medium Sqrt

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 = 8
Output: 2
Explanation: The square root of 8 is 2.8284, and since we need to return the floor of the square root (integer), hence we returned 2.

Example 2:

Input: x = 4
Output: 2
Explanation: The square root of 4 is 2.

Example 3:

Input: x = 2
Output: 1
Explanation: The square root of 2 is 1.414, and since we need to return the floor of the square root (integer), hence we returned 1.

Constraints:

0 <= x <= 2³¹ - 1

Try it yourself

Try solving this question here:

For a detailed solution, see the next chapter.

✅ Solution Sqrt

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 = 8
Output: 2
Explanation: The square root of 8 is 2.8284, and since we need to return the floor of the square root (integer), hence we returned 2.

Example 2:

Input: x = 4
Output: 2
Explanation: The square root of 4 is 2.

Example 3:

Input: x = 2
Output: 1
Explanation: The square root of 2 is 1.414, and since we need to return the floor of the square root (integer), hence we returned 1.

Constraints:

0 <= x <= 2³¹ - 1

Solution

We can use a Binary Search approach to calculate the square root of an integer x without using any in-built sqrt function.

For any integer x, the square root of x will lie between 0 and x/2 (inclusive) for x > 2. Therefore, we can efficiently use binary search within this range to find the integer part of the square root (i.e., the floor value). By narrowing down the potential values step-by-step, binary search allows us to determine the largest integer y such that y*y is less than or equal to x.

This approach ensures an efficient and accurate computation of the square root, especially for large values of x, due to the logarithmic nature of binary search.

Step-by-Step Algorithm

Handle Base Cases: If x is less than 2, return x directly (since the square root of 0 is 0, and the square root of 1 is 1).
Initialize Pointers: Set left to 2 and right to x / 2.
Binary Search Loop:
- While left is less than or equal to right:
  - Calculate mid as left + (right - left) // 2.
  - Calculate num as mid * mid.
  - If num is greater than x, move right to mid - 1.
  - If num is less than x, move left to mid + 1.
  - If num equals x, return mid.
Return Result: Return right as the integer part of the square root of x.

Algorithm Walkthrough

Let's consider the input n = 8:

Initialize:
- Input x = 8.
- Since x is not less than 2, proceed to the next step.
- Set left = 2 and right = 4 (8 // 2).
First Iteration:
- Calculate mid = 2 + (4 - 2) // 2 = 3.
- Calculate num = 3 * 3 = 9.
- Since num (9) is greater than x (8), move right to mid - 1 = 2.
Second Iteration:
- Now left = 2 and right = 2.
- Calculate mid = 2 + (2 - 2) // 2 = 2.
- Calculate num = 2 * 2 = 4.
- Since num (4) is less than x (8), move left to mid + 1 = 3.
End Loop:
- Now left = 3 and right = 2.
- Since left (3) is greater than right (2), exit the loop.
Return Result:
- Return right = 2 as the integer part of the square root of x = 8.

Code

Here is the code for this algorithm:

java

class Solution {

  public int mySqrt(int x) {
    if (x < 2) return x; // return x if it is 0 or 1

    int left = 2,
      right = x / 2;
    int mid;
    long num;
    while (left <= right) {
      // binary search for the square root
      mid = left + (right - left) / 2;
      num = (long) mid * mid;
      if (num > x) right = mid - 1;
      // if mid * mid is greater than x, set right to mid - 1
      else if (num < x) left = mid + 1;
      // if mid * mid is less than x, set left to mid + 1
      else return mid; // if mid * mid is equal to x, return mid
    }

    return right;
  }

  public static void main(String[] args) {
    Solution solution = new Solution();

    int input1 = 4;
    int expectedOutput1 = 2;
    int result1 = solution.mySqrt(input1);
    System.out.println(result1 == expectedOutput1); // Expected output: true

    int input2 = 8;
    int expectedOutput2 = 2;
    int result2 = solution.mySqrt(input2);
    System.out.println(result2 == expectedOutput2); // Expected output: true

    int input4 = 2;
    int expectedOutput4 = 1;
    int result4 = solution.mySqrt(input4);
    System.out.println(result4 == expectedOutput4); // Expected output: true

    int input5 = 3;
    int expectedOutput5 = 1;
    int result5 = solution.mySqrt(input5);
    System.out.println(result5 == expectedOutput5); // Expected output: true

    int input6 = 15;
    int expectedOutput6 = 3;
    int result6 = solution.mySqrt(input6);
    System.out.println(result6 == expectedOutput6); // Expected output: true
  }
}

Complexity Analysis

Time Complexity

Binary Search Algorithm: The key part of this algorithm is the binary search, which repeatedly divides the search interval in half. The time complexity of binary search is , where is the size of the search space. In this case, the search space is initially from 2 to x/2.
Search Space: The maximum size of the search space is x/2 (when ). For smaller values of x, the function immediately returns x, as it's either 0 or 1.
Overall Time Complexity: Considering the binary search on a range up to x/2, the time complexity is , which simplifies to .

Space Complexity

Constant Extra Space: The algorithm uses a fixed number of integer variables (left, right, pivot, num), regardless of the input size.
No Recursive Calls or Dynamic Allocation: The implementation does not use recursion or allocate additional data structures that grow with the input size.
Overall Space Complexity: Given the constant amount of extra space, the space complexity is , meaning it's constant.

🤖 Don't fully get this? Learn it with Claude

Stuck on Sqrt? Open Claude, copy a block below, and it'll teach you this exact concept — visually and interactively.

🪜 Hint ladder (no spoilers)

Progressively stronger hints — you still solve it.

I'm working on the problem **Sqrt** (DSA). Give me a HINT LADDER: start with the tiniest nudge, then wait. Only reveal the next, stronger hint when I ask. Do NOT show the full solution unless I type 'show solution'. Keep me doing the thinking. If you're unsure or a claim isn't standard, say so and reason from first principles instead of guessing.

🎨 Explain the approach visually

See the technique, not just code.

Explain the optimal approach to **Sqrt** with a VISUAL walkthrough: trace it on a small concrete example using ASCII art / a step-by-step diagram, narrate what changes each step, then give time & space complexity with a one-line derivation. If you're unsure or a claim isn't standard, say so and reason from first principles instead of guessing.

🔍 Review my solution

Catch bugs, edge cases, sub-optimality.

I'll paste my solution to **Sqrt**. Review it for correctness, missed edge cases, and time/space complexity, then coach me toward the optimal — don't just rewrite it. Ask me to paste my code now. If you're unsure or a claim isn't standard, say so and reason from first principles instead of guessing.

🔁 Drill the pattern

Lock in recognition with look-alikes.

Give me 2 problems that use the SAME underlying pattern as **Sqrt**. For each, let me attempt first, then review my answer and name the trigger signal that reveals the pattern. If you're unsure or a claim isn't standard, say so and reason from first principles instead of guessing.

📝 My notes

← Search a 2D Matrix II