I just wanted to suggest a more subtle way of explaining it that may be a bit better. So, binary search is applied here because we are searching for a value in a specific range of values, which is a range of sorted numbers. That value we are looking for is the largest value n such that n * n <= x. So, we can use a variable to store the result, and if we ever find a value whose square is <= x, update the result to that. However, since our goal is to find the largest, toss that pivot out and set left = mid + 1. If the square is too large, then we want to lower our right bound and set right = mid - 1.

class Solution {
  public int mySqrt(int x) {
    int left = 0;
    int right = x;
    int res = 0;

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

      if((long) mid * mid <= x){
        res = mid;
        left = mid + 1;
      }
      else{
        right = mid - 1;
      }
    }
    return res;
  }
}