Sqrt(x)
Sqrt(x) is a easy binary search problem solved with the search on answer range pattern.
The best approach, optimal (binary search for the last true), runs in O(log x) time and O(1) space.
Below are 2 approaches in Java, from linear search up.
Problem
Given a non-negative integer x, return its square root rounded down to an integer, without using a built-in power or square root function.
Examples
Example 1
- Input
x = 16- Output
4
Example 2
- Input
x = 20- Output
4- Why
- sqrt(20) ≈ 4.47, rounded down.
Constraints
0 <= x <= 2^31 - 1- No built-in pow or sqrt.
Animated walkthrough
A narrated, step-by-step animation that builds the solution from the idea up. Press play, or step through it at your own pace.
Concepts first, then the problem and every approach, step by step.
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Solutions
Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.
| Approach | Time | Space |
|---|---|---|
| Linear search | O(√x) | O(1) |
| Optimal (binary search for the last true) | O(log x) | O(1) |
1Linear search
O(√x)O(1)Increase r while (r + 1)^2 <= x.
- r = 0; while (r + 1)^2 <= x: r++.
class Solution {
public int mySqrt(int x) {
long r = 0;
while ((r + 1) * (r + 1) <= x) r++;
return (int) r;
}
}2Optimal (binary search for the last true)
O(log x)O(1)r · r <= x is true for small r and false for large r. Binary search for the last r where it is true, using long for the square.
- lo = 0, hi = x.
- If mid * mid <= x: ans = mid, lo = mid + 1. Else hi = mid - 1.
class Solution {
public int mySqrt(int x) {
long lo = 0, hi = x, ans = 0;
while (lo <= hi) {
long mid = (lo + hi) / 2;
if (mid * mid <= x) { ans = mid; lo = mid + 1; }
else hi = mid - 1;
}
return (int) ans;
}
}Edge cases to test
- x = 0 and x = 1
- mid * mid overflows int; use long
Hints
Hint 1
Find the largest r with r · r <= x.
FAQ
What is the best time complexity for Sqrt(x)?
Optimal (binary search for the last true) runs in O(log x) time and O(1) extra space.
Which pattern does Sqrt(x) use?
It is a binary search problem that uses the search on answer range pattern. Other problems with the same pattern: Pow(x, n) (Binary Exponentiation), Find Nth root of M, Find position of an element in a sorted array of infinite numbers (Galloping Search).
Is there a brute force solution for Sqrt(x)?
Yes. Linear search takes O(√x) time and O(1) space. Increase r while (r + 1)^2 <= x.
Which edge cases should I test for Sqrt(x)?
x = 0 and x = 1; mid mid overflows int; use long.