LeetCode #69 — EASY

Sqrt(x)

Build confidence with an intuition-first walkthrough focused on math fundamentals.

The Problem

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 integer, 2 is returned.

Constraints:

0 <= x <= 2³¹ - 1

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Binary Search

Example 1

Example 2

Core Insight

What unlocks the optimal approach

Try exploring all integers. (Credits: @annujoshi)
Use the sorted property of integers to reduced the search space. (Credits: @annujoshi)

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

class Solution {
    public int mySqrt(int x) {
        if (x < 2) return x;

        int left = 1, right = x / 2;
        int ans = 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if ((long) mid * mid <= x) {
                ans = mid;
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return ans;
    }
}

func mySqrt(x int) int {
    if x < 2 {
        return x
    }

    left, right := 1, x/2
    ans := 1
    for left <= right {
        mid := left + (right-left)/2
        if mid <= x/mid {
            ans = mid
            left = mid + 1
        } else {
            right = mid - 1
        }
    }
    return ans
}

class Solution:
    def mySqrt(self, x: int) -> int:
        if x < 2:
            return x

        left, right = 1, x // 2
        ans = 1
        while left <= right:
            mid = (left + right) // 2
            if mid * mid <= x:
                ans = mid
                left = mid + 1
            else:
                right = mid - 1
        return ans

impl Solution {
    pub fn my_sqrt(x: i32) -> i32 {
        if x < 2 {
            return x;
        }

        let mut left = 1;
        let mut right = x / 2;
        let mut ans = 1;

        while left <= right {
            let mid = left + (right - left) / 2;
            if mid <= x / mid {
                ans = mid;
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        ans
    }
}

function mySqrt(x: number): number {
  if (x < 2) return x;

  let left = 1;
  let right = Math.floor(x / 2);
  let ans = 1;

  while (left <= right) {
    const mid = left + Math.floor((right - left) / 2);
    if (mid * mid <= x) {
      ans = mid;
      left = mid + 1;
    } else {
      right = mid - 1;
    }
  }
  return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log x)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.