Check if a given number is a Perfect square using Binary Search

Check if a given number N is a perfect square or not. If yes then return the number of which it is a perfect square, Else print -1.

Examples:

Input: N = 4900
Output 70
Explanation:
4900 is a perfect square number of 70 because 70 * 70 = 4900



Input: N = 81
Output: 9
Explanation:
81 is a perfect square number of 9 because 9 * 9 = 81

Approach: To solve the problem mentioned above we will use the Binary Search Algorithm.

  • Find the mid element from the start and last value and compare the value of the square of mid(mid*mid) with N.
  • If it is equal then return the mid otherwise check if the square(mid*mid) is greater than N then recursive call with the same start value but changed last to mid-1 value and if the square(mid*mid) is less than the N then recursive call with the same last value but changed start value.
  • If the N is not a square root then return -1.

Below is the implementation of above approach:

C++

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// C++ program to check if a
// given number is Perfect
// square using Binary Search
  
#include <iostream>
using namespace std;
  
// function to check for
// perfect square number
int checkPerfectSquare(
    long int N,
    long int start,
    long int last)
{
    // Find the mid value
    // from start and last
    long int mid = (start + last) / 2;
  
    if (start > last) {
        return -1;
    }
  
    // check if we got the number which
    // is square root of the perfect
    // square number N
    if (mid * mid == N) {
        return mid;
    }
  
    // if the square(mid) is greater than N
    // it means only lower values then mid
    // will be possibly the square root of N
    else if (mid * mid > N) {
        return checkPerfectSquare(
            N, start, mid - 1);
    }
  
    // if the square(mid) is less than N
    // it means only higher values then mid
    // will be possibly the square root of N
    else {
        return checkPerfectSquare(
            N, mid + 1, last);
    }
}
  
// Driver code
int main()
{
    long int N = 65;
  
    cout << checkPerfectSquare(N, 1, N);
    return 0;
}

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Java

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// Java program to check if a
// given number is Perfect
// square using Binary Search
import java.util.*;
  
class GFG {
  
// Function to check for
// perfect square number
static int checkPerfectSquare(long N, 
                              long start,
                              long last)
{
    // Find the mid value
    // from start and last
    long mid = (start + last) / 2;
  
    if (start > last)
    {
        return -1;
    }
  
    // Check if we got the number which
    // is square root of the perfect
    // square number N
    if (mid * mid == N)
    {
        return (int)mid;
    }
  
    // If the square(mid) is greater than N
    // it means only lower values then mid
    // will be possibly the square root of N
    else if (mid * mid > N)
    {
        return checkPerfectSquare(N, start, 
                                  mid - 1);
    }
  
    // If the square(mid) is less than N
    // it means only higher values then mid
    // will be possibly the square root of N
    else 
    {
        return checkPerfectSquare(N, mid + 1
                                  last);
    }
}
  
// Driver code
public static void main(String[] args)
{
    long N = 65;
    System.out.println(checkPerfectSquare(N, 1, N));
}
}
  
// This code is contributed by offbeat

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Python3

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# Python3 program to check if a 
# given number is perfect 
# square using Binary Search 
  
# Function to check for 
# perfect square number 
def checkPerfectSquare(N, start, last): 
  
    # Find the mid value 
    # from start and last 
    mid = int((start + last) / 2
  
    if (start > last): 
        return -1
  
    # Check if we got the number which 
    # is square root of the perfect 
    # square number N 
    if (mid * mid == N):
        return mid
  
    # If the square(mid) is greater than N 
    # it means only lower values then mid 
    # will be possibly the square root of N 
    elif (mid * mid > N):
        return checkPerfectSquare(N, start, 
                                  mid - 1
  
    # If the square(mid) is less than N 
    # it means only higher values then mid 
    # will be possibly the square root of N 
    else:
        return checkPerfectSquare(N, mid + 1
                                  last)
  
# Driver code 
N = 65
print (checkPerfectSquare(N, 1, N)) 
  
# This code is contributed by PratikBasu

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Output:

-1

Time Complexity: O(Logn)

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Improved By : PratikBasu, offbeat