <= x / mid` helps mitigate this.

✗Common mistake 2: Incorrect binary search implementation: Off-by-one errors in the binary search logic can lead to incorrect results. Carefully checking the boundary conditions is essential.

✅

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: Game development: Calculating the distance between two points on a grid often requires the square root operation. An integer approximation using this algorithm is suitable when pixel-perfect accuracy isn't critical.

🏦Real-world scenario 2: Image processing: Many image processing algorithms (e.g., those involving distances or magnitudes) can use this algorithm for fast integer-based approximations of square roots.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

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

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: August 9, 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

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadMay 29

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

@DevExpert92May 22

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

@AlgoMasterMay 15

❓

Frequently Asked Questions

Can binary search be applied to non-sorted arrays?

Binary search requires a monotonic property, not necessarily sorted data. It can be applied to any search space where you can determine if the answer is in the left or right half. Examples include searching rotated arrays or finding minimum in optimization problems.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Sqrt(x)"?

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

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