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Valid Perfect Square

Certification: Basic Level Accuracy: 57.14% Submissions: 7 Points: 5

Write a Java program to determine if a given positive integer is a perfect square without using the Math.sqrt() function or any other built-in square root function. A perfect square is an integer that is the square of an integer. For example, 16 is a perfect square because it is 4^2.

Example 1
    

  • Input: num = 16
    • 

  • Output: true
    • 

  • Explanation:

      

    • Step 1: 16 is a perfect square because it is equal to 4^2.
      • 

    • Step 2: 4 * 4 = 16
      • 

    • Step 3: Therefore, 16 is a perfect square.
      • 

    •
Example 2
    

  • Input: num = 14
    • 

  • Output: false
    • 

  • Explanation:

      

    • Step 1: 14 is not a perfect square.
      • 

    • Step 2: The closest perfect squares are 9 (3^2) and 16 (4^2).
      • 

    • Step 3: Since 14 falls between these two perfect squares, it is not a perfect square itself.
      • 

    •
Constraints
    

  • 1 ≤ num ≤ 2^31 - 1
    • 

  • You are not allowed to use any built-in library function such as Math.sqrt()
    • 

  • Time Complexity: O(log n)
    • 

  • Space Complexity: O(1)
    •
NumberWalmartD. E. Shaw
Editorial

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Hint
Solution

Solution Hints

    
  
  • Use binary search to find the square root of the number
    • 
  
  • Start with a search range from 1 to num/2
    • 
  
  • For each mid value, check if mid * mid equals the given number
    • 
  
  • If mid * mid is greater than the number, search in the left half
    • 
  
  • If mid * mid is less than the number, search in the right half
    •

Steps to solve by this approach:

 Step 1: Handle the edge case where num < 2, as 0 and 1 are perfect squares.
 Step 2: Initialize the search range with left = 1 and right = num/2 (since the square root can't be greater than num/2 for num > 1).
 Step 3: Use binary search to find a potential square root.
 Step 4: For each mid value, calculate mid * mid and compare it with num.
 Step 5: If mid * mid equals num, return true as we found the square root.
 Step 6: If mid * mid > num, search in the left half; otherwise, search in the right half.
 Step 7: If we exit the loop without finding a square root, return false.

Submitted Code :