Check if given number is perfect square
Last Updated :
17 Sep, 2024
Given a number n, check if it is a perfect square or not.
Examples :
Input : n = 36
Output : Yes
Input : n = 2500
Output : Yes
Explanation: 2500 is a perfect square of 50
Input : n = 8
Output : No
Using sqrt()
- Take the floor()ed square root of the number.
- Multiply the square root twice.
- Use boolean equal operator to verify if the product of square root is equal to the number given.
Code Implementation:
C++
// CPP program to find if x is a
// perfect square.
#include <bits/stdc++.h>
using namespace std;
bool isPerfectSquare(long long x)
{
// Find floating point value of
// square root of x.
if (x >= 0) {
long long sr = sqrt(x);
// if product of square root
//is equal, then
// return T/F
return (sr * sr == x);
}
// else return false if n<0
return false;
}
int main()
{
long long x = 49;
if (isPerfectSquare(x))
cout << "Yes";
else
cout << "No";
return 0;
}
// C program to find if x is a
// perfect square.
#include <stdio.h>
#include <math.h>
int isPerfectSquare(long long x)
{
// Find floating point value of
// square root of x.
if (x >= 0) {
long long sr = sqrt(x);
// if product of square root
// is equal, then
// return T/F
return (sr * sr == x);
}
// else return false if n<0
return 0;
}
int main()
{
long long x = 49;
if (isPerfectSquare(x))
printf("Yes");
else
printf("No");
return 0;
}
// Java program to find if x is a perfect square.
public class GfG {
public static boolean isPerfectSquare(long x)
{
// Find floating point value of
// square root of x.
if (x >= 0) {
long sr = (long)Math.sqrt(x);
// if product of square root
// is equal, then
// return T/F
return (sr * sr == x);
}
// else return false if n<0
return false;
}
public static void main(String[] args)
{
long x = 49;
if (isPerfectSquare(x))
System.out.println("Yes");
else
System.out.println("No");
}
}
# Python program to find if x is a
# perfect square.
def is_perfect_square(x):
# Find floating point value of
# square root of x.
if x >= 0:
sr = int(x ** 0.5)
# if product of square root
# is equal, then
# return T/F
return (sr * sr == x)
# else return false if n<0
return False
x = 49
if is_perfect_square(x):
print("Yes")
else:
print("No")
// C# program to find if x is a
// perfect square.
using System;
class GfG {
static bool IsPerfectSquare(long x) {
// Find floating point value of
// square root of x.
if (x >= 0) {
long sr = (long) Math.Sqrt(x);
// if product of square root
// is equal, then
// return T/F
return (sr * sr == x);
}
// else return false if n<0
return false;
}
static void Main() {
long x = 49;
if (IsPerfectSquare(x))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
// JavaScript program to find if x is a
// perfect square.
function isPerfectSquare(x) {
// Find floating point value of
// square root of x.
if (x >= 0) {
let sr = Math.floor(Math.sqrt(x));
// if product of square root
// is equal, then
// return T/F
return (sr * sr === x);
}
// else return false if n<0
return false;
}
const x = 49;
if (isPerfectSquare(x))
console.log("Yes");
else
console.log("No");
Time Complexity: O(log(x))
Auxiliary Space: O(1)
- Use the floor and ceil and sqrt() function.
- If they are equal that implies the number is a perfect square.
