EASY Binary Search
69. Sqrt(x)
📖 Problem
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.
🧠 Visual Learning Aid
1 Model the input into the right structure
2 Choose the core technique and invariant
3 Execute step-by-step with a sample
4 Validate complexity and edge cases
JS/TS Refreshers
- •Array methods (`push`, `pop`, `shift`, `slice`)
- •Object/Map/Set usage patterns
- •Function parameter and return typing
- •Midpoint overflow-safe math
- •Loop invariants
- •Monotonic condition design
Logical Thinking Concepts
- •Define invariants before coding
- •Check edge cases first (`[]`, single element, duplicates)
- •Estimate time/space before implementation
- •Apply Binary Search reasoning pattern
💡 Approach
- → Use binary search between 1 and x
- → Square mid and compare with x
- → If mid*mid > x, search left half
- → If mid*mid < x, search right half
- → Time: O(log x), Space: O(1)
🛠️ Hints & Pitfalls
Hints
- •Use binary search between 1 and x
- •Square mid and compare with x
- •If mid*mid > x, search left half
Common Pitfalls
- •If mid*mid < x, search right half
- •Time: O(log x), Space: O(1)
🧪 Test Cases
Hidden tests on submit: 4
Test Case 1
Not run Input:
mySqrt(4); Expected:
2 Test Case 2
Not run Input:
mySqrt(8); Expected:
2 Test Case 3
Not run Input:
mySqrt(0); Expected:
0 📝 Code Editor
📤 Output