Documentation for the Integer Square Root Calculation Code

Problem Statement

The objective is to find the integer part of the square root of a non-negative integer x, denoted as sqrt(x). The result should be the largest integer n such that n * n <= x. We are not allowed to use any built-in square root functions.

Thought Process / Intuition / Analogy

To solve this problem, we can use a straightforward approach based on the definition of square roots:

1. Understanding Square Roots:

   • The square root of a number x is a value n such that when n is multiplied by itself (i.e., n*n), it yields x or the closest integer lower than x.
   • For example, the square root of 16 is 4, because (4 \times 4 = 16). For 20, the square root is also 4, since (4 \times 4 = 16) is the largest perfect square less than 20.

2. Iterative Search:

   • We can use a loop to iterate through integers starting from 1, checking whether squaring the integer gives a result that is less than or equal to x. This method ensures we find the largest integer whose square does not exceed x.

3. Early Return for Edge Cases:

   • If x is less than 2, we can immediately return x because the square roots of 0 and 1 are themselves.

Designing the Optimal Solution

The design of the solution focuses on simplicity and efficiency:

• Handling Special Cases: The function checks if x is less than 2, allowing for a quick return for these specific cases.
• Looping from 1: The loop starts from 1 and continues until i * i exceeds x. This ensures that we find the largest integer i whose square is less than or equal to x.

Explanation for Using long long

The use of long long is crucial to prevent overflow during calculations. The int type in C++ has a maximum value of (2,147,483,647); hence, squaring larger integers could exceed this limit and cause overflow.

For example, if i becomes 46,341, then (i * i) would exceed (2,147,483,647) (the max value for an int), resulting in incorrect calculations.

By using long long, we can safely perform calculations for larger values and ensure that the results remain accurate, allowing us to properly determine the integer square root for values of x up to (2,147,483,647).

The choice of long long is indeed influenced by the constraints provided, specifically the range of values for x.

Explanation of Constraints

The constraints state that:

[ 0 <= x <= 2^{31} - 1 ]

This means that x can take on values from 0 up to (2,147,483,647) (which is (2^{31} - 1)).

Implications for Choosing Data Types

1. Maximum Value of x:

   • Since x can be as large as (2,147,483,647), when calculating the integer square root, we need to consider the maximum possible value of i that we might square.
   • The integer square root of (2,147,483,647) is approximately (46,340) (since (46,340^2 = 2,149,303,600) which exceeds (2,147,483,647)).

2. Calculating (i * i):

   • If we were to use int for i, squaring it when i approaches (46,341) would lead to overflow:
     • (46,341 * 46,341 = 2,148,484,481) (which is greater than the maximum value for int).
   • This overflow would result in incorrect calculations, potentially yielding negative or unexpected values.

3. Using long long:

   • By using long long, we can safely handle the calculations without risk of overflow, as long long can represent values up to (9,223,372,036,854,775,807), which is far beyond the range we need for this problem.
   • This ensures that even at the upper limits of x, the calculations remain accurate.

The constraints directly inform the choice of data types in the implementation. Given that x can be as large as (2^{31} - 1), using long long for the variable i in the loop is a prudent decision to avoid overflow and ensure the correctness of the integer square root calculation. This consideration is essential in robust software development, especially when dealing with numerical computations.

Step-by-Step Dry Run

Let’s dry run the code for x = 20 to illustrate how it processes this input.

Code Implementation:

class Solution {  
public:  
    int mySqrt(int x) {  
        if (x < 2) return x; // Handle 0 and 1 directly  

        long long ans = 0;  

        for (long long i = 1; i * i <= x; i++) {  
            ans = i; // Update ans to the current i  
        }  

        return ans; // Return the last valid integer  
    }  
};

Dry Run Example: x = 20

• Initialization:

  • Start with x = 20
  • ans = 0

• Loop Execution:

  1. Iteration 1 (i = 1):

     • Check if (1 * 1 <= 20) (True)
     • Update ans to 1.

  2. Iteration 2 (i = 2):

     • Check if (2 * 2 <= 20) (True)
     • Update ans to 2.

  3. Iteration 3 (i = 3):

     • Check if (3 * 3 <= 20) (True)
     • Update ans to 3.

  4. Iteration 4 (i = 4):

     • Check if (4 * 4 <= 20) (True)
     • Update ans to 4.

  5. Iteration 5 (i = 5):

     • Check if (5 * 5 <= 20) (False)
     • Exit the loop.

• Final Output:

  • The loop ends, and ans is 4, which is returned as the integer square root of 20 (since (5 * 5 = 25) is greater than 20).

Conclusion

This documentation provides a detailed overview of how to compute the integer part of the square root of x using an iterative approach. The thought process focuses on understanding the definition of square roots and establishing conditions for breaking out of the loop early when the square exceeds x. The dry run clarifies how the code executes step by step, demonstrating its correctness.

By using long long, we prevent overflow issues, ensuring accurate calculations for larger inputs. This solution is efficient for a wide range of inputs and serves as a good example of how to implement basic algorithms in C++. If you have further questions or need clarifications, feel free to ask! 😊