Code Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
bool isPerfectSquare(long long n)
{
// If ceil and floor are equal
// the number is a perfect
// square
if (ceil((double)sqrt(n)) == floor((double)sqrt(n))){
return true;
}
else{
return false;
}
}
int main()
{
long long x = 49;
if (isPerfectSquare(x))
cout << "Yes";
else
cout << "No";
return 0;
}
#include <stdio.h>
#include <math.h>
// If ceil and floor are equal
// the number is a perfect
// square
int isPerfectSquare(long long n) {
if (ceil((double)sqrt(n)) == floor((double)sqrt(n))) {
return 1; // true
}
else {
return 0; // false
}
}
int main() {
long long x = 49;
if (isPerfectSquare(x))
printf("Yes");
else
printf("No");
return 0;
}
public class GfG {
public static boolean isPerfectSquare(long n) {
// If ceil and floor are equal
// the number is a perfect
// square
if (Math.ceil(Math.sqrt(n)) == Math.floor(Math.sqrt(n))) {
return true;
} else {
return false;
}
}
public static void main(String[] args) {
long x = 49;
if (isPerfectSquare(x))
System.out.println("Yes");
else
System.out.println("No");
}
}
import math
def is_perfect_square(n):
# If ceil and floor are equal
# the number is a perfect
# square
if math.ceil(math.sqrt(n)) == math.floor(math.sqrt(n)):
return True
else:
return False
x = 49
if is_perfect_square(x):
print("Yes")
else:
print("No")
using System;
class GfG {
public static bool IsPerfectSquare(long n) {
// If ceil and floor are equal
// the number is a perfect
// square
if (Math.Ceiling(Math.Sqrt(n)) == Math.Floor(Math.Sqrt(n))) {
return true;
} else {
return false;
}
}
static void Main() {
long x = 49;
if (IsPerfectSquare(x))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
function isPerfectSquare(n) {
// If ceil and floor are equal
// the number is a perfect
// square
if (Math.ceil(Math.sqrt(n)) === Math.floor(Math.sqrt(n))) {
return true;
} else {
return false;
}
}
const x = 49;
if (isPerfectSquare(x)) {
console.log("Yes");
} else {
console.log("No");
}
Time Complexity : O(sqrt(n))
Auxiliary space: O(1)
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
// Function to check if a number is a perfect square using
// binary search
bool isPerfectSquare(long long n)
{
// Base case: 0 and 1 are perfect squares
if (n <= 1) {
return true;
}
// Initialize boundaries for binary search
long long left = 1, right = n;
while (left <= right) {
// Calculate middle value
long long mid = left + (right - left) / 2;
// Calculate square of the middle value
long long square = mid * mid;
// If the square matches n, n is a perfect square
if (square == n) {
return true;
}
// If the square is smaller than n, search the right
// half
else if (square < n) {
left = mid + 1;
}
// If the square is larger than n, search the left
// half
else {
right = mid - 1;
}
}
// If the loop completes without finding a perfect
// square, n is not a perfect square
return false;
}
int main()
{
long long x = 49;
if (isPerfectSquare(x))
cout << "Yes";
else
cout << "No";
return 0;
}
#include <stdio.h>
// Function to check if a number is a perfect square using
// binary search
int isPerfectSquare(long long n) {
// Base case: 0 and 1 are perfect squares
if (n <= 1) {
return 1;
}
// Initialize boundaries for binary search
long long left = 1, right = n;
while (left <= right) {
// Calculate middle value
long long mid = left + (right - left) / 2;
// Calculate square of the middle value
long long square = mid * mid;
// If the square matches n, n is a perfect square
if (square == n) {
return 1;
}
// If the square is smaller than n, search the right half
else if (square < n) {
left = mid + 1;
}
// If the square is larger than n, search the left half
else {
right = mid - 1;
}
}
// If the loop completes without finding a perfect square, n is not a perfect square
return 0;
}
int main() {
long long x = 49;
if (isPerfectSquare(x))
printf("Yes");
else
printf("No");
return 0;
}
public class GfG {
// Function to check if a number is a perfect square using
// binary search
public static boolean isPerfectSquare(long n) {
// Base case: 0 and 1 are perfect squares
if (n <= 1) {
return true;
}
// Initialize boundaries for binary search
long left = 1, right = n;
while (left <= right) {
// Calculate middle value
long mid = left + (right - left) / 2;
// Calculate square of the middle value
long square = mid * mid;
// If the square matches n, n is a perfect square
if (square == n) {
return true;
}
// If the square is smaller than n, search the right
// half
else if (square < n) {
left = mid + 1;
}
// If the square is larger than n, search the left
// half
else {
right = mid - 1;
}
}
// If the loop completes without finding a perfect
// square, n is not a perfect square
return false;
}
public static void main(String[] args) {
long x = 49;
if (isPerfectSquare(x))
System.out.println("Yes");
else
System.out.println("No");
}
}
def is_perfect_square(n):
# Base case: 0 and 1 are perfect squares
if n <= 1:
return True
# Initialize boundaries for binary search
left, right = 1, n
while left <= right:
# Calculate middle value
mid = left + (right - left) // 2
# Calculate square of the middle value
square = mid * mid
# If the square matches n, n is a perfect square
if square == n:
return True
# If the square is smaller than n, search the right half
elif square < n:
left = mid + 1
# If the square is larger than n, search the left half
else:
right = mid - 1
# If the loop completes without finding a perfect square, n is not a perfect square
return False
x = 49
if is_perfect_square(x):
print("Yes")
else:
print("No")
using System;
class GfG {
// Function to check if a number is a perfect square using
// binary search
static bool IsPerfectSquare(long n) {
// Base case: 0 and 1 are perfect squares
if (n <= 1) {
return true;
}
// Initialize boundaries for binary search
long left = 1, right = n;
while (left <= right) {
// Calculate middle value
long mid = left + (right - left) / 2;
// Calculate square of the middle value
long square = mid * mid;
// If the square matches n, n is a perfect square
if (square == n) {
return true;
}
// If the square is smaller than n, search the right half
else if (square < n) {
left = mid + 1;
}
// If the square is larger than n, search the left half
else {
right = mid - 1;
}
}
// If the loop completes without finding a perfect square, n is not a perfect square
return false;
}
static void Main() {
long x = 49;
if (IsPerfectSquare(x))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
function isPerfectSquare(n) {
// Base case: 0 and 1 are perfect squares
if (n <= 1) {
return true;
}
// Initialize boundaries for binary search
let left = 1, right = n;
while (left <= right) {
// Calculate middle value
let mid = left + Math.floor((right - left) / 2);
// Calculate square of the middle value
let square = mid * mid;
// If the square matches n, n is a perfect square
if (square === n) {
return true;
}
// If the square is smaller than n, search the right half
else if (square < n) {
left = mid + 1;
}
// If the square is larger than n, search the left half
else {
right = mid - 1;
}
}
// If the loop completes without finding a perfect square, n is not a perfect square
return false;
}
const x = 49;
if (isPerfectSquare(x)) {
console.log("Yes");
} else {
console.log("No");
}
Time Complexity: O(log n)
Auxiliary Space: O(1)
Using Mathematical Properties:
The idea is based on the fact that perfect squares are always some of first few odd numbers.
1 + 3 = 4
1 + 3 + 5 = 9
1 + 3 + 5 + 7 = 16
1 + 3 + 5 + 7 + 9 = 25
1 + 3 + 5 + 7 + 9 + 11 = 36
......................................................
Code Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
bool isPerfectSquare(long long n) {
// 0 is considered as a perfect
// square
if (n == 0) return true;
long long odd = 1;
while (n > 0) {
n -= odd;
odd += 2;
}
return n == 0;
}
int main()
{
long long x = 49;
if (isPerfectSquare(x))
cout << "Yes";
else
cout << "No";
return 0;
}
#include <stdio.h>
// 0 is considered as a perfect
// square
int isPerfectSquare(long long n) {
if (n == 0) return 1; // true
long long odd = 1;
while (n > 0) {
n -= odd;
odd += 2;
}
return n == 0;
}
int main() {
long long x = 49;
if (isPerfectSquare(x))
printf("Yes");
else
printf("No");
return 0;
}
public class GfG {
public static boolean isPerfectSquare(long n) {
// 0 is considered as a perfect
// square
if (n == 0) return true;
long odd = 1;
while (n > 0) {
n -= odd;
odd += 2;
}
return n == 0;
}
public static void main(String[] args) {
long x = 49;
if (isPerfectSquare(x))
System.out.println("Yes");
else
System.out.println("No");
}
}
def is_perfect_square(n):
# 0 is considered as a perfect
# square
if n == 0:
return True
odd = 1
while n > 0:
n -= odd
odd += 2
return n == 0
x = 49
if is_perfect_square(x):
print("Yes")
else:
print("No")
using System;
class GfG {
public static bool IsPerfectSquare(long n) {
// 0 is considered as a perfect
// square
if (n == 0) return true;
long odd = 1;
while (n > 0) {
n -= odd;
odd += 2;
}
return n == 0;
}
static void Main() {
long x = 49;
if (IsPerfectSquare(x))
Console.WriteLine("Yes");
else
Console.WriteLine("No");
}
}
function isPerfectSquare(n) {
// 0 is considered as a perfect
// square
if (n === 0) return true;
let odd = 1;
while (n > 0) {
n -= odd;
odd += 2;
}
return n === 0;
}
const x = 49;
if (isPerfectSquare(x)) {
console.log("Yes");
} else {
console.log("No");
}
How does this work?
1 + 3 + 5 + ... (2n-1) = Sum(2*i - 1) where 1<=i<=n
= 2*Sum(i) - Sum(1) where 1<=i<=n
= 2n(n+1)/2 - n
= n(n+1) - n
= n2'
Time Complexity Analysis
Let us assume that the above loop runs i times. The following series would have i terms.
1 + 3 + 5 ........ = n [Let there be i terms]
1 + (1 + 2) + (1 + 2 + 2) + ........ = n [Let there be i terms]
(1 + 1 + ..... i-times) + (2 + 4 + 6 .... (i-1)-times) = n
i + (2^(i-1) - 1) = n
2^(i-1) = n + 1 - i
Using this expression, we can say that i is upper bounded by Log n
